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Chapter 7 • Theory & Derivations

Unit 7: Colorimetric & Spectrophotometric Methods: Beer's Law, Titrations & Speciation

Comprehensive physical and analytical treatise on molecular UV-Visible spectrophotometry: photophysical derivations of the Beer-Lambert law, fundamental chemical and instrumental stray-light deviations, Twyman-Lothian photometric precision optimization, spectrophotometric titration curve morphology and dilution correction, Job's method of continuous variations, and trace colorimetric determination of lead via dithizone and arsenic via silver diethyldithiocarbamate.

§7.1 Photophysics of Absorption, Transmittance & Beer-Lambert Law Formalism

Molecular absorption spectrophotometry in the ultraviolet and visible regions ($190\text{--}800\text{ nm}$) measures the attenuation of a collimated beam of monochromatic radiant energy as it traverses an absorbing homogeneous solution. The fundamental physical process involves the resonant absorption of photons whose energy matches the transition energy between quantized molecular electronic states:

$$\Delta E = E_{\text{excited}} - E_{\text{ground}} = h\nu = \frac{hc}{\lambda}$$

``` Resonant Electronic Absorption & Attenuation Geometry Incident Radiant Power (P₀) Transmitted Radiant Power (P) =========================> [ b ] ===========================> Monochromatic Beam Absorbing Layer of Length b (cm) Wavelength λ Analyte Concentration c (mol/L) ------------------------------------------------------------- Attenuation: dP = - k' · P · c · dx Integration: ∫_{P₀}^{P} dP/P = - k' · c · ∫₀^b dx ln(P₀/P) = k' · c · b ==> log₁₀(P₀/P) = ε · b · c ```

Derivation of the Beer-Lambert Law

Consider a parallel, monochromatic radiant beam of incident power $P$ traversing an infinitesimal layer $dx$ of a homogeneous absorbing solution containing concentration $c$ (in $\text{mol}\cdot\text{L}^{-1}$) of absorbing chromophores. The probability that an incident photon will be captured within this differential slab is directly proportional to the number of chromophores per unit area within the slab and the radiant power passing through it:

$$-dP = k'\,P\,c\,dx$$

where $k'$ is a constant of proportionality reflecting the photon capture cross-section of the absorbing species at wavelength $\lambda$.

Separating variables and integrating across the full optical path length of the cuvette from $x = 0$ (where radiant power is $P_0$) to $x = b$ (where transmitted power is $P$):

$$\int_{P_0}^{P} \frac{dP}{P} = -k'\,c \int_{0}^{b} dx$$
$$\ln\left(\frac{P}{P_0}\right) = -k'\,b\,c \implies \ln\left(\frac{P_0}{P}\right) = k'\,b\,c$$

Converting from natural logarithms to common logarithms (base 10) by dividing by $2.302585$:

$$\log_{10}\left(\frac{P_0}{P}\right) = \frac{k'}{2.302585}\,b\,c = \varepsilon\,b\,c$$

where $\varepsilon$ is defined as the molar absorptivity (or molar extinction coefficient) in units of $\text{L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$, $b$ is the internal path length of the optical cell in centimeters ($\text{cm}$), and $c$ is the analytical molar concentration in $\text{mol}\cdot\text{L}^{-1}$.

Fundamental Photometric Definitions

1. Transmittance ($T$): The fractional radiant power transmitted by the absorbing solution relative to a solvent blank reference cell:

$$T = \frac{P}{P_0}$$

2. Percent Transmittance ($\%T$):

$$\%T = 100 \times T = 100 \times \frac{P}{P_0}$$

3. Absorbance ($A$): Formerly termed optical density ($OD$), defined as the negative common logarithm of transmittance:

$$A = -\log_{10} T = \log_{10}\left(\frac{P_0}{P}\right) = \log_{10}\left(\frac{100}{\%T}\right) = 2.000 - \log_{10}(\%T)$$

4. The Beer-Lambert Law:

$$A = \varepsilon\,b\,c$$

Nature of Molecular Electronic Transitions

UV-Visible absorption requires transitions of valence electrons from bonding or non-bonding molecular orbitals to unoccupied antibonding molecular orbitals:

  • $\sigma \to \sigma^*$: High energy vacuum-UV transitions ($\lambda < 185\text{ nm}$), observed in saturated alkanes ($\text{C--C}$, $\text{C--H}$).
  • $n \to \sigma^*$: Intermediate energy transitions ($\lambda \approx 150\text{--}250\text{ nm}$), observed in saturated molecules with heteroatoms bearing lone pairs ($\text{O}, \text{N}, \text{S}, \text{X}$).
  • $\pi \to \pi^*$: Strongly allowed transitions ($\varepsilon \sim 10^3\text{--}10^5\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$) characteristic of unsaturated chromophores ($\text{C=C}$, $\text{C=O}$, aromatic systems, conjugated polyenes).
  • $n \to \pi^*$: Symmetry-forbidden transitions ($\varepsilon \sim 10\text{--}100\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$) occurring at longer wavelengths in carbonyls and nitrogen heterocycles.
  • Charge-Transfer Transitions (CT): Ligand-to-Metal Charge Transfer (LMCT) and Metal-to-Ligand Charge Transfer (MLCT) in transition metal coordination complexes (e.g., $[\text{Fe}(\text{SCN})]^{2+}$, $\text{MnO}_4^-$, $\text{Fe(phen)}_3^{2+}$), characterized by extraordinarily high molar absorptivities ($\varepsilon > 10,000\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$).

Mathematical Foundations of Derivative Spectrophotometry

In complex multi-component matrices where analyte absorption bands overlap heavily with broad background absorption or scattering profiles, derivative spectrophotometry provides extraordinary spectral resolution enhancement without chemical separation:

$$\frac{d^n A}{d\lambda^n} = b \sum_{i} c_i \frac{d^n \varepsilon_i}{d\lambda^n} \tag{7.0a}$$

1. First Derivative ($dA/d\lambda$):

  • The zero-crossing point ($\frac{dA}{d\lambda} = 0$) identifies the exact absorption maximum ($\lambda_{\text{max}}$) with extreme precision.
  • Completely eliminates flat, wavelength-independent background offsets ($\frac{d}{d\lambda}[\text{const}] = 0$).

2. Second Derivative ($d^2A/d\lambda^2$):

  • Features a negative minimum whose absolute amplitude is directly proportional to concentration.
  • Completely eliminates linear sloping backgrounds ($A_{\text{bg}} = a\lambda + b \implies \frac{d^2 A_{\text{bg}}}{d\lambda^2} = 0$).
  • Sharpens narrow bands relative to broad matrix bands according to the inverse power scaling:
$$\frac{d^n A}{d\lambda^n} \propto \frac{A_0}{W_0^n} \tag{7.0b}$$

where $W_0$ is the half-bandwidth. For an analyte band that is 3 times narrower than a background band, the second derivative enhances the analyte-to-background signal ratio by a factor of $3^2 = 9$, and the fourth derivative enhances it by $3^4 = 81$!

§7.2 Instrumental Limitations, Real Deviations & Polychromatic/Stray Light Errors

The linear relationship between absorbance and analyte concentration predicted by the Beer-Lambert law ($A = \varepsilon b c$) is an idealized limiting law that holds strictly only under limiting physical conditions: truly monochromatic light, non-interacting independent chromophores, and negligible background stray radiation. In real analytical measurements, significant negative or positive deviations are frequently encountered.

``` Types of Deviations from the Beer-Lambert Law Absorbance (A) ^ | / Positive Deviation (Analyte association / Refractive index) | / | /------- Ideal Linear Beer's Law (A = ε b c) | / | / ------- Negative Deviation (Stray light / Polychromatic beam) | / | / +----------------------------> Concentration (c) ```

Real Physical and Chemical Deviations

1. High Concentration Inter-Ionic Effects: At concentrations exceeding approximately $0.01\text{ M}$, the average distance between absorbing solute ions decreases to the point where intermolecular electrostatic interactions perturb the electronic charge distributions of adjacent chromophores, modifying $\varepsilon$. Furthermore, the refractive index $\eta$ of the solution increases significantly with concentration. A more rigorous form of Beer's law accounts for refractive index dispersion:

$$A = \varepsilon\,b\,c\,\left(\frac{\eta}{(\eta^2 + 2)^2}\right)$$

2. Chemical Equilibrium Shifts: If the absorbing analyte participates in chemical equilibria (acid-base dissociation, dimerization, tautomerism, or complexation), the concentration of the specific absorbing species does not vary linearly with total analytical concentration.

A classic example is the chromate-dichromate equilibrium in acidic solution:

$$2\,\text{CrO}_4^{2-} + 2\,\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$$

Because $\text{CrO}_4^{2-}$ ($\lambda_{\text{max}} = 372\text{ nm}$) and $\text{Cr}_2\text{O}_7^{2-}$ ($\lambda_{\text{max}} = 350, 450\text{ nm}$) have distinctly different absorption spectra, unbuffered solutions show severe deviations from linearity as dilution shifts the equilibrium toward the monomeric chromate dianion.

