Chapter 14, Problem 32QAP

Enough water is added to the buffer in Question 30 to make the total volume 5.00 L.

(a) Calculate the pH of the buffer.

(b) Calculate the pH of the buffer after adding 0.0250 mol of HCl to 0.376 L of the buffer.

(c) Calculate the pH of the buffer after adding 0.0250 mol of KOH to 0.376 L of the buffer.

(d) Compare your answers to Question 30 (a-c) with your answers to (a-c) of this problem.

(e) Comment on the effect of dilution on the pH of a buffer and on its buffer capacity.

 

Interpretation Introduction

(a)

Interpretation:

The pH of the given buffer should be calculated.

Concept introduction:

The pH of a buffer solution is given by Henderson Hasselbalch’s equation which is:

$\text{pH = p}{K}_{\text{a}}+\text{log}\frac{\text{[salt]}}{\text{[acid]}}$

Where,

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

Answer to Problem 32QAP

The pH of the buffer solution is 4.29.

Explanation of Solution

The given values are listed below.

• Volume of potassium hydrogen tartrate ( ${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6$) is 239 mL
• Molarity of potassium hydrogen tartrate ( ${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6$) = 0.187 M
• Volume of potassium tartrate ( ${\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6$) = 137 mL
• Molarity of potassium tartrate ( ${\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6$) =0.288 M
• $K_{\text{a}}$ of ${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6$ is $\text{4}\text{.55}\times {\text{10}}^{-\text{5}}$.

The moles of the salt and acid is calculated as follows:

$\begin{array}{c}\text{Moles of}{\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6=\text{ molarity}\times \text{volume (L) }\\ \text{= 0}\text{.288 mol/L}\times \text{0}\text{.137 L}\\ \text{= 0}\text{.0395 mol}\end{array}$

Also,

$\begin{array}{c}{\text{Moles of KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6=\text{ molarity}\times \text{volume (L)}\\ \text{= 0}\text{.187 mol/L}\times \text{0}\text{.239 L }\\ \text{= 0}\text{.0447 mol}\end{array}$

It is given that 5 L is further added to the solution.

The final concentration of the salt and acid is calculated as follows:

$\begin{array}{c}\left[{\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6\right]\text{= }\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}\\ =\frac{\text{0}\text{.0395 mol}}{5\text{ L}}\\ =\text{0}\text{.0079 mol/L}\end{array}$

Similarly,

$\begin{array}{c}\left[{\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6\right]\text{ = }\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}\\ =\frac{\text{0}\text{.0447 mol}}{5\text{ L}}\\ =\text{0}\text{.0089 mol/L}\end{array}$

The relation between $\text{p}{K}_{\text{a}}$ and $K_{\text{a}}$ is given by the formula,

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

Subsitute the value of $K_{\text{a}}$ for ${\text{H}}_{\text{2}}{\text{C}}_4{\text{H}}_4{\text{O}}_6$ in above formula.

$\begin{array}{c}\text{p}{K}_{\text{a}}=-\text{log}\left[\text{4}\text{.55}\times {\text{10}}^{-\text{5}}\right]\\ =-\left[\log 4.55-5\log 10\right]\\ =4.34\end{array}$

The pH of the buffer is calculated by using Henderson Hasselbalch’s equation as shown below.

$\text{pH = p}{K}_{\text{a}}+\text{log}\frac{\text{[salt]}}{\text{[acid]}}$

Subsitute the value of $\text{p}{K}_{\text{a}}$, concentration of salt and acid in above formula.

$\begin{array}{l}\text{pH = 4}\text{.34}+\text{log}\left(\frac{0.0079}{0.0089}\right)\\ \text{pH = 4}\text{.34}+\text{log}\left(\text{0}\text{.89}\right)\\ \text{pH = 4}\text{.34}+\left(-\text{0}\text{.051}\right)\\ \text{pH = 4}\text{.29}\end{array}$

Therefore, the pH of the buffer solution is 4.29.

