# Determination of pKa by absorbance and pH of buffer solutions

I've been tasked (as part of a report) to calculate the $\mathrm pK_\mathrm a$ of 4-nitrophenol by measuring the pH and absorbance of six 'buffer solutions' containing the 4-nitrophenol:

$$\begin{array}{cccc}\hline \text{Solution} & \text{Volume of base / }\pu{cm^3} & \text{Volume of acid / }\pu{cm^3} & \text{Volume of phenol / }\pu{cm^3} \\ \hline 1 & 2.80 & 7.20 & 1.00 \\ 2 & 5.00 & 5.00 & 1.00 \\ 3 & 6.20 & 3.80 & 1.00 \\ 4 & 7.20 & 2.80 & 1.00 \\ 5 & 8.70 & 1.30 & 1.00 \\ 6 & 10.00 & 0.00 & 1.00 \\ \hline \end{array}$$

We have to then plot a graph of $\mathrm{pH}$ versus $\log_{10}[A/(A_f-A)]$ and use that graph to calculate $\mathrm pK_\mathrm a$. $\text{A}_{f}$ is the absorbance of Solution 6 - we assume that the weak acid in the buffer is present only as its conjugate base - and A is the absorbance of the buffer solution.

1) Where does the expression $\log_{10}[A/(A_f-A)]$ come from? I thought the derivation would be as follows, which clearly gives a different result ($\text{A}_{f}$ on the numerator rather than $\text{A}$):

Using the Beer-Lambert Law,

\begin{align} A &= \epsilon cl \\ c_\text{base} &= \frac{A_{f}}{\epsilon l} \\ \end{align}

And substituting into the Henderson–Hasselbalch equation:

\begin{align} \mathrm{pH} &= \mathrm pK_\mathrm a + \log_{10}\left(\frac{[\text{conj. base}]}{[\text{acid}]}\right) \\ \mathrm{pH} &= \mathrm pK_\mathrm a + \log_{10}\left(\frac{A_{f}}{A_{f}-A}\right) \\ \end{align}

2) Once I have the graph, how do I calculate $\mathrm pK_\mathrm a$ from it? Assumed it would be using the gradient of the line of best fit, but that won't work since $\mathrm pK_\mathrm a$ is given by the difference of $\mathrm{pH}$ and $\log_{10}[A/(A_f-A)]$ rather than the quotient.

Make the plot $10^{\mathrm{pH}}(A_{\ce{L-}} - A)$ versus $A$, the adjusted equation will be of the form $$10^{\mathrm{pH}}(A_{\ce{L-}} - A) = 10^{\mathrm{p}K_\mathrm{a}}A - 10^{\mathrm{p}K_\mathrm{a}}A_{\ce{HL}},$$ the slope is $10^{\mathrm{p}K_\mathrm{a}}$, that is, $\mathrm{p}K_\mathrm{a} = \log_{10}(\text{slope})$.