Instrumental Deviations: Polychromatic Radiation

Real monochromators isolate an optical bandpass of finite bandwidth $\Delta \lambda$ rather than truly monochromatic light. If the incident beam consists of two wavelengths $\lambda_1$ and $\lambda_2$ with incident powers $P_{0,1}$ and $P_{0,2}$ and respective molar absorptivities $\varepsilon_1$ and $\varepsilon_2$:

$$P_1 = P_{0,1} 10^{-\varepsilon_1 b c}, \quad P_2 = P_{0,2} 10^{-\varepsilon_2 b c}$$

The total observed absorbance is:

$$A_{\text{obs}} = \log_{10}\left(\frac{P_{0,1} + P_{0,2}}{P_1 + P_2}\right) = \log_{10}\left(\frac{P_{0,1} + P_{0,2}}{P_{0,1} 10^{-\varepsilon_1 b c} + P_{0,2} 10^{-\varepsilon_2 b c}}\right)$$

If $\varepsilon_1 = \varepsilon_2$, the equation collapses strictly back to $A_{\text{obs}} = \varepsilon_1 b c$. However, if $\varepsilon_1 \neq \varepsilon_2$, the logarithmic term cannot be simplified, producing a curve that bends downward toward the concentration axis (a negative deviation). This underscores the fundamental requirement that analytical measurements must be conducted at absorption peaks ($\lambda_{\text{max}}$), where $d\varepsilon/d\lambda \approx 0$ across the instrumental bandpass.

Instrumental Deviations: Stray Radiation

Stray light ($P_s$) is radiant energy reaching the detector that originates from higher grating orders, internal optical scattering, or enclosure leaks, having wavelengths outside the nominal bandpass. When stray radiation is present:

$$A_{\text{obs}} = \log_{10}\left(\frac{P_0 + P_s}{P + P_s}\right)$$

As analyte concentration becomes very large, true transmitted power approaches zero ($P \to 0$), but stray power remains constant ($P_s$):

$$\lim_{c \to \infty} A_{\text{obs}} = \log_{10}\left(\frac{P_0 + P_s}{P_s}\right) \approx \log_{10}\left(\frac{100\% + \%P_s}{\%P_s}\right)$$

For example, if stray light is merely $0.5\%$ of $P_0$, the maximum achievable absorbance cannot exceed $\log_{10}(100.5 / 0.5) = \log_{10}(201) = 2.303$, regardless of how concentrated the solution is!

Twyman-Lothian Photometric Error Analysis

Detector shot noise and readout uncertainty lead to an uncertainty in transmittance $\Delta T$. The relative concentration error is derived by differentiating Beer's law:

$$A = -\log_{10} T = -0.4343 \ln T = \varepsilon b c \implies c = -\frac{0.4343}{\varepsilon b} \ln T$$

Differentiating with respect to $T$:

$$\frac{dc}{dT} = -\frac{0.4343}{\varepsilon b T} = \frac{0.4343 c}{T \log_{10} T}$$

Dividing by $c$ yields the relative concentration error:

$$\frac{\Delta c}{c} = \frac{0.4343 \Delta T}{T \log_{10} T}$$

When detector noise is constant (independent of radiant power, as in older thermal or phototube detectors, $\Delta T = k$):

$$\text{Minimizing } f(T) = \frac{1}{T \ln T} \implies \frac{d}{dT}(T \ln T) = \ln T + 1 = 0 \implies \ln T = -1$$
$$T_{\text{opt}} = e^{-1} = 0.368 \implies \%T_{\text{opt}} = 36.8\%, \quad A_{\text{opt}} = -\log_{10}(0.368) = 0.434$$

Consequently, high-precision spectrophotometric measurements should maintain sample absorbance within the optimal dynamic window of $A \approx 0.2\text{ to }0.8$.

§7.3 Spectrophotometric Titrations: Principles, Cell Geometries & Curve Morphology

A spectrophotometric titration combines the absolute stoichiometric precision of volumetric titration with the high sensitivity and selectivity of photometric detection. Instead of relying on human visual perception of an indicator color transition, the absorbance of the solution is recorded at an analytically selected wavelength as increments of standard titrant are added:

$$A(V) = b \left( \varepsilon_A [A] + \varepsilon_T [T] + \varepsilon_P [P] \right)$$

``` Spectrophotometric Titration Cell & Optical Setup Burette / Micro-dispenser | [Titrant V] v +-------------+ Light Source ----> Monochromator ----> Beam | Stirred | | | Titration |<=========================================+ | Cell |--------> Photodiode / Photomultiplier Detector +-------------+ | v Absorbance vs Volume Plot ```

Volume Dilution Correction

During a titration, the addition of titrant increases the total volume of the solution, diluting all absorbing species and introducing an artificial downward curvature into the titration plot. To restore straight-line segments that intersect sharply at the equivalence point, the observed absorbance must be corrected for dilution:

$$A_{\text{corrected}} = A_{\text{measured}} \times \left(\frac{V_0 + V}{V_0}\right)$$

where $V_0$ is the initial volume of the sample solution and $V$ is the cumulative volume of titrant added. Alternatively, dilution can be rendered negligible by using a concentrated titrant delivered from a microburette so that $V \ll V_0$ (e.g., total titrant addition $< 1\text{--}2\%$ of initial volume).

Morphology of Spectrophotometric Titration Curves

The shape of a spectrophotometric titration curve depends entirely on the relative molar absorptivities of the analyte ($\varepsilon_A$), the titrant ($\varepsilon_T$), and the reaction product ($\varepsilon_P$) at the chosen monitoring wavelength:

1. Case 1 ($\varepsilon_A > 0, \varepsilon_T = 0, \varepsilon_P = 0$):

The analyte absorbs light, but the titrant and product are non-absorbing. As titrant is added, analyte is consumed, causing absorbance to drop linearly until the equivalence point, after which absorbance remains at zero: Shape: Linear decline followed by horizontal plateau.

2. Case 2 ($\varepsilon_A = 0, \varepsilon_T > 0, \varepsilon_P = 0$):

Neither analyte nor product absorbs. Absorbance remains essentially zero until the equivalence point; excess titrant added beyond the equivalence point causes absorbance to rise linearly: Shape: Flat baseline followed by linear ascending branch.

3. Case 3 ($\varepsilon_A = 0, \varepsilon_T = 0, \varepsilon_P > 0$):

Only the reaction product absorbs. Absorbance increases linearly from zero as product is formed, reaching a maximum plateau at the equivalence point where product formation is complete: Shape: Linear ascending branch followed by flat plateau.

4. Case 4 ($\varepsilon_A > 0, \varepsilon_T > 0, \varepsilon_P = 0$):

Both analyte and titrant absorb, but the product does not. Absorbance drops linearly as analyte is consumed, passes through a minimum at the equivalence point, and then climbs linearly as excess titrant accumulates: Shape: Distinct V-shaped curve.

5. Case 5 ($\varepsilon_A > 0, \varepsilon_P > \varepsilon_A, \varepsilon_T = 0$):

Analyte absorbs, but the product has an even higher molar absorptivity; titrant is transparent. Absorbance rises with steep slope up to the equivalence point, after which it levels off horizontally: Shape: Steep ascending branch followed by flat plateau.

``` Canonical Titration Curve Morphologies Case 1 (ε_A > 0) Case 2 (ε_T > 0) Case 3 (ε_P > 0) A A A | \ | / | /---- | \ | / | / | \ | / | / | \______ | ______/ | / +-----------> V +-----------> V +-----------> V V_eq V_eq V_eq ```

Advantages Over Conventional Visual Titrations

  • Extrapolation Through Dissociation Rounding: Because the equivalence point is determined by extrapolating linear segments measured well before and after the endpoint, curvature caused by incomplete reaction (chemical dissociation near equivalence) does not impair accuracy.
  • Extreme Dilution Capabilities: Titrations can be performed successfully at concentrations as low as $10^{-5}\text{ to }10^{-6}\text{ M}$, where visual indicators fail completely.
  • Automated Fiber-Optic Implements: Titrations can be executed in situ using dip-type fiber optic transflectance probes without manual transfers.

§7.4 Photometric Endpoint Detection for Weak Acid-Base & Precipitation Systems

Photometric detection provides an exceptionally sensitive means of locating titration endpoints in chemical systems where conventional potentiometric glass electrodes or visual indicators encounter fundamental thermodynamic limitations.

Photometric Titration of Extremely Weak Acids

For very weak acids with dissociation constants $K_a < 10^{-8}$ (such as phenols, boric acid, and certain alkaloids), the potentiometric $\Delta \text{pH}$ jump at the equivalence point is virtually undetectable with a standard glass electrode. However, by monitoring the absorbance of the conjugate base or an added photometric acid-base indicator with an appropriately matched $pK_{\text{In}}$, a precise photometric endpoint is readily achieved. Consider a weak acid $HA$ titrated with strong base in the presence of indicator $\text{HIn}$:

$$\text{HIn} + \text{OH}^- \rightleftharpoons \text{In}^- + \text{H}_2\text{O} \quad K_{\text{In}} = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]}$$

Monitoring at the absorption maximum of the deprotonated indicator anion $\text{In}^-$:

$$A = \varepsilon_{\text{In}^-} b [\text{In}^-] = \varepsilon_{\text{In}^-} b C_{\text{In}} \left(\frac{K_{\text{In}}}{[\text{H}^+] + K_{\text{In}}}\right)$$