Interpretation Introduction

(b)

Interpretation:

The pH of the given buffer should be calculated after addition of 0.0250 moles of HCl.

Concept introduction:

The pH of a buffer solution is given by Henderson Hasselbalch’s equation which is:

$\text{pH = p}{K}_{\text{a}}+\text{log}\frac{\text{[salt]}}{\text{[acid]}}$

Where,

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

Answer to Problem 32QAP

The pH of the solution is 1.23 after adding the 0.0250 moles of HCl.

Explanation of Solution

The given values are listed below.

• Volume of potassium hydrogen tartrate ( ${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6$) is 239 mL
• Molarity of potassium hydrogen tartrate ( ${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6$) = 0.187 M
• Volume of potassium tartrate ( ${\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6$) = 137 mL
• Molarity of potassium tartrate ( ${\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6$) =0.288 M
• $K_{\text{a}}$ of ${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6$ is $\text{4}\text{.55}\times {\text{10}}^{-\text{5}}$.

The moles of the salt and acid is calculated as follows:

$\begin{array}{c}\text{Moles of}{\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6=\text{ molarity}\times \text{volume (L) }\\ \text{= 0}\text{.288 mol/L}\times \text{0}\text{.137 L}\\ \text{= 0}\text{.0395 mol}\end{array}$

Also,

$\begin{array}{c}{\text{Moles of KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6=\text{ molarity}\times \text{volume (L)}\\ \text{= 0}\text{.187 mol/L}\times \text{0}\text{.239 L }\\ \text{= 0}\text{.0447 mol}\end{array}$

It is given that 5 L is further added to the solution.

The final concentration of the salt and acid is calculated as follows:

$\begin{array}{c}\left[{\text{K}}_2{\text{C}}_4{\text{H}}_{\text{4}}{\text{O}}_6\right]\text{= }\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}\\ =\frac{\text{0}\text{.0395 mol}}{5\text{ L}}\\ =\text{0}\text{.0079 mol/L}\end{array}$

And,

$\begin{array}{c}\left[{\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6\right]\text{ = }\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}\\ =\frac{\text{0}\text{.0447 mol}}{5\text{ L}}\\ =\text{0}\text{.0089 mol/L}\end{array}$

It is given that the 0.376 L of above solution is taken. Therefore, the new moles of salt and acid are calculated as shown below.

$\begin{array}{c}\text{Moles of}{\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6=\text{molarity}\times \text{volume (L) }\\ \text{= 0}\text{.0079 mol/L}\times \text{0}\text{.376 L}\\ \text{= 0}\text{.003 mol}\end{array}$

$\begin{array}{c}{\text{Moles of KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6=\text{molarity}\times \text{volume(L)}\\ \text{= 0}\text{.0089 mol/L}\times \text{0}\text{.376 L}\\ \text{= 0}\text{.00335 mol}\end{array}$

Therefore, the new moles of salt and acid are 0.003 mol and 0.00335 mol respectively.

It is given that the 0.0250 moles of HCl is added to the 0.376 L of buffer solution. The proton will react with the salt to produce weakly dissociated acid. The reaction involved during the mixing of acid to the buffer solution as shown below.

${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6{}^{2-}\text{+ 2HCl}\stackrel{}{\to }{\text{H}}_2{\text{C}}_4{\text{H}}_{\text{4}}{\text{O}}_6+2{\text{Cl}}^{-}$

In the above reaction, the salt is a limiting agent. Therefore, all amount of salt would be reacted to produce weak acid. The concentration of each species is shown below.

$\begin{array}{ccccccc}{\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6& +& \text{2HCl}& \to & {\text{H}}_2{\text{C}}_4{\text{H}}_{\text{4}}{\text{O}}_6& +& 2{\text{Cl}}^{-}\\ 0.003\text{mol}& & 0.0250\text{mol}& & 0.00335\text{mol}& & -\\ 0& & \left(0.0250-0.003\right)=0.022\text{mol}& & \left(0.00335+0.003\right)=0.00635\text{mol}& & \end{array}$

The final solution of strong acid and weak acid is obtained. It is not a buffer solution. ${\text{H}}_2{\text{C}}_4{\text{H}}_{\text{4}}{\text{O}}_6$ is a weak acid and will not dissociate completely. Therefore, the pH of the solution is due to the presence of HCl. The final moles of HCl are 0.022 mol.

