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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.

My two questions about this are:

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.

I hope I have described the question properly.

Many thanks,

Sensei_Stig

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.

My two questions about this are:

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.

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.

My two questions about this are:

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.

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.

My two questions about this are:

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.

I hope I have described the question properly.

Many thanks,

Sensei_Stig

I've been tasked (as part of a report) to calculate the pKa of 4-nitrophenol by measuring the pH and absorbance of six 'buffer solutions' containing the 4-nitrophenol (described in Table 1 - my apologies on the formatting, can't figure out how to make stack exchange recognise the $\LaTeX$ code for it). We have to then plot a graph of $\text{pH}$ versus $\log_{10}(\frac{A}{A_{f}-A})$ and use that graph to calculate $\text{pK}_{a}$. $\text{A}_{f}$ 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.

My two questions about this are:

1. Where does the expression $\log_{10}(\frac{A}{A_{f}-A})$ 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,

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

And substituting into the Henderson-Hasselbalch equation:

$$ \begin{align} pH &= pK_{a}+log_{10}\frac{[conj. base]}{[acid]} \\ pH &= pK_{a}+log_{10}\frac{A_{f}}{A_{f}-A} \\ \end{align} $$

2. Once I have the graph, how do I calculate $\text{pK}_{a}$ from it? Assumed it would be using the gradient of the line of best fit, but that won't work since $\text{pK}_{a}$ is given by the difference of $\text{pH}$ and $log_{10}\frac{A}{A_{f}-A}$ rather than the quotient.

I hope I have described the issue properly.

Many thanks,

Sensei_Stig

Table 1: Volumes of components in each solution $(cm^{3})$
-----------------base --- acid --- phenol
Solution 1 ---- 2.80 --- 7.20 --- 1.00
Solution 2 ---- 5.00 --- 5.00 --- 1.00
Solution 3 ---- 6.20 --- 3.80 --- 1.00
Solution 4 ---- 7.20 --- 2.80 --- 1.00
Solution 5 ---- 8.70 --- 1.30 --- 1.00
Solution 6 --- 10.00 ------ 0 --- 1.00

(Again, my apologies for the formatting)

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.

My two questions about this are:

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.