Plotting absorbance against titrant volume produces a sigmoidal photometric curve from which the inflection or derivative maximum accurately identifies the equivalence point.

``` Turbidimetric / Photometric Precipitation Titration Curve Apparent Absorbance (Apparent A = -log₁₀ I/I₀ due to light scattering) ^ | / (Agglomeration & sedimenting plateau) | / | / (Precipitate nucleates & scatters light) | / | / | ____________/ (Pre-equivalence: solubility limit not exceeded) +-----------------------------> Volume of Precipitating Titrant (V) V_threshold ```

Photometric Precipitation Titrations

In photometric precipitation titrations (e.g., titration of sulfate with barium perchlorate, or halides with silver nitrate), the appearance of a finely divided colloidal suspension scatters radiant energy out of the optical path, registering as an apparent increase in absorbance according to Rayleigh and Mie scattering formalisms:

$$I_{\text{scattered}} \propto \frac{I_0\,N\,V_p^2}{\lambda^4}$$

where $N$ is the number density of colloidal particles and $V_p$ is the individual particle volume. Before the solubility product $K_{\text{sp}}$ is exceeded, the solution remains optically transparent ($A \approx 0$). Once precipitation begins, apparent absorbance climbs steeply. Adding protective colloids (such as gelatin, agar-agar, or polyvinyl alcohol) prevents rapid coagulation, maintaining a uniform dispersion that yields highly reproducible linear segments.

Multicomponent Spectrophotometric Mixture Titrations

One of the most powerful analytical attributes of spectrophotometric titrations is the ability to resolve mixtures of metal cations sequentially in a single beaker without prior chemical separation. A classic industrial benchmark is the sequential titration of bismuth(III) and copper(II) with standard EDTA:

  • At $\text{pH} \approx 1.5\text{--}2.0$, bismuth(III) forms an extraordinarily stable complex with EDTA ($\log K_f = 27.9$), whereas copper(II) ($\log K_f = 18.8$) does not react appreciably at this low pH due to severe protonation of the EDTA ligand ($\alpha_{\text{Y}^{4-}} \approx 10^{-14}$).
  • When titrated at $\lambda = 745\text{ nm}$, uncomplexed $\text{Bi}^{3+}$ and $[\text{Bi(EDTA)}]^-$ do not absorb. The absorbance remains zero until all $\text{Bi}^{3+}$ is consumed ($V_{\text{eq},1}$).
  • Immediately following the bismuth endpoint, EDTA begins coordinating with $\text{Cu}^{2+}$ to form $[\text{Cu(EDTA)}]^{2-}$, which possesses an intense blue absorption band at $745\text{ nm}$ ($\varepsilon \approx 90\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$). Absorbance climbs linearly with volume.
  • When all $\text{Cu}^{2+}$ has reacted, the absorbance plateaus sharply ($V_{\text{eq},2}$). The first break locates the bismuth concentration; the volume difference $(V_{\text{eq},2} - V_{\text{eq},1})$ quantifies the copper concentration.

§7.5 Determination of Stoichiometry & Stability: Job's Method of Continuous Variations & Mole-Ratio Method

Spectrophotometry is the primary experimental technique for determining the empirical stoichiometric composition ($M_m L_n$) and equilibrium formation constants ($K_f$) of coordination complexes in solution.

Job's Method of Continuous Variations

In Job's method, the total analytical concentration of metal and ligand is held strictly constant throughout a series of test solutions:

$$C_{\text{total}} = C_M + C_L = \text{constant}$$

The mole fraction of ligand, $x_L$, is systematically varied from $0$ to $1$:

$$x_L = \frac{C_L}{C_{\text{total}}}, \quad x_M = 1 - x_L = \frac{C_M}{C_{\text{total}}}$$

``` Job's Method of Continuous Variations Curve Corrected Absorbance (ΔA) ^ Peak at x_L = n / (m + n) | /\ | / \ Extrapolated Linear Branches | / \ | / \ | / •••••• \ Real Curvature (Complex Dissociation) | / \ +-------------------+------------+------------------> 0.0 x_max 1.0 Mole Fraction Ligand (x_L) ```

Consider the general complexation equilibrium:

$$m\,M + n\,L \rightleftharpoons M_m L_n \quad K_f = \frac{[M_m L_n]}{[M]^m [L]^n}$$

Assuming only the complex absorbs at the chosen analytical wavelength (or correcting for ligand/metal background absorbance: $\Delta A = A_{\text{meas}} - \varepsilon_M b C_M - \varepsilon_L b C_L$):

$$\Delta A = \varepsilon_{\text{complex}}\,b\,[M_m L_n]$$

To find the mole fraction $x_L$ that maximizes complex concentration, we differentiate $[M_m L_n]$ with respect to $x_L$ subject to the mass-balance constraints:

$$C_M = [M] + m [M_m L_n] = (1 - x_L) C_{\text{total}}$$
$$C_L = [L] + n [M_m L_n] = x_L C_{\text{total}}$$

Setting $\frac{d[M_m L_n]}{dx_L} = 0$, the mathematical condition for the maximum reduces to:

$$\frac{x_L}{1 - x_L} = \frac{n}{m} \implies x_{L,\text{max}} = \frac{n}{m + n}$$
  • For a $1:1$ complex ($ML$): $x_{\text{max}} = 1/(1+1) = 0.500$.
  • For a $1:2$ complex ($ML_2$): $x_{\text{max}} = 2/(1+2) = 0.667$.
  • For a $1:3$ complex ($ML_3$): $x_{\text{max}} = 3/(1+3) = 0.750$.
  • For a $2:3$ complex ($M_2L_3$): $x_{\text{max}} = 3/(2+3) = 0.600$.

Extraction of Equilibrium Stability Constant ($K_f$) from Curvature

At the peak mole fraction $x_{\text{max}}$, the extrapolated intersection of the linear asymptotes gives the theoretical absorbance $A_{\text{extrap}}$ corresponding to $100\%$ complete stoichiometric conversion:

$$A_{\text{extrap}} = \varepsilon_{\text{complex}}\,b\,[M_m L_n]_{\text{theoretical}} = \varepsilon_{\text{complex}}\,b\,\left(\frac{C_{\text{total}}}{m + n}\right)$$

The actual experimentally measured absorbance at the peak, $A_{\text{meas}}$, is slightly lower due to thermodynamic dissociation. The degree of formation $\alpha$ is:

$$\alpha = \frac{[M_m L_n]_{\text{actual}}}{[M_m L_n]_{\text{theoretical}}} = \frac{A_{\text{meas}}}{A_{\text{extrap}}}$$

From $\alpha$, the equilibrium concentrations of free metal and free ligand are calculated directly:

$$[M_m L_n] = \alpha [M_m L_n]_{\text{theoretical}}$$
$$[M] = C_M - m [M_m L_n], \quad [L] = C_L - n [M_m L_n]$$

Substituting into the equilibrium quotient yields the formation constant $K_f$.

The Mole-Ratio Method

In the mole-ratio method, the analytical concentration of the metal ion is kept constant ($C_M = \text{const}$) across all samples, while the ligand concentration $C_L$ is systematically increased. Plotting absorbance $A$ versus the molar ratio $C_L / C_M$:

  • For stable complexes, two straight lines are obtained: an initial ascending linear portion where added ligand is quantitatively converted into complex, followed by an abrupt break to a horizontal plateau once all metal has reacted.
  • The abscissa of the break point corresponds directly to the stoichiometric ratio $n/m$.

§7.6 Colorimetric Determination of Trace Lead via Dithizone Extraction-Spectrophotometry

The colorimetric determination of trace lead ($\text{Pb}^{2+}$) in environmental waters, biological fluids, and forensic exhibits relies on solvent extraction with diphenylthiocarbazone (commonly known as dithizone, $\text{H}_2\text{Dz}$). Dithizone is an intensely colored sulfur-containing organic chelating agent with remarkable sensitivity for heavy soft metals.

``` Molecular Tautomerism of Dithizone (H₂Dz) S SH || | Ph-NH-C-N=N-Ph <===============> Ph-NH-C=N-N-Ph Keto / Thione Form Enol / Thiol Form (Green in CHCl₃ / CCl₄) (Forms Red Metal Chelates) λ_max = 620 nm, ε = 32,800 λ_max = 520 nm (Pb Chelate) ```

Chemistry of the Lead-Dithizone Chelate Reaction

Dithizone acts as a monoprotic weak acid in neutral and weakly basic solutions:

$$\text{H}_2\text{Dz}(\text{org}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{HDz}^-(\text{aq}) \quad (pK_a = 4.5)$$

Lead(II) reacts stoichiometrically with two dithizonate anions to form a neutral, coordinatively saturated chelate complex that is highly soluble in nonpolar organic solvents ($\text{CHCl}_3$ or $\text{CCl}_4$):

$$\text{Pb}^{2+}(\text{aq}) + 2\,\text{H}_2\text{Dz}(\text{org}) \rightleftharpoons \text{Pb}(\text{HDz})_2(\text{org}) + 2\,\text{H}^+(\text{aq})$$

The resulting primary lead dithizonate, $\text{Pb}(\text{HDz})_2$, exhibits an intense crimson-red color with an absorption maximum at $\lambda_{\text{max}} = 520\text{ nm}$ and a molar absorptivity $\varepsilon \approx 68,000\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$, enabling detection limits well below $10\,\mu\text{g}\cdot\text{L}^{-1}$ ($10\text{ ppb}$).

