Chapter 14, Problem 10QAP

Calculate [OH^-] and pH in a solution in which the hydrogen sulfite ion, HSO₃^-, is 0.429 M and the sulfite ion is

(a) 0.0249 M (b) 0.247 M

(c) 0.504 M (d) 0.811 M

(e) 1.223 M

Interpretation Introduction

Interpretation:

The molality of HSO₃- ion is 0.429 M and that of ${\text{SO}}_3^{2-}$ ion is 0.0249 M. The concentration of hydroxide ion and pH of the solution needs to be determined.

Concept Introduction :

From the hydrogen ion concentration, the concentration of hydroxide ion can be calculated as follows:

  $\left[{\text{OH}}^{-}\right]=\frac{K_w}{\left[{\text{H}}^{+}\right]}$

Here, $K_w$ is ionic product of water and equals to ${10}^{-14}$ .

From the hydrogen ion concentration, the pH of the solution can be calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Answer to Problem 10QAP

  $9.7\times {10}^{-9}$ and 5.99.

Explanation of Solution

For, a buffer solution, the concentration of H+ can be calculated using the following equation:

  $\left[H^{+}\right]=\frac{K_a\times \left[HB\right]}{\left[B^{-}\right]}$.....(1)

K_a = equilibrium constant,

[HB] = concentration of weak acid.

[B^-] = concentration of conjugate base.

Since,

  $\left[H^{+}\right]\left[O{H}^{-}\right]=1.0\times {10}^{-14}$ (at 25⁰C) ….. (2)

The pH of a solution can be calculated using the formula:

pH = -log[H⁺]

In a buffer, the weak acid HB is the hydrogen sulfite while the conjugate base is sulfite ion. $K_a=6.0\times {10}^{-8}$ and HSO₃- is 0.429 M.

The concentration of the sulfite ion [SO₃^2-] is 0.0249 M. substituting the given values in equation (1)

  $\left[H^{+}\right]=\frac{K_a\times \left[HS{O}_4{}^{3-}\right]}{\left[S{O}_3{}^{2-}\right]}$

  $\begin{array}{l}6.0\times {10}^{-8}\times \frac{0.429}{0.0249}\\ =1.03\times {10}^{-6}\end{array}$

pH of the buffer is calculated using equation (3) as follows:

  $\begin{array}{l}\text{pH = -log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \text{ = -log }\left(\text{1}{\text{.03×10}}^{\text{-6}}\right)\hfill \\ \text{ = 5}\text{.99}\hfill \end{array}$

[OH^-] is calculated using equation (2) as follows:

  $\begin{array}{l}1.03\times {10}^{-6}\times \left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}\hfill \\ \left[{\text{OH}}^{-}\right]=9.7\times {10}^{-9}\hfill \end{array}$

So, the pH of the solution is 5.99 and [OH^-] is $9.7\times {10}^{-9}\text{M}$ .

Interpretation Introduction

Interpretation:

The molality of HSO₃- ion is 0.429 M and that of ${\text{SO}}_3^{2-}$ ion is 0.247 M. The concentration of hydroxide ion and pH of the solution needs to be determined.

Concept Introduction :

From the hydrogen ion concentration, the concentration of hydroxide ion can be calculated as follows:

  $\left[{\text{OH}}^{-}\right]=\frac{K_w}{\left[{\text{H}}^{+}\right]}$

Here, $K_w$ is ionic product of water and equals to ${10}^{-14}$ .

From the hydrogen ion concentration, the pH of the solution can be calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Answer to Problem 10QAP

  $9.6\times {10}^{-8}\text{M}$ and 6.98.

Explanation of Solution

The concentration of the sulfite ion [SO₃^2-] is 0.0247 M. substituting the given values in equation (1)

  $\left[H^{+}\right]=\frac{K_a\times \left[HS{O}_4{}^{3-}\right]}{\left[S{O}_3{}^{2-}\right]}$

  $\begin{array}{l}6.0\times {10}^{-8}\times \frac{0.429}{0.0247}\\ =1.04\times {10}^{-6}\end{array}$

pH of the buffer is calculated using equation (3) as follows:

  $\begin{array}{l}\text{pH = -log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \text{ = -log }\left(\text{1}{\text{.04 x10}}^{\text{-6}}\right)\hfill \\ \text{ = 6}\text{.98}\hfill \end{array}$

[OH^-] is calculated using equation (2) as follows:

  $\begin{array}{l}\text{1}\text{.04}\times {\text{10}}^{\text{-6}}\times \left[{\text{OH}}^{\text{-}}\right]\text{ = 1}\text{.0}\times {\text{10}}^{\text{-14}}\hfill \\ \left[{\text{OH}}^{\text{-}}\right]\text{ = 9}\text{.6}\times {\text{10}}^{\text{-9}}\hfill \end{array}$

So, the pH of the solution is 6.98 and [OH^-] is $\text{9}\text{.6}\times {\text{10}}^{\text{-9}}$ M

Interpretation Introduction

Interpretation:

The molality of HSO₃- ion is 0.429 M and that of ${\text{SO}}_3^{2-}$ ion is 0.504 M. The concentration of hydroxide ion and pH of the solution needs to be determined.

Concept Introduction :

From the hydrogen ion concentration, the concentration of hydroxide ion can be calculated as follows:

  $\left[{\text{OH}}^{-}\right]=\frac{K_w}{\left[{\text{H}}^{+}\right]}$

Here, $K_w$ is ionic product of water and equals to ${10}^{-14}$ .

From the hydrogen ion concentration, the pH of the solution can be calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Answer to Problem 10QAP

  $1.96\times {10}^{-7}\text{ M}$ and 7.29.