The concentration of HCl is calculated as follows:

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{\text{number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}\\ =\frac{\text{0}\text{.022}\text{mol}}{\text{0}\text{.376}\text{L}}\\ =0.0585\text{ mol/L}\end{array}$

The pH of the solution is calculated as follows:

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.0585\right)\\ =-\left(-1.23\right)\\ =1.23\end{array}$

Therefore, the pH of the solution is 1.23.

Interpretation Introduction

(c)

Interpretation:

The pH of the given buffer should be calculated after addition of 0.0250 moles of KOH.

Concept introduction:

The pH of a buffer solution is given by Henderson Hasselbalch’s equation which is:

$\text{pH = p}{K}_{\text{a}}+\text{log}\frac{\text{[salt]}}{\text{[acid]}}$

Where,

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

Answer to Problem 32QAP

The pH of the buffer solution is 12.76 after adding the 0.250 moles of KOH.

Explanation of Solution

The given values are listed below.

• Volume of potassium hydrogen tartrate ( ${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6$) is 239 mL
• Molarity of potassium hydrogen tartrate ( ${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6$) = 0.187 M
• Volume of potassium tartrate ( ${\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6$) = 137 mL
• Molarity of potassium tartrate ( ${\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6$) =0.288 M
• $K_{\text{a}}$ of ${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6$ is $\text{4}\text{.55}\times {\text{10}}^{-\text{5}}$.

The moles of the salt and acid is calculated as follows:

$\begin{array}{c}\text{Moles of}{\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6=\text{ molarity}\times \text{volume (L) }\\ \text{= 0}\text{.288}\times \text{0}\text{.137}\\ \text{= 0}\text{.0395 mol}\end{array}$

And,

$\begin{array}{c}{\text{Moles of KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6=\text{molarity}\times \text{volume (L)}\\ \text{= 0}\text{.187}\times \text{0}\text{.239 }\\ \text{= 0}\text{.0447 mol}\end{array}$

It is given that 5 L is further added to the solution.

The final concentration of the salt and acid is calculated as follows:

$\begin{array}{c}\left[{\text{K}}_2{\text{C}}_4{\text{H}}_{\text{4}}{\text{O}}_6\right]\text{= }\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}\\ =\frac{\text{0}\text{.0395 mol}}{5\text{ L}}\\ =\text{0}\text{.0079 mol/L}\end{array}$

Also,

$\begin{array}{c}\left[{\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6\right]\text{ = }\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}\\ =\frac{\text{0}\text{.0447 mol}}{5\text{ L}}\\ =\text{0}\text{.0089 mol/L}\end{array}$

It is given that the 0.376 L of above solution is taken. Therefore, the new moles of salt and acid are calculated as shown below.

$\begin{array}{c}\text{Moles of}{\text{K}}_{\text{2}}{\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6=\text{molarity}\times \text{volume (L) }\\ \text{= 0}\text{.0079 mol/L}\times \text{0}\text{.376 L}\\ \text{= 0}\text{.003 mol}\end{array}$

And,

$\begin{array}{c}{\text{Moles of KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6=\text{molarity}\times \text{volume(L)}\\ \text{= 0}\text{.0089 mol/L}\times \text{0}\text{.376 L}\\ \text{= 0}\text{.00335 mol}\end{array}$

Therefore, the new moles of salt and acid are 0.003 and 0.00335 moles respectively.

It is given that the 0.0250 moles of KOH is added to the 0.376 L of buffer solution. The hydroxide ion abstract a proton from acid. The reaction involved during the mixing of base to the buffer solution as shown below.