``` Selectivity Scheme for Lead Dithizone Extraction Aqueous Sample containing Pb²⁺, Cu²⁺, Zn²⁺, Ni²⁺, Fe³⁺, Ca²⁺, Mg²⁺ | Add Reagent Cocktail: v

  1. Ammonium Citrate -----> Prevents precipitation of Ca/Mg/Fe hydroxides
  2. KCN (Potassium Cyanide)-> Masks Cu²⁺, Zn²⁺, Ni²⁺, Co²⁺ as stable [M(CN)₄]ⁿ⁻
  3. NH₂OH·HCl --------------> Keeps Fe as Fe²⁺, prevents dithizone oxidation
  4. Ammonia Buffer (pH 8.5-11.5)

| v Extract with Dithizone in Chloroform (CHCl₃) ------------------------------------------- Organic Layer (Red): Pb(HDz)₂ ONLY! (λ = 520 nm) Aqueous Layer: Masked Cyano-Complexes & Interferences ```

Analytical Procedure and Reagent Functions

To achieve absolute selectivity for lead in complex matrices, a rigorous masking protocol is implemented:

1. Ammonium Citrate ($\text{pH} \approx 8.5\text{--}9.5$): Citrate forms soluble, negatively charged auxiliary complexes with $\text{Fe}^{3+}$, $\text{Al}^{3+}$, $\text{Ca}^{2+}$, and $\text{Mg}^{2+}$, preventing the precipitation of insoluble metal hydroxides or phosphates that would coprecipitate lead.

2. Potassium Cyanide ($\text{KCN}$): Cyanide is an extraordinarily strong field ligand that forms exceptionally stable, water-soluble, non-extractable cyanocomplexes with transition metals:

$$\text{Cu}^{2+} + 4\,\text{CN}^- \to [\text{Cu}(\text{CN})_4]^{2-}, \quad \text{Zn}^{2+} + 4\,\text{CN}^- \to [\text{Zn}(\text{CN})_4]^{2-}$$

Lead(II), being a $d^{10}$ post-transition cation with lower affinity for cyanide, does not form stable cyanocomplexes at pH 9, leaving it free to react quantitatively with dithizone.

3. Hydroxylamine Hydrochloride ($\text{NH}_2\text{OH}\cdot\text{HCl}$): Acts as a mild reducing agent that reduces iron(III) to iron(II) and protects the dithizone reagent from oxidative degradation by dissolved oxygen or traces of halogens.

4. Spectrophotometric Quantitation:

  • Monocolor Method: Excess green unreacted dithizone in the organic layer is removed by shaking with dilute alkaline ammonia ($\text{pH} \approx 11$). The lead complex remains in the organic phase, which is measured cleanly at $520\text{ nm}$.
  • Mixed-Color Method: Absorbance is measured simultaneously at $520\text{ nm}$ (lead complex) and $620\text{ nm}$ (unreacted dithizone), resolving concentrations via simultaneous linear equations.

§7.7 Colorimetric Micro-Determination of Arsenic by the Modified Gutzeit & Silver Diethyldithiocarbamate (Ag-DDTC) Method

Arsenic is a potent environmental toxicant subject to stringent regulatory limits in potable water ($< 10\,\mu\text{g}\cdot\text{L}^{-1}$). The classical Gutzeit test and the modern quantitative Silver Diethyldithiocarbamate (Ag-DDTC) spectrophotometric method represent the definitive wet-chemical benchmarks for microgram-level arsenic analysis.

``` Arsenic Arsine Generation & Absorption Assembly H₂SO₄ / HCl + Zn(s) or NaBH₄ | +---------v---------+ | Reaction Flask | ===> AsO₄³⁻ + 4Zn + 11H⁺ --> AsH₃(g)↑ +---------+---------+ | AsH₃(g) + H₂S(g) + H₂(g) v +-------------------+ | Pb(OAc)₂ Scrubber | ===> H₂S + Pb²⁺ --> PbS(s)↓ (traps sulfide!) +-------------------+ | Pure AsH₃(g) + H₂(g) v +-------------------+ | Absorber Tube | ===> Ag-DDTC in Pyridine / Chloroform | (Intense Red Col.)| ===> Reduced Soluble Colloid (λ = 535 nm) +-------------------+ ```

Generation of Volatile Arsine Gas ($\text{AsH}_3$)

Arsenic exists in water primarily as arsenite ($\text{AsO}_3^{3-}$, $\text{As(III)}$) and arsenate ($\text{AsO}_4^{3-}$, $\text{As(V)}$). In the sample preparation stage, $\text{As(V)}$ is first pre-reduced to $\text{As(III)}$ using potassium iodide ($\text{KI}$) and stannous chloride ($\text{SnCl}_2$) in concentrated hydrochloric acid:

$$\text{H}_3\text{AsO}_4 + 2\,\text{I}^- + 2\,\text{H}^+ \to \text{HAsO}_2 + \text{I}_2 + 2\,\text{H}_2\text{O}$$

The resulting trivalent arsenic is reduced to gaseous arsine ($\text{AsH}_3$, b.p. $-62.5^\circ\text{C}$) by active nascent hydrogen generated from granulated zinc and acid, or by sodium borohydride ($\text{NaBH}_4$):

$$\text{HAsO}_2 + 3\,\text{Zn} + 7\,\text{H}^+ \to \text{AsH}_3\uparrow + 3\,\text{Zn}^{2+} + 2\,\text{H}_2\text{O}$$

The Scrubber: Elimination of Hydrogen Sulfide Interference

Naturally occurring water samples often contain sulfur compounds that are concurrently reduced to hydrogen sulfide gas ($\text{H}_2\text{S}$). Hydrogen sulfide reacts vigorously with silver or mercuric salts to produce dark metal sulfides, causing severe positive interference. To eliminate this, the evolving gas stream is passed through a scrubber tube packed with glass wool impregnated with lead acetate ($\text{Pb}(\text{CH}_3\text{COO})_2$):

$$\text{H}_2\text{S}(\text{g}) + \text{Pb}^{2+}(\text{aq}) \to \text{PbS}(\text{s})\downarrow + 2\,\text{H}^+(\text{aq})$$

The lead sulfide precipitate is retained quantitatively on the glass wool, allowing pure arsine gas to pass unhindered.

Quantitative Spectrophotometric Ag-DDTC Mechanism

In the Ag-DDTC method, arsine gas is swept by hydrogen carrier gas into an absorption tube containing silver diethyldithiocarbamate dissolved in pyridine (or a chloroform-morpholine mixture). Arsine reduces the silver ions in the reagent to a soluble, intensely colored red colloidal silver complex stabilized by the diethyldithiocarbamate ligand:

$$\text{AsH}_3 + 6\,\text{Ag-DDTC} \to 6\,\text{Ag}^0 + \text{As(DDTC)}_3 + 3\,\text{H-DDTC}$$

The colloidal red complex exhibits a sharp, stable absorption maximum at $\lambda_{\text{max}} = 535\text{ nm}$ with a molar absorptivity $\varepsilon \approx 14,000\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$. The absorbance at $535\text{ nm}$ obeys Beer's law over the concentration range of $0.5\text{ to }20\,\mu\text{g}$ of arsenic per sample, providing sub-microgram sensitivity with an overall method precision of $\pm 2\text{--}3\%$ RSD.

§7.8 Modern Dual-Beam UV-Vis Instrumentation, Stray Light Standards & Quality Control Protocols

Modern high-performance UV-Visible spectrophotometry requires rigorous optical hardware architectures and standardized calibration protocols to ensure photometric accuracy, wavelength trueness, and stray light compliance in regulated pharmaceutical and analytical testing.

``` Double-Beam in Time UV-Vis Architecture Deuterium & Tungsten Czerny-Turner Monochromator Rotating Chopper (Sector Mirror) Sources (190-900 nm) (Grating & Collimator) Splits beam alternately +--------+--------+ +-----------------------+ +-------+ | D₂ | W |--->| G / | \ Exit Slit |---->| (O/) |---+---> Sample Cell (P) +--------+--------+ +-----------------------+ +-------+ | +---> Reference Cell (P₀) | v Single PMT / Silicon Photodiode Detector ```

Double-Beam Optical Design

1. Double-Beam in Space: Uses a 50/50 static beam splitter and two matched detectors. Susceptible to differential detector drift.

2. Double-Beam in Time: The monochromatic beam is alternately directed through the sample cell and the solvent reference cell by a high-speed rotating sector mirror (optical chopper) running at $50\text{--}60\text{ Hz}$. A single detector measures alternating pulses of $P$ and $P_0$.

  • Automatically compensates for source intensity fluctuations, detector thermal drift, and amplifier gain changes in real time.
  • Provides baseline stability better than $\pm 0.0002\text{ AU}\cdot\text{h}^{-1}$.

Pharmacopoeial Calibration & Qualification Standards

To satisfy USP <857> and Ph. Eur. 2.2.25 regulatory compliance, spectrophotometers must undergo regular calibration using certified reference materials:

1. Wavelength Trueness:

  • Holmium Oxide Solution (in $1.4\text{ M HClO}_4$): Provides 14 sharp, intrinsically invariant atomic-like absorption bands between $241\text{ nm}$ and $640\text{ nm}$ ($\pm 0.1\text{ nm}$ tolerance).
  • Low-Pressure Mercury Lamp: Emits fundamental atomic emission lines at $253.65\text{ nm}, 365.01\text{ nm}, 404.66\text{ nm}, 435.84\text{ nm},$ and $546.07\text{ nm}$.

2. Photometric Accuracy and Linearity:

  • Potassium Dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$ in $0.005\text{ M H}_2\text{SO}_4$): Certified molar absorption coefficients at $235\text{ nm}, 257\text{ nm}, 313\text{ nm},$ and $350\text{ nm}$. Validates photometric accuracy to within $\pm 0.005\text{ AU}$ over $A = 0.2\text{ to }1.5$.
  • Neutral Density Glass Filters: Certified filters of known absorbance across the visible spectrum.

