Chapter 1, Problem 20P

Interpretation Introduction

Interpretation: The $\text{pH}$ of the solution of sodium acetate with concentration of $0.1\text{ M}$ and $0.01\text{ M}$ is to be determined with varying volume of HCl added to it.

Concept introduction: To calculate the $\text{pH}$ of the solution Henderson and Hasselbalch equation is used. The expression is as given below,

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]}$

Where,

• $\text{p}{K}_{\text{a}}$ is dissociation constant.
• $\left[\text{Base}\right]$ is the concentration of base.
• $\left[\text{Acid}\right]$ is the concentration of acid.

Answer to Problem 20P

The $\text{pH}$ values when different concentrations of $\text{HCl}$ is added to the $0.1\text{ M}$ sodium acetate is 6.33, 6.02, 5.69, and 4.74. While in case of $0.01\text{ M}$ sodium acetate it is 5.22, 4.74, 3.54 and 1.35.

Explanation of Solution

In the given condition there are two cases which is mentioned as below,

Case I-

Concentration of first solution is $0.1\text{ M}$ and the concentrations of $\text{HCl}$ added to this solution is $0.0025\text{ M}$ , $0.005\text{ M}$ , $0.01\text{ M}$ and $0.05\text{ M}$ .

Case II-

Concentration of the other solution is $0.01\text{ M}$ and the concentrations of $\text{HCl}$ added to this solution is $0.0025\text{ M}$ , $0.005\text{ M}$ , $0.01\text{ M}$ and $0.05\text{ M}$ .

The volume is not provided; consider the volume of the buffer to be $1\text{ L}$ .

To calculate $\text{pH}$ we need the Henderson−Hasselbalch equation which is as given below,

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}$   ....... (1)

Where,

• $\text{p}{K}_{\text{a}}$ is dissociation constant.
• $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of ${\text{CH}}_3{\text{COO}}^{-}$ ion.
• $\left[{\text{CH}}_3\text{COOH}\right]$ is the concentration of ${\text{CH}}_3\text{COOH}$ acid.

For the first case,

   (a) Concentration of sodium acetate is $0.1\text{ M}$ and $0.0025\text{ M}$HCl is added to it.

The reaction can be given as,

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+HCl}}_{\left(\text{aq}\right)}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$ after the reaction will be,

  $\begin{array}{c}\left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.1-0.0025\\ \left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.0975\end{array}$

Substitute the values to equation (1),

  $\begin{array}{c}\text{pH}=4.74+\log \frac{\left[0.0975\right]}{\left[0.0025\right]}\\ \text{pH}=4.74+\log 39\\ =4.74+1.591\\ =6.33\end{array}$

The $\text{pH}$ of the solution will be 6.33.

For the first case,

   (b)Concentration of sodium acetate is $0.1\text{ M}$ and $0.005\text{ M}$HCl is added to it.

The reaction can be given as,

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+HCl}}_{\left(\text{aq}\right)}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ after the reaction will be,

  $\begin{array}{c}\left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.1-0.005\\ \left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.095\end{array}$

Substitute the value to equation (1),

  $\begin{array}{c}\text{pH}=4.74+\log \frac{\left[0.095\right]}{\left[0.005\right]}\\ \text{pH}=4.74+\log 19\\ =4.74+1.2787\\ =6.02\end{array}$

The $\text{pH}$ of the solution will be 6.02.

For the first case,

   (c) Concentration of sodium acetate is $0.1\text{ M}$ and $0.01\text{ M}$HCl is added to it.

The reaction can be given as,

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+HCl}}_{\left(\text{aq}\right)}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ after the reaction will be,

  $\begin{array}{c}\left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.1-0.01\\ \left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.09\end{array}$

Substitute the value to equation (1),

  $\begin{array}{c}\text{pH}=4.74+\log \frac{\left[0.09\right]}{\left[0.01\right]}\\ \text{pH}=4.74+\log 9\\ =4.74+0.9542\\ =5.69\end{array}$

The $\text{pH}$ of the solution will be 5.69.

For the first case,

   (d) Concentration of sodium acetate is $0.1\text{ M}$ and $0.05\text{ M}$HCl is added to it.

The reaction can be given as,

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+HCl}}_{\left(\text{aq}\right)}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ after the reaction will be,

  $\begin{array}{c}\left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.1-0.05\\ \left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.05\end{array}$

Substitute the value to equation (1),

  $\begin{array}{c}\text{pH}=4.74+\log \frac{\left[0.05\right]}{\left[0.05\right]}\\ \text{pH}=4.74+\log 1\\ =4.74+0\\ =4.74\end{array}$

The $\text{pH}$ of the solution will be 4.74.

