Chapter 16, Problem 1KSP

Calculate the pH of a solution that is 0.22 M in nitrite ion ${\text{(NO}}_2{}^{-})$ at $\text{25°C}\text{. }{K}_{\text{a}}$ for nitrous acid ${\text{(HNO}}_2\text{) is 4}\text{.5}\times {\text{10}}^{-4}$.

(a)

11.79

(b)

8.34

(0

5.65

(d)

7.00

(e)

2.21

Interpretation Introduction

Interpretation:

The pH of the solution that contains the nitrite ion is to be determined.

Concept introduction:

The ionic reaction that takes place is

${\text{A}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons \text{HA}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

Here, ${\text{A}}^{-}$ is the ion that forms the weak acid $\text{HA}$. The pH of this solution is determined by the concentration of the $\left[{\text{OH}}^{-}\right]$ ions.

The relationship between ${\text{K}}_{\text{b}}$, ${\text{K}}_{\text{a}}$, and ${\text{K}}_{\text{w}}$ gives the quantitative information of the relationship of the strength of an acid with its conjugate base, or vice-versa.

${\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}={\text{K}}_{\text{w}}$ …… (1)

${\text{K}}_{\text{b}}$ is the measure of dissociation of a base and is known as the base-ionization constant, which is specific at a particular temperature.

${\text{K}}_{\text{b}}=\frac{\left[{\text{OH}}^{-}\right]\left[\text{HA}\right]}{\left[{\text{A}}^{-}\right]}$ …… (2)

The formula to calculate the pOH of the solution from the concentration of hydroxide ions is

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$ …… (3)

pH is the measure of acidity of a solution and it depends on the concentration of hydronium ions and temperature of the solution. The relationship between pH and pOH is

$\text{pH}+\text{pOH}=14$ …… (4)

Answer to Problem 1KSP

Solution: $8.34$.

Explanation of Solution

Given information:

The concentration of nitrite ion $\left({\text{NO}}_{\text{2}}{}^{-}\right)$ at 25°Cis $0.22M$. The ${\text{K}}_{\text{a}}$ value for nitrous acid is given as $4.5\times {10}^{-4}$.

The anion ${\text{NO}}_{\text{2}}{}^{-}$ comes from the weak acid ${\text{HNO}}_{\text{2}}$. Thus, the anion recombines with water to produce the acid and hydroxide ions according to the reaction:

${\text{NO}}_{\text{2}}{{}^{-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{HNO}}_{\text{2}}{}_{\left(\text{aq}\right)}+{\text{OH}}^{-}{}_{\left(\text{aq}\right)}$

${\text{K}}_{\text{b}}$ of ${\text{NO}}_{\text{2}}{}^{-}$, which is the conjugate base of nitrous acid, is calculated using equation (1) and substitute the values of ${\text{K}}_{\text{a}}$ and ${\text{K}}_{\text{w}}$ as

$\begin{array}{c}\left(4.5\times {10}^{-4}\right)\times {\text{K}}_{\text{b}}=1.0\times {10}^{-14}\\ {\text{K}}_{\text{b}}=\frac{1.0\times {10}^{-14}}{4.5\times {10}^{-4}}\\ {\text{K}}_{\text{b}}=2.2\times {10}^{-11}\end{array}$

Now, prepare an equilibrium table and represent each of the species in terms of $x$ as

$\begin{array}{ccccc}& {\text{NO}}_{\text{2}}{}^{-}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{HNO}}_{\text{2}}\left(aq\right)\\ \text{Initial concentration}\left(M\right)& 0.22& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.22-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (2) as

${\text{K}}_{\text{b}}=\frac{\left(x\right)\left(x\right)}{\left(0.22-x\right)}$

Since the value of ${\text{K}}_{\text{b}}$ is very small, the quantity of base that recombines to form acid is less. Therefore, $\left(0.22-x\right)$ can be approximated as $0.22$. Now, substitute the value of ${\text{K}}_{\text{b}}$ in the above equation as

$\begin{array}{c}2.2\times {10}^{-11}=\frac{\left(x\right)\left(x\right)}{\left(0.22\right)}\\ x^2=4.89\times {10}^{-12}\\ x=\sqrt{4.89\times {10}^{-12}}\\ x=2.2\times {10}^{-6}\end{array}$

Thus,

$\begin{array}{c}\left[{\text{HNO}}_2\right]=2.2\times {10}^{-6}M\\ \left[{\text{OH}}^{-}\right]=2.2\times {10}^{-6}M\end{array}$

Now, substitute the value of $\left[{\text{OH}}^{-}\right]$ in equation (3) as

$\begin{array}{c}\text{pOH}=-\log \left[2.2\times {10}^{-6}\right]\\ =5.66\end{array}$

Substitute this value of pOH in equation (4) to calculate the pH of the solution as

$\begin{array}{c}\text{pH}+\text{5}\text{.66}=14\\ \text{pH}=14-5.66\\ \text{pH}=8.34\end{array}$

Conclusion

Therefore, the pH of the solution is $8.34$.