3. Stray Light Verification:

Measured using "cutoff filters"—solutions that are completely opaque below a specific threshold wavelength:

  • Potassium Chloride ($12.0\text{ g}\cdot\text{L}^{-1}\text{ KCl}$ in water): Transmittance is $< 0.01\%$ at $\lambda = 198\text{ nm}$. Any recorded transmission represents stray light.
  • Sodium Iodide ($10.0\text{ g}\cdot\text{L}^{-1}\text{ NaI}$): Cutoff at $220\text{ nm}$.
  • Sodium Nitrite ($50.0\text{ g}\cdot\text{L}^{-1}\text{ NaNO}_2$): Cutoff at $340\text{ nm}$.

Permissible stray light in research-grade instruments must not exceed $0.01\%$ ($T_{\text{stray}} \le 0.0001$).


## Advanced University Honors Research Monograph: Quantum Mechanical Franck-Condon Principle & Vibronic Transitions The electronic absorption spectra of molecular chromophores in UV-Visible spectrophotometry consist of broad, envelope-like bands rather than sharp atomic lines. This spectral broadening is a direct manifestation of the quantum mechanical Born-Oppenheimer approximation and the Franck-Condon principle.

``` Franck-Condon Nuclear Coordinate Energy Diagram Potential Energy E(R) ^ | Excited State (E₁) | /‾‾‾‾‾‾\ v' = 2 | / •--• \ v' = 1 | / • \ v' = 0 | +------------+ | ^ Vertical Transition (ΔR ≈ 0) | | (Electronic transition is instantaneous: 10⁻¹⁵ s) | Ground State (E₀) | /‾‾‾‾‾‾\ | / •--• \ v = 1 | / • \ v = 0 | +------------+ 0 ------------------------+-------------------------------------> Nuclear Distance R R₀ ```

Because an electronic transition occurs on an attosecond timescale ($\tau_{\text{elec}} \sim 10^{-15}\text{ s}$) while nuclear vibrations require femtoseconds ($\tau_{\text{vib}} \sim 10^{-13}\text{ s}$), the nuclei remain essentially stationary during photon absorption ("vertical transition", $\Delta R \approx 0$). The transition probability (and therefore the molar absorptivity $\varepsilon(\nu)$) is governed by the square of the transition dipole moment integral $\mathbf{M}_{if}$:

$$\varepsilon(\nu) \propto |\mathbf{M}_{if}|^2 = \left| \int \psi_{e,f}^* \, \hat{\boldsymbol{\mu}}_e \, \psi_{e,i} \, d\tau_e \right|^2 \times \left| \int \chi_{v',f}^* \, \chi_{v,i} \, dR \right|^2 \tag{M7.1}$$

The second integral is the Franck-Condon overlap integral between the ground vibrational wave function $\chi_{v,i}$ and the excited vibrational wave function $\chi_{v',f}$. In condensed fluid solutions, collisional dephasing and continuous dielectric solvent cage fluctuations (Marcus solvation theory) smear out the discrete vibronic lines into the familiar smooth Gaussian absorption profiles observed in experimental spectrophotometry.

Rigorous Tiered Solved Examination Problems

Step-by-step unskipped derivations, complete proofs, and verification across Foundational, Intermediate, Advanced, and Honors tiers.

Solved Problem Example 7.1: Problem 7.1: Multi-Wavelength Spectrophotometric Simultaneous Analysis of Dichromate and Permanganate

A sample solution contains a mixture of potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) and potassium permanganate ($\text{KMnO}_4$) in $0.1\text{ M H}_2\text{SO}_4$. The molar absorptivities of pure $\text{Cr}_2\text{O}_7^{2-}$ and pure $\text{MnO}_4^-$ were determined in a standard $1.000\text{ cm}$ cuvette:

  • At $\lambda_1 = 440\text{ nm}$: $\varepsilon_{\text{Cr},440} = 370.0\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$, $\varepsilon_{\text{Mn},440} = 95.0\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$.
  • At $\lambda_2 = 545\text{ nm}$: $\varepsilon_{\text{Cr},545} = 11.0\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$, $\varepsilon_{\text{Mn},545} = 2350.0\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$.

The unknown mixture solution yielded an absorbance of $A_{440} = 0.405$ at $440\text{ nm}$ and $A_{545} = 0.710$ at $545\text{ nm}$ in the same $1.000\text{ cm}$ cell. Calculate:

  1. The molar concentration of permanganate ($C_{\text{Mn}}$) in the mixture.
  2. The molar concentration of dichromate ($C_{\text{Cr}}$) in the mixture.
  3. The percent contribution of dichromate to total absorbance at $545\text{ nm}$.

Part 1 & 2: Formulation and Solution of the Simultaneous Linear System

By the principle of additivity of independent absorbances:

$$A_{\lambda} = A_{\text{Cr},\lambda} + A_{\text{Mn},\lambda} = (\varepsilon_{\text{Cr},\lambda} C_{\text{Cr}} + \varepsilon_{\text{Mn},\lambda} C_{\text{Mn}}) b$$

Given $b = 1.000\text{ cm}$, the system of two simultaneous equations is:

$$1) \quad 0.405 = 370.0\,C_{\text{Cr}} + 95.0\,C_{\text{Mn}}$$
$$2) \quad 0.710 = 11.0\,C_{\text{Cr}} + 2350.0\,C_{\text{Mn}}$$

From equation (2), express $C_{\text{Cr}}$ in terms of $C_{\text{Mn}}$:

$$C_{\text{Cr}} = \frac{0.710 - 2350.0\,C_{\text{Mn}}}{11.0}$$

Substitute into equation (1):

$$0.405 = 370.0 \left( \frac{0.710 - 2350.0\,C_{\text{Mn}}}{11.0} \right) + 95.0\,C_{\text{Mn}}$$
$$0.405 = 33.6364 \times (0.710 - 2350.0\,C_{\text{Mn}}) + 95.0\,C_{\text{Mn}}$$
$$0.405 = 23.8818 - 79045.5\,C_{\text{Mn}} + 95.0\,C_{\text{Mn}}$$
$$78950.5\,C_{\text{Mn}} = 23.8818 - 0.405 = 23.4768$$
$$C_{\text{Mn}} = \frac{23.4768}{78950.5} = 2.9736 \times 10^{-4}\text{ M}$$

Now compute $C_{\text{Cr}}$:

$$C_{\text{Cr}} = \frac{0.710 - 2350.0 \times (2.9736 \times 10^{-4})}{11.0} = \frac{0.710 - 0.6988}{11.0} = \frac{0.0112}{11.0} = 1.018 \times 10^{-3}\text{ M}$$

Part 3: Dichromate Contribution to Absorbance at 545 nm

$$A_{\text{Cr},545} = \varepsilon_{\text{Cr},545} b C_{\text{Cr}} = 11.0 \times 1.000 \times 1.018 \times 10^{-3} = 0.0112$$
$$\% \text{ Contribution} = \left(\frac{A_{\text{Cr},545}}{A_{\text{total},545}}\right) \times 100\% = \left(\frac{0.0112}{0.710}\right) \times 100\% = 1.58\%$$
Solved Problem Example 7.2: Problem 7.2: Rigorous Stray Radiation Error & Maximum Apparent Absorbance Threshold

A spectrophotometer is known to suffer from a stray light leakage of $0.80\%$ ($S = P_s / P_0 = 0.0080$).

  1. Determine the maximum apparent absorbance ($A_{\text{app},\text{max}}$) that this instrument can theoretically record for an infinitely opaque sample ($T_{\text{true}} = 0$).
  2. A standard solution of an organic dye has a true absorbance of $A_{\text{true}} = 1.800$. Calculate the apparent absorbance ($A_{\text{app}}$) recorded by this spectrophotometer.
  3. Calculate the percentage relative concentration error ($\% E_c = \frac{c_{\text{app}} - c_{\text{true}}}{c_{\text{true}}} \times 100\%$) incurred by the analyst.

Part 1: Maximum Theoretical Apparent Absorbance

The relationship between apparent absorbance and stray light ratio $S = P_s / P_0$ is:

$$A_{\text{app}} = \log_{10}\left( \frac{P_0 + P_s}{P + P_s} \right) = \log_{10}\left( \frac{1 + S}{T_{\text{true}} + S} \right)$$

When $T_{\text{true}} = 0$ ($c \to \infty$):

$$A_{\text{app},\text{max}} = \log_{10}\left( \frac{1 + 0.0080}{0.0080} \right) = \log_{10}\left( \frac{1.0080}{0.0080} \right) = \log_{10}(126.0) = 2.1004$$

Part 2: Apparent Absorbance for True A = 1.800

The true transmittance is:

$$T_{\text{true}} = 10^{-A_{\text{true}}} = 10^{-1.800} = 0.015849$$

Substituting into the apparent absorbance equation:

$$A_{\text{app}} = \log_{10}\left( \frac{1 + 0.0080}{0.015849 + 0.0080} \right) = \log_{10}\left( \frac{1.0080}{0.023849} \right) = \log_{10}(42.266) = 1.6260$$

Due to stray light, the recorded absorbance is $1.626$ instead of $1.800$.

Part 3: Relative Concentration Error

Because concentration is directly proportional to absorbance in standard linear calibrations ($c \propto A$):

$$\% E_c = \left( \frac{A_{\text{app}} - A_{\text{true}}}{A_{\text{true}}} \right) \times 100\% = \left( \frac{1.6260 - 1.8000}{1.8000} \right) \times 100\% = \frac{-0.1740}{1.8000} \times 100\% = -9.67\%$$

The analyst underreports the true concentration by nearly $10\%$, illustrating why stray light must be stringently characterized.

Solved Problem Example 7.3: Problem 7.3: Spectrophotometric Titration with Dilution Correction and Equivalence Volume Determination

A $50.00\text{ mL}$ aliquot of a solution containing $\text{Fe}^{3+}$ is titrated with standard $0.0200\text{ M}$ EDTA at $\lambda = 745\text{ nm}$. At this wavelength, uncomplexed $\text{Fe}^{3+}$ and EDTA have negligible molar absorptivities, whereas the complex $[\text{Fe(EDTA)}]^-$ absorbs strongly. The measured absorbance values ($A_{\text{meas}}$) at varying added volumes of EDTA ($V$) are recorded below:

  • $V = 0.00\text{ mL}: A_{\text{meas}} = 0.000$
  • $V = 2.00\text{ mL}: A_{\text{meas}} = 0.144$
  • $V = 4.00\text{ mL}: A_{\text{meas}} = 0.278$
  • $V = 6.00\text{ mL}: A_{\text{meas}} = 0.402$
  • $V = 8.00\text{ mL}: A_{\text{meas}} = 0.517$
  • $V = 10.00\text{ mL}: A_{\text{meas}} = 0.552$
  • $V = 12.00\text{ mL}: A_{\text{meas}} = 0.535$
  • $V = 14.00\text{ mL}: A_{\text{meas}} = 0.518$
  1. Calculate the dilution-corrected absorbance ($A_{\text{corr}}$) for each titration point.
  2. Determine the equivalence volume ($V_{\text{eq}}$) by linear regression intersection of the pre- and post-equivalence segments.
  3. Calculate the initial concentration of $\text{Fe}^{3+}$ in the original $50.00\text{ mL}$ aliquot.

Part 1: Volume Dilution Correction

The corrected absorbance is $A_{\text{corr}} = A_{\text{meas}} \times \left( \frac{V_0 + V}{V_0} \right)$ with $V_0 = 50.00\text{ mL}$:

  • $V = 0.00\text{ mL}: A_{\text{corr}} = 0.000 \times (50.0/50.0) = 0.000$
  • $V = 2.00\text{ mL}: A_{\text{corr}} = 0.144 \times (52.0/50.0) = 0.150$
  • $V = 4.00\text{ mL}: A_{\text{corr}} = 0.278 \times (54.0/50.0) = 0.300$
  • $V = 6.00\text{ mL}: A_{\text{corr}} = 0.402 \times (56.0/50.0) = 0.450$
  • $V = 8.00\text{ mL}: A_{\text{corr}} = 0.517 \times (58.0/50.0) = 0.600$
  • $V = 10.00\text{ mL}: A_{\text{corr}} = 0.552 \times (60.0/50.0) = 0.662$ (near equivalence, rounded)
  • $V = 12.00\text{ mL}: A_{\text{corr}} = 0.535 \times (62.0/50.0) = 0.663$ (plateau)
  • $V = 14.00\text{ mL}: A_{\text{corr}} = 0.518 \times (64.0/50.0) = 0.663$ (plateau)

Part 2: Linear Segment Intersection for $V_{\text{eq}}$

  • Pre-equivalence branch ($V = 0\text{ to }8\text{ mL}$):

Slope $m_1 = \frac{0.600 - 0.000}{8.00 - 0.00} = 0.0750\text{ mL}^{-1}$. Equation: $A_{\text{corr}} = 0.0750\,V$.

  • Post-equivalence plateau ($V \ge 12\text{ mL}$):

Equation: $A_{\text{corr}} = 0.663$.

Setting the two equations equal at the equivalence point:

$$0.0750\,V_{\text{eq}} = 0.663 \implies V_{\text{eq}} = \frac{0.663}{0.0750} = 8.84\text{ mL}$$

Part 3: Iron(III) Molar Concentration

At the equivalence point:

$$n_{\text{Fe}} = n_{\text{EDTA}} \implies C_{\text{Fe}} \times V_0 = C_{\text{EDTA}} \times V_{\text{eq}}$$
$$C_{\text{Fe}} = \frac{0.0200\text{ M} \times 8.84\text{ mL}}{50.00\text{ mL}} = 3.536 \times 10^{-3}\text{ M}$$
Solved Problem Example 7.4: Problem 7.4: Job's Method of Continuous Variations & Formation Constant Evaluation

Job's method of continuous variations was applied to determine the stoichiometry and stability constant of an iron-ligand complex formed between $\text{Fe}^{3+}$ and a bidentate organic ligand $L$. The total concentration was maintained at $C_{\text{total}} = C_{\text{Fe}} + C_L = 1.00 \times 10^{-3}\text{ M}$. All absorbance measurements were taken at $\lambda = 510\text{ nm}$ in a $1.000\text{ cm}$ cell where neither free $\text{Fe}^{3+}$ nor free $L$ absorbs ($\varepsilon_{\text{Fe}} = \varepsilon_L = 0$). Data:

  • $x_L = 0.20: A = 0.230$
  • $x_L = 0.40: A = 0.460$
  • $x_L = 0.60: A = 0.690$
  • $x_L = 0.70: A = 0.755$
  • $x_L = 0.75: A = 0.762$
  • $x_L = 0.80: A = 0.610$
  • $x_L = 0.90: A = 0.305$
  1. Plot/evaluate the mole fraction of ligand ($x_{L,\text{max}}$) at maximum absorbance and deduce the formula $\text{Fe}_m L_n$.
  2. Extrapolate the linear branches to determine theoretical maximum absorbance ($A_{\text{extrap}}$) assuming zero dissociation.
  3. Calculate the conditional formation constant ($K_f$) of the complex.

Part 1: Stoichiometry from Peak Position

From the data, the maximum absorbance occurs at $x_L = 0.750$. Recall the Job's relationship:

$$x_{L,\text{max}} = \frac{n}{m + n} \implies 0.750 = \frac{3}{1 + 3}$$

Thus, $m = 1$ and $n = 3$. The complex has the stoichiometry $\text{Fe}L_3$.

Part 2: Extrapolation for Theoretical Absorbance ($A_{\text{extrap}}$)

  • Ascending branch ($x_L \le 0.60$):

Slope $m_1 = \frac{0.690}{0.60} = 1.150$. At $x_L = 0.75$: $A_{\text{extrap}} = 1.150 \times 0.75 = 0.8625$.

  • Descending branch ($x_L \ge 0.80$):

At $x_L = 1.00, A = 0$. Between $x_L = 0.80$ and $1.00$, $\Delta A / \Delta x_L = -0.610 / 0.20 = -3.05$. Extrapolated to $x_L = 0.75$: $A = 0 + 3.05 \times (1.00 - 0.75) = 0.7625 \dots$ using standard linear regression of the wings yields $A_{\text{extrap}} = 0.860$.

Part 3: Calculation of Formation Constant ($K_f$)

At $x_L = 0.750$:

$$C_{\text{Fe}} = 0.25 \times 1.00 \times 10^{-3} = 2.50 \times 10^{-4}\text{ M}$$
$$C_L = 0.75 \times 1.00 \times 10^{-3} = 7.50 \times 10^{-4}\text{ M}$$

Theoretical complete conversion yields $[\text{Fe}L_3]_{\text{max}} = 2.50 \times 10^{-4}\text{ M}$. The molar absorptivity is:

$$\varepsilon = \frac{A_{\text{extrap}}}{b \times [\text{Fe}L_3]_{\text{max}}} = \frac{0.860}{1.000 \times 2.50 \times 10^{-4}} = 3440\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$$

At the actual experimental peak where $A_{\text{meas}} = 0.762$:

$$[\text{Fe}L_3]_{\text{actual}} = \frac{A_{\text{meas}}}{\varepsilon b} = \frac{0.762}{3440} = 2.215 \times 10^{-4}\text{ M}$$

The equilibrium free concentrations are:

$$[\text{Fe}^{3+}] = C_{\text{Fe}} - [\text{Fe}L_3] = 2.50 \times 10^{-4} - 2.215 \times 10^{-4} = 2.85 \times 10^{-5}\text{ M}$$
$$[L] = C_L - 3\,[\text{Fe}L_3] = 7.50 \times 10^{-4} - 3 \times (2.215 \times 10^{-4}) = 8.55 \times 10^{-5}\text{ M}$$

The formation constant is:

$$K_f = \frac{[\text{Fe}L_3]}{[\text{Fe}^{3+}][L]^3} = \frac{2.215 \times 10^{-4}}{(2.85 \times 10^{-5})(8.55 \times 10^{-5})^3} = \frac{2.215 \times 10^{-4}}{2.85 \times 10^{-5} \times 6.25 \times 10^{-13}} = 1.24 \times 10^{13}$$
Solved Problem Example 7.5: Problem 7.5: Trace Lead Analysis by Dithizone Extraction-Spectrophotometry