${\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6{\text{+ OH}}^{-}\to {\text{KC}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6{}^{-}{\text{+ H}}_{\text{2}}\text{O}$

In the above reaction, the acid is a limiting agent. Therefore, all amount of acid would be reacted to produce strong base. The concentration of each specie is shown below.

$\begin{array}{ccccccc}{\text{KHC}}_4{\text{H}}_{\text{4}}{\text{O}}_6& +& {\text{OH}}^{-}& \to & {\text{KC}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_6{}^{-}& +& {\text{H}}_{\text{2}}\text{O}\\ 0.00335\text{mol}& & 0.0250\text{mol}& & 0.003\text{mol}& & -\\ 0& & \left(0.0250-0.00335\right)=0.02165\text{mol}& & \left(0.003+0.00335\right)=0.00635\text{mol}& & \end{array}$

The final solution of strong base and salt is obtained. It is not a buffer solution. Therefore, the pOH of the solution is due to the presence of ${\text{OH}}^{-}$. The final moles of ${\text{OH}}^{-}$ are 0.02165 mol.

The concentration of ${\text{OH}}^{-}$ is calculated as follows:

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{\text{number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}\\ =\frac{\text{0}\text{.02165}\text{mol}}{\text{0}\text{.376}\text{L}}\\ =0.0575\end{array}$

The pOH of the solution is calculated as follows:

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.0575\right)\\ =-\left(-1.24\right)\\ =1.24\end{array}$

The pH is calculated as follows:

$\begin{array}{c}\text{pH}=14-\text{pOH}\\ =14-1.24\\ =12.76\end{array}$

Therefore, the pH of the solution is 12.76.

Interpretation Introduction

(d)

Interpretation:

The pH value from (a-c) in question 30 should be compared with the of pH value from (a-c) in question 32.

Concept introduction:

The pH of a buffer solution is given by Henderson Hasselbalch’s equation which is:

$\text{pH = p}{K}_{\text{a}}+\text{log}\frac{\text{[salt]}}{\text{[acid]}}$

Where,

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

Answer to Problem 32QAP

The pH value of part (a) in exercise 30 and 32 are same. The pH value of part (b) and part (c) in exercise 30 and 32 are different. In part (b) and (c) of exercise 32, the final solution after adding an acid or a base are not buffer. The calculations are different.

Explanation of Solution

The pH value of part (a) in exercise 30 and 32 are same that is 4.29. This is because the ratio of log of concentration of salt to the concentration of an acid is same.

The pH value of part (b) in exercise 30 and 32 are different. In exercise 30, the final solution is obtained as buffer solution. The pH of the buffer solution is calculated by using the Henderson Hasselbalch’s equation. In exercise 32, the final solution is not buffer. Therefore, the pH of the solution is calculated by the concentration of the hydrogen ions.

The pH value of part (c) in exercise 30 and 32 are different. In exercise 30, the final solution is obtained as buffer solution. The pH of the buffer solution is calculated by using the Henderson Hasselbalch’s equation. In exercise 32, the final solution is not buffer. Firstly, the pOH of the solution is calculated by the concentration of the hydroxide ions and then the pH is calculated by substracting the pOH from 14.

Interpretation Introduction

(e)

Interpretation:

The effect of dilution on the pH of a buffer and on its buffer capacity should be stated.

Concept introduction:

The pH of a buffer solution is given by Henderson Hasselbalch’s equation which is:

$\text{pH = p}{K}_{\text{a}}+\text{log}\frac{\text{[salt]}}{\text{[acid]}}$

Where,

$\text{p}{K}_{\text{a}}=-\log K_{\text{a}}$

Answer to Problem 32QAP

The dilution does not affect the pH of the buffer solution because the ratio of the concentration of the conjugate base to the concentration of conjugate acid remains constant. It reduces the buffer capacity.

Explanation of Solution

Buffer capacity tells the efficiency of the buffer to stop the change in the pH. Dilution reduces the buffer capacity.

There is no effect of the dilution on the pH of a buffer solution, because the ratio of the concentration of the conjugate base to the concentration of conjugate acid remains constant.