For the second case,

   (a) Concentration of sodium acetate is $0.01\text{ M}$ and $0.0025\text{ M}$HCl is added to it.

The reaction can be given as,

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+HCl}}_{\left(\text{aq}\right)}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ after the reaction will be,

  $\begin{array}{c}\left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.01-0.0025\\ =0.0075\end{array}$

Substitute the value to equation (1),

  $\begin{array}{c}\text{pH}=4.74+\log \frac{\left[0.0075\right]}{\left[0.0025\right]}\\ =4.74+\log 3\\ =5.22\end{array}$

The $\text{pH}$ of the solution will be 5.22.

For the second case,

   (b) Concentration of sodium acetate is $0.01\text{ M}$ and $0.005\text{ M}$is added to it.

The reaction can be given as,

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+HCl}}_{\left(\text{aq}\right)}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ after the reaction will be,

  $\begin{array}{c}\left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.01-0.005\\ =0.005\end{array}$

Substitute the value to equation (1),

  $\begin{array}{c}\text{pH}=4.74+\log \frac{\left[0.005\right]}{\left[0.005\right]}\\ =4.74+\log \left(1\right)\\ =4.74\end{array}$

The $\text{pH}$ of the solution will be 4.74.

For the second case,

   (c) Concentration of sodium acetate is $0.01\text{ M}$ and $0.01\text{ M}$HCl is added to it.

The reaction can be given as,

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+HCl}}_{\left(\text{aq}\right)}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ after the reaction will be,

  $\begin{array}{c}\left[{\text{CH}}_3{\text{COO}}^{-}\right]=0.01-0.01\\ =0\end{array}$

So, moles of acetic acids are 0.01 mole.

Concentration of acetic acid is:

  $\left[{\text{CH}}_3\text{COOH}\right]=\frac{\text{0}\text{.01 mole}}{\text{1 L+1 L}}\text{= 0}\text{.005 M}$ (volume of salt and acid is the volume of solution)

The equation of ${\text{K}}_{\text{a}}$ is given as,

  $K_{\text{a}}=\frac{\left[{\text{CH}}_3{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}$   ....... (2)

Where,

• $K_{\text{a}}$ is dissociation constant.
• $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of ${\text{CH}}_3{\text{COO}}^{-}$ ion.
• $\left[{\text{CH}}_3\text{COOH}\right]$ is the concentration of ${\text{CH}}_3\text{COOH}$ acid.
• $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ion.

Substitute the values in the equation (2),

  $1.8\times {10}^{-5}=\frac{\left[\text{x}\right]\left[\text{x}\right]}{\left[0.005-\text{x}\right]}$

Rearrange for x,

  $\begin{array}{c}{\text{x}}^2-1.8\times {10}^{-5}\left(0.005-\text{x}\right)=0\\ \text{x}=2.9\times {10}^{-4}\end{array}$

Substitute the value of x in equation of $\text{pH}$ ,

  $\begin{array}{c}\text{pH}=-\log \left(2.9\times {10}^{-4}\right)\\ =3.54\end{array}$

The $\text{pH}$ of the solution will be 3.54.

For the second case,

   (d) Concentration of sodium acetate is 0.01 M and 0.05 M HCl is added to it.

The reaction can be given as,

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+HCl}}_{\left(\text{aq}\right)}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

In this case, the HCl is a strong acid and the formed acetic acid is a weak acid. Thus, HCl dominates the acidity. (Also the concentration of acetic acid is small).

The concentration of ${\text{H}}^{+}$ after the reaction will be,

  $\begin{array}{c}\left[{\text{H}}^{+}\right]=0.05-0.005\\ =0.045\end{array}$

Substitute the value of concentration of hydrogen ion into $\text{pH}$ expression we get,

  $\begin{array}{c}\text{pH}=-\log \left(0.045\right)\\ =1.35\end{array}$

The $\text{pH}$ of the solution will be 1.35

Conclusion

Thus the $\text{pH}$ values when different concentrations of $\text{HCl}$ is added to the $0.1\text{ M}$ sodium acetate is 6.33, 6.03, 5.69, and 4.74. While in case of $0.01\text{ M}$ sodium acetate it is 5.21, 4.74, 3.54 and 1.35.