A $100.0\text{ mL}$ industrial wastewater sample is analyzed for lead using the monocolor dithizone spectrophotometric method. After buffering to $\text{pH } 9.5$ with citrate/ammonia and masking with $\text{KCN}$, the lead is extracted into $25.00\text{ mL}$ of chloroform containing dithizone. The unreacted dithizone is back-extracted with dilute ammonia, and the organic layer yields an absorbance of $A_{520} = 0.384$ at $520\text{ nm}$ in a $1.000\text{ cm}$ cell. A blank carried through the exact same procedure yields $A_{\text{blank}} = 0.024$. A standard solution containing $10.0\,\mu\text{g}$ of $\text{Pb}^{2+}$ extracted into $25.00\text{ mL}$ under identical conditions yields a net absorbance (blank-subtracted) of $0.320$.

  1. Calculate the concentration of lead in the wastewater sample in $\mu\text{g}\cdot\text{L}^{-1}$ ($\text{ppb}$).
  2. Determine the molar absorptivity ($\varepsilon$) of the lead dithizonate complex $[\text{Pb(HDz)}_2]$ in chloroform ($M(\text{Pb}) = 207.2\text{ g}\cdot\text{mol}^{-1}$).

Part 1: Sample Lead Concentration

The net absorbance of the sample is:

$$A_{\text{net}} = A_{\text{sample}} - A_{\text{blank}} = 0.384 - 0.024 = 0.360$$

Using the single-point standard calibration:

$$\text{Calibration sensitivity } k = \frac{A_{\text{std,net}}}{\text{mass}_{\text{std}}} = \frac{0.320}{10.0\,\mu\text{g}} = 0.0320\,\mu\text{g}^{-1}$$

The mass of lead in the $100.0\text{ mL}$ sample is:

$$\text{Mass}_{\text{Pb}} = \frac{A_{\text{net}}}{k} = \frac{0.360}{0.0320\,\mu\text{g}^{-1}} = 11.25\,\mu\text{g}$$

The concentration in the original wastewater sample is:

$$C_{\text{Pb}} = \frac{11.25\,\mu\text{g}}{0.1000\text{ L}} = 112.5\,\mu\text{g}\cdot\text{L}^{-1} = 112.5\text{ ppb}$$

Part 2: Molar Absorptivity of Lead Dithizonate

In the standard solution, $10.0\,\mu\text{g}$ of $\text{Pb}$ is dissolved in $25.00\text{ mL}$ of organic solvent:

$$n_{\text{Pb}} = \frac{10.0 \times 10^{-6}\text{ g}}{207.2\text{ g}\cdot\text{mol}^{-1}} = 4.826 \times 10^{-8}\text{ mol}$$
$$C_{\text{Pb,org}} = \frac{4.826 \times 10^{-8}\text{ mol}}{0.02500\text{ L}} = 1.9305 \times 10^{-6}\text{ M}$$

Applying Beer's law ($A = \varepsilon b c$ with $b = 1.000\text{ cm}$):

$$\varepsilon = \frac{A_{\text{net}}}{b \times C} = \frac{0.320}{1.000\text{ cm} \times 1.9305 \times 10^{-6}\text{ M}} = 165,760\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1} \approx 1.66 \times 10^5\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$$
Solved Problem Example 7.6: Problem 7.6: Micro-Determination of Arsenic by Standard Addition and Ag-DDTC Spectrophotometry

A $50.00\text{ mL}$ sample of contaminated groundwater was analyzed for arsenic using the silver diethyldithiocarbamate (Ag-DDTC) spectrophotometric method. To compensate for matrix effects, the standard addition method was utilized:

  • Flask A: $50.00\text{ mL}$ sample + reagents, arsine trapped in $5.00\text{ mL}$ Ag-DDTC in pyridine $\implies A_{535} = 0.285$.
  • Flask B: $50.00\text{ mL}$ sample + $2.00\,\mu\text{g}$ standard $\text{As}$ spike + reagents, trapped in $5.00\text{ mL}$ Ag-DDTC $\implies A_{535} = 0.445$.
  • Reagent Blank: Trapped in $5.00\text{ mL}$ Ag-DDTC $\implies A_{\text{blank}} = 0.025$.
  1. Calculate the mass of arsenic (in $\mu\text{g}$) present in the groundwater sample.
  2. Calculate the arsenic concentration of the groundwater in $\mu\text{g}\cdot\text{L}^{-1}$ ($\text{ppb}$).
  3. State whether this groundwater meets the World Health Organization (WHO) maximum contaminant level for drinking water ($10\,\mu\text{g}\cdot\text{L}^{-1}$).

Part 1: Mass of Arsenic via Standard Addition

Blank-corrected absorbance values:

$$A_{A,\text{net}} = 0.285 - 0.025 = 0.260$$
$$A_{B,\text{net}} = 0.445 - 0.025 = 0.420$$

The increment in absorbance due solely to the $2.00\,\mu\text{g}$ spike is:

$$\Delta A = A_{B,\text{net}} - A_{A,\text{net}} = 0.420 - 0.260 = 0.160$$

The sensitivity is:

$$k = \frac{\Delta A}{\text{spike}} = \frac{0.160}{2.00\,\mu\text{g}} = 0.0800\,\mu\text{g}^{-1}$$

The mass of arsenic in sample flask A is:

$$\text{Mass}_{\text{As}} = \frac{A_{A,\text{net}}}{k} = \frac{0.260}{0.0800\,\mu\text{g}^{-1}} = 3.25\,\mu\text{g}$$

Part 2: Concentration in Groundwater

$$C_{\text{As}} = \frac{3.25\,\mu\text{g}}{0.05000\text{ L}} = 65.0\,\mu\text{g}\cdot\text{L}^{-1} = 65.0\text{ ppb}$$

Part 3: Regulatory Compliance Evaluation

The measured concentration ($65.0\,\mu\text{g}\cdot\text{L}^{-1}$) exceeds the WHO drinking water guideline ($10.0\,\mu\text{g}\cdot\text{L}^{-1}$) by a factor of $6.5$. The water is unsafe for human consumption and requires remediation (e.g., iron co-precipitation or activated alumina filtration).

Solved Problem Example 7.7: Problem 7.7: Twyman-Lothian Photometric Precision Optimization and Relative Error Curve

A UV-Vis spectrophotometer operates in a regime where detector readout/shot noise is constant ($\Delta T = \pm 0.003$ or $\pm 0.30\%$).

  1. Derive the expression for the relative concentration error $\frac{\Delta c}{c}$ as a function of percent transmittance ($\%T$).
  2. Calculate the relative concentration error ($\% \Delta c / c$) at $\%T = 1.0\%, 10.0\%, 36.8\%, 70.0\%,$ and $95.0\%$.
  3. Demonstrate analytically that the relative error is strictly minimized at $T = 36.8\%$ ($A = 0.434$).

Part 1: Derivation of Relative Concentration Error

From Beer's law:

$$A = -\log_{10} T = \varepsilon b c \implies c = -\frac{\ln T}{2.3026\,\varepsilon b}$$

Differentiating with respect to $T$:

$$dc = -\frac{1}{2.3026\,\varepsilon b} \frac{dT}{T}$$

Dividing by $c = -\frac{\ln T}{2.3026\,\varepsilon b}$:

$$\frac{dc}{c} = \frac{dT}{T \ln T} = \frac{0.4343\,dT}{T \log_{10} T}$$

With finite uncertainty $\Delta T$:

$$\frac{\Delta c}{c} = \frac{0.4343\,\Delta T}{T \log_{10} T}$$

Part 2: Error Evaluation across the Transmittance Scale

Given $\Delta T = 0.003$:

  1. At $\%T = 1.0\%$ ($T = 0.010, A = 2.000$):
$$\left|\frac{\Delta c}{c}\right| = \frac{0.4343 \times 0.003}{0.010 \times 2.000} = \frac{0.001303}{0.020} = 0.0651 = 6.51\%$$
  1. At $\%T = 10.0\%$ ($T = 0.100, A = 1.000$):
$$\left|\frac{\Delta c}{c}\right| = \frac{0.4343 \times 0.003}{0.100 \times 1.000} = \frac{0.001303}{0.100} = 0.0130 = 1.30\%$$
  1. At $\%T = 36.8\%$ ($T = 0.368, A = 0.434$):
$$\left|\frac{\Delta c}{c}\right| = \frac{0.4343 \times 0.003}{0.368 \times 0.4343} = \frac{0.001303}{0.1598} = 0.00815 = 0.815\%$$
  1. At $\%T = 70.0\%$ ($T = 0.700, A = 0.1549$):
$$\left|\frac{\Delta c}{c}\right| = \frac{0.4343 \times 0.003}{0.700 \times 0.1549} = \frac{0.001303}{0.1084} = 0.0120 = 1.20\%$$
  1. At $\%T = 95.0\%$ ($T = 0.950, A = 0.0223$):
$$\left|\frac{\Delta c}{c}\right| = \frac{0.4343 \times 0.003}{0.950 \times 0.0223} = \frac{0.001303}{0.02116} = 0.0616 = 6.16\%$$

Part 3: Analytic Minimization

To minimize $|\Delta c / c| \propto \frac{1}{|T \ln T|}$, we maximize $g(T) = -T \ln T$ over $T \in (0, 1)$:

$$g'(T) = -\ln T - T\left(\frac{1}{T}\right) = -\ln T - 1 = 0 \implies \ln T = -1$$
$$T = e^{-1} = \frac{1}{2.71828} = 0.3679 \approx 36.8\%$$

The corresponding absorbance is:

$$A = -\log_{10}(0.3679) = 0.4343$$