Explanation of Solution

The concentration of the sulfite ion [SO₃^2-] is 0.504 M. substituting the given values in equation (1)

  $\left[H^{+}\right]=\frac{K_a\times \left[HS{O}_4{}^{3-}\right]}{\left[S{O}_3{}^{2-}\right]}$

  $\begin{array}{l}6.0\times {10}^{-8}\times \frac{0.429}{0.504}\\ =5.11\times {10}^{-8}\end{array}$

pH of the buffer is calculated using equation (3) as follows:

  $\begin{array}{l}pH=-log\left[H^{+}\right]\hfill \\ =-log\left(5.11x10^{-8}\right)\hfill \\ =7.29\hfill \end{array}$

[OH^-] is calculated using equation (2) as follows:

  $\begin{array}{l}\text{5}{\text{.11× 10}}^{\text{-8}}\text{× }\left[{\text{OH}}^{\text{-}}\right]\text{ = 1}{\text{.0× 10}}^{\text{-14}}\hfill \\ \left[{\text{OH}}^{\text{-}}\right]\text{ = 1}{\text{.96 × 10}}^{\text{-7}}\hfill \end{array}$

So, the pH of the solution is 7.29 and [OH^-] is $\text{1}{\text{.96 × 10}}^{\text{-7}}\text{ M}$ .

Interpretation Introduction

Interpretation:

The molality of HSO₃- ion is 0.429 M and that of ${\text{SO}}_3^{2-}$ ion is 0.811 M. The concentration of hydroxide ion and pH of the solution needs to be determined.

Concept Introduction :

From the hydrogen ion concentration, the concentration of hydroxide ion can be calculated as follows:

  $\left[{\text{OH}}^{-}\right]=\frac{K_w}{\left[{\text{H}}^{+}\right]}$

Here, $K_w$ is ionic product of water and equals to ${10}^{-14}$ .

From the hydrogen ion concentration, the pH of the solution can be calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Answer to Problem 10QAP

  $3.15\times {10}^{-7}\text{M}$ and 7.50

Explanation of Solution

The concentration of the sulfite ion [SO₃^2-] is 0.811 M. substituting the given values in equation (1)

  $\left[H^{+}\right]=\frac{K_a\times \left[HS{O}_4{}^{3-}\right]}{\left[S{O}_3{}^{2-}\right]}$

  $\begin{array}{l}6.0\times {10}^{-8}\times \frac{0.429}{0.811}\\ =3.17\times {10}^{-8}\end{array}$

pH of the buffer is calculated using equation (3) as follows:

  $\begin{array}{l}\text{pH = -log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \text{ = -log }\left(\text{3}\text{.17}\times {\text{10}}^{\text{-8}}\right)\hfill \\ \text{ = 7}\text{.50}\hfill \end{array}$

[OH^-] is calculated using equation (2) as follows:

  $\begin{array}{l}\text{3}\text{.17}\times {\text{10}}^{\text{-8}}\times \left[{\text{OH}}^{\text{-}}\right]\text{ = 1}\text{.0 }\times {\text{ 10}}^{\text{-14}}\hfill \\ \left[{\text{OH}}^{\text{-}}\right]\text{ = 3}\text{.15}\times {\text{10}}^{\text{-7}}\hfill \end{array}$

So, the pH of the solution is 7.50 and [OH^-] is $\text{3}\text{.15}\times {\text{10}}^{\text{-7}}\text{ M}$ .

Interpretation Introduction

Interpretation:

The molality of HSO₃- ion is 0.429 M and that of ${\text{SO}}_3^{2-}$ ion is 1.223 M. The concentration of hydroxide ion and pH of the solution needs to be determined.

Concept Introduction :

From the hydrogen ion concentration, the concentration of hydroxide ion can be calculated as follows:

  $\left[{\text{OH}}^{-}\right]=\frac{K_w}{\left[{\text{H}}^{+}\right]}$

Here, $K_w$ is ionic product of water and equals to ${10}^{-14}$ .

From the hydrogen ion concentration, the pH of the solution can be calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Answer to Problem 10QAP

  $4.76\times {10}^{-7}\text{M}$ and 7.68.

Explanation of Solution

The concentration of the sulfite ion [SO₃^2-] is 1.223 M. substituting the given values in equation (1)

  $\left[H^{+}\right]=\frac{K_a\times \left[HS{O}_4{}^{3-}\right]}{\left[S{O}_3{}^{2-}\right]}$

  $\begin{array}{l}6.0\times {10}^{-8}\times \frac{0.429}{1.223}\\ =2.10\times {10}^{-8}\end{array}$

pH of the buffer is calculated using equation (3) as follows:

  $\begin{array}{l}\text{pH = -log }\left[{\text{H}}^{\text{+}}\right]\hfill \\ \text{ = -log }\left(\text{2}\text{.10}\times {\text{10}}^{\text{-8}}\right)\hfill \\ \text{ = 7}\text{.68}\hfill \end{array}$

[OH^-] is calculated using equation (2) as follows:

  $\begin{array}{l}\text{2}{\text{.10 x 10}}^{\text{-8}}\times \left[{\text{OH}}^{\text{-}}\right]\text{ = 1}\text{.0 }\times {\text{ 10}}^{\text{-14}}\hfill \\ \left[{\text{OH}}^{\text{-}}\right]\text{ = 4}\text{.76 }\times {\text{10}}^{\text{-7}}\text{M}\hfill \end{array}$

So, the pH of the solution is 7.68and [OH^-] is $\text{4}\text{.76 }\times {\text{10}}^{\text{-7}}\text{M}$ .