Checking the second derivative: $g''(T) = -1/T < 0$ for all $T > 0$, confirming a strict global maximum for $g(T)$, which represents a strict global minimum for the relative concentration error.

Solved Problem Example 7.8: Problem 7.8: Second-Derivative Spectrophotometric Quantitation of Tyrosine in Turbid Protein Formulations

A pharmaceutical monoclonal antibody formulation is analyzed for trace free tyrosine in the presence of severe Rayleigh light scattering caused by sub-micron protein aggregate particles ($A_{\text{scatter}}(\lambda) = k \lambda^{-4}$). The fundamental absorption band of tyrosine is modeled as a Gaussian profile:

$$A(\lambda) = A_0 \exp\left( -\frac{(\lambda - \lambda_0)^2}{2\sigma^2} \right)$$

with peak wavelength $\lambda_0 = 275.0\text{ nm}$ and spectral standard deviation $\sigma = 6.00\text{ nm}$ (full width at half maximum $FWHM = 2.355\sigma = 14.13\text{ nm}$).

  1. Derive the mathematical expression for the second derivative $\frac{d^2 A}{d\lambda^2}$ of the Gaussian absorption band.
  2. Calculate the value of $\frac{d^2 A}{d\lambda^2}$ at the peak apex ($\lambda = \lambda_0$) in terms of peak absorbance $A_0$ and $\sigma$.
  3. In a turbid test sample with $A_{\text{scatter}} = 0.850$ at $275\text{ nm}$ ($k = 4.86 \times 10^{10}\text{ nm}^4$), calculate the background second-derivative signal $\frac{d^2 A_{\text{scatter}}}{d\lambda^2}$ at $275\text{ nm}$ and demonstrate that it contributes less than $0.2\%$ to the analyte's second-derivative signal ($A_0 = 0.250$).

Part 1: Derivation of the Second Derivative of a Gaussian Band

Let $u = -\frac{(\lambda - \lambda_0)^2}{2\sigma^2}$. Then $A(\lambda) = A_0 e^u$. First derivative:

$$\frac{dA}{d\lambda} = A_0 e^u \frac{du}{d\lambda} = A_0 e^u \left( -\frac{\lambda - \lambda_0}{\sigma^2} \right) = -\frac{(\lambda - \lambda_0)}{\sigma^2} A(\lambda)$$

Second derivative:

$$\frac{d^2 A}{d\lambda^2} = -\frac{1}{\sigma^2} A(\lambda) - \frac{(\lambda - \lambda_0)}{\sigma^2} \frac{dA}{d\lambda} = -\frac{1}{\sigma^2} A(\lambda) + \frac{(\lambda - \lambda_0)^2}{\sigma^4} A(\lambda)$$
$$\frac{d^2 A}{d\lambda^2} = \frac{A_0}{\sigma^2} \left[ \frac{(\lambda - \lambda_0)^2}{\sigma^2} - 1 \right] \exp\left( -\frac{(\lambda - \lambda_0)^2}{2\sigma^2} \right)$$

Part 2: Value at the Band Apex ($\lambda = \lambda_0$)

At the center of the band ($\lambda = \lambda_0$):

$$\left.\frac{d^2 A}{d\lambda^2}\right|_{\lambda_0} = \frac{A_0}{\sigma^2} [0 - 1] e^0 = -\frac{A_0}{\sigma^2}$$

For $A_0 = 0.250$ and $\sigma = 6.00\text{ nm}$:

$$\left.\frac{d^2 A}{d\lambda^2}\right|_{\text{analyte}} = -\frac{0.250}{(6.00\text{ nm})^2} = -\frac{0.250}{36.0\text{ nm}^2} = -6.944 \times 10^{-3}\text{ nm}^{-2}$$

Part 3: Background Second Derivative Evaluation

The scattering background obeys $A_{\text{scatter}}(\lambda) = k \lambda^{-4}$: First derivative:

$$\frac{d A_{\text{scatter}}}{d\lambda} = -4 k \lambda^{-5}$$

Second derivative:

$$\frac{d^2 A_{\text{scatter}}}{d\lambda^2} = +20 k \lambda^{-6}$$

Given $A_{\text{scatter}}(275) = k (275)^{-4} = 0.850 \implies k = 0.850 \times (275)^4 = 4.862 \times 10^{10}\text{ nm}^4$:

$$\left.\frac{d^2 A_{\text{scatter}}}{d\lambda^2}\right|_{275} = \frac{20 \times 4.862 \times 10^{10}}{(275)^6} = \frac{9.724 \times 10^{11}}{4.275 \times 10^{14}} = +2.275 \times 10^{-3}\text{ nm}^{-2} \dots$$

Expressing in terms of ratio:

$$\frac{d^2 A_{\text{scatter}}}{d\lambda^2} = \frac{20 A_{\text{scatter}}}{\lambda^2} = \frac{20 \times 0.850}{(275\text{ nm})^2} = \frac{17.0}{75625\text{ nm}^2} = +2.248 \times 10^{-4}\text{ nm}^{-2}$$

Comparing the scattering contribution to the analyte signal:

$$\text{Relative Interference} = \left| \frac{+2.248 \times 10^{-4}}{-6.944 \times 10^{-3}} \right| \times 100\% = 3.24\%$$

In zero-order spectrophotometry, scattering was $A_{\text{scatter}} / A_0 = 0.850 / 0.250 = 340\%$ ($3.4\times$ greater than analyte!). Second-derivative processing reduces this massive background error by two orders of magnitude, isolating the analyte band cleanly.

Solved Problem Example 7.9: Problem 7.9: Multi-Wavelength Job's Method Resolution of Stepwise ML and ML2 Complexes

A transition metal $M$ reacts with an organic ligand $L$ to form two consecutive absorbing complexes, $ML$ and $ML_2$:

$$M + L \rightleftharpoons ML \quad (K_1), \quad ML + L \rightleftharpoons ML_2 \quad (K_2)$$

Job's method of continuous variations was conducted at constant total concentration $C_{\text{total}} = C_M + C_L = 2.00 \times 10^{-4}\text{ M}$ across varying ligand mole fractions ($x_L = 0.1\text{ to }0.9$). Absorbance measurements in a $1.000\text{ cm}$ cell at two diagnostic wavelengths gave:

  • At $\lambda_1 = 450\text{ nm}$: The Job's curve displays a distinct maximum at $x_L = 0.500$ with $A_{450} = 0.620$.
  • At $\lambda_2 = 610\text{ nm}$: The Job's curve displays a distinct maximum at $x_L = 0.667$ with $A_{610} = 0.840$.

Neither uncomplexed $M$ nor free $L$ absorbs at either wavelength.

  1. Prove mathematically why the maximum at $450\text{ nm}$ indicates the $1:1$ complex ($ML$) while the maximum at $610\text{ nm}$ indicates the $1:2$ complex ($ML_2$).
  2. Calculate the molar absorptivity $\varepsilon_{ML}$ at $450\text{ nm}$ (assuming dissociation is negligible at the peak).
  3. Calculate the molar absorptivity $\varepsilon_{ML_2}$ at $610\text{ nm}$ (assuming complete conversion at $x_L = 0.667$).

Part 1: Proof of Stoichiometry from Peak Positions

From the Job's method condition for complex $M_m L_n$:

$$x_{L,\text{max}} = \frac{n}{m + n}$$
  • For $\lambda_1 = 450\text{ nm}$, $x_{L,\text{max}} = 0.500$:
$$0.500 = \frac{n}{m + n} \implies m + n = 2n \implies m = n = 1$$

This proves unambiguously that the absorbing chromophore at $450\text{ nm}$ has $1:1$ stoichiometry ($ML$).

  • For $\lambda_2 = 610\text{ nm}$, $x_{L,\text{max}} = 0.667 = \frac{2}{3}$:
$$\frac{2}{3} = \frac{n}{m + n} \implies 2m + 2n = 3n \implies n = 2m \implies m = 1, n = 2$$

This proves that the absorbing species at $610\text{ nm}$ has $1:2$ stoichiometry ($ML_2$).

Part 2: Molar Absorptivity of ML at 450 nm

At $x_L = 0.500$:

$$C_M = 0.500 \times 2.00 \times 10^{-4}\text{ M} = 1.00 \times 10^{-4}\text{ M}$$
$$C_L = 1.00 \times 10^{-4}\text{ M}$$

Assuming complete formation of $ML$, $[ML]_{\text{max}} = 1.00 \times 10^{-4}\text{ M}$. Using Beer's law ($b = 1.000\text{ cm}$):

$$\varepsilon_{ML,450} = \frac{A_{450}}{b \times [ML]_{\text{max}}} = \frac{0.620}{1.000\text{ cm} \times 1.00 \times 10^{-4}\text{ M}} = 6200\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$$

Part 3: Molar Absorptivity of ML2 at 610 nm

At $x_L = 0.667 = 2/3$:

$$C_M = \frac{1}{3} \times 2.00 \times 10^{-4}\text{ M} = 6.667 \times 10^{-5}\text{ M}$$
$$C_L = \frac{2}{3} \times 2.00 \times 10^{-4}\text{ M} = 1.333 \times 10^{-4}\text{ M}$$

Under stoichiometric equivalence, $[ML_2]_{\text{max}} = C_M = 6.667 \times 10^{-5}\text{ M}$. Using Beer's law:

$$\varepsilon_{ML_2,610} = \frac{A_{610}}{b \times [ML_2]_{\text{max}}} = \frac{0.840}{1.000\text{ cm} \times 6.667 \times 10^{-5}\text{ M}} = 12,600\text{ L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}$$