Chapter 15, Problem 7SAQ

Interpretation Introduction

Interpretation:

The option corresponding to the correct $\text{pH}$ value of a given ${\text{HClO}}_{\text{2}}\left(aq\right)$ solution of $\text{0}\text{.155}\text{M}$ concentration is to be identified.

Concept introduction: The $\text{pH}$ value determines the solution is acidic, basic and neutral in nature.

The $\text{pH}$ value of the given solution is calculated by the formula,

$\text{pH}=-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$

To determine: The option corresponding to the correct $\text{pH}$ value of a given ${\text{HClO}}_{\text{2}}\left(aq\right)$ solution of $\text{0}\text{.155}\text{M}$ concentration.

Correct answer: The correct option is (c).

The given concentration of ${\text{HClO}}_{\text{2}}\left(aq\right)$ solution is $\text{0}\text{.155}\text{M}$ .

The given value of $K_{\text{a}}$ for ${\text{HClO}}_{\text{2}}$ is $0.011$ .

The ionization of ${\text{HClO}}_{\text{2}}\left(aq\right)$ is,

${\text{HClO}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{ClO}}_2^{-}\left(aq\right)+{\text{H}}_3{\text{O}}^{+}\left(aq\right)$

The ICE Table for ionization of ${\text{HClO}}_{\text{2}}\left(aq\right)$ is given below.

  $\begin{array}{cccc}& \left[{\text{HClO}}_{\text{2}}\right]\text{M}& \left[{\text{ClO}}_2^{-}\right]\text{M}& \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\text{M}\\ \text{Intial}& 0.155& 0& 0\\ \text{Change}& -x& +x& +x\\ \text{Equlibrium}& 0.155-x& x& x\end{array}$

The equilibrium constant for the above reaction is,

$K_{\text{a}}=\frac{\left[{\text{ClO}}_2^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{HClO}}_{\text{2}}\right]}$

Where,

• $K_{\text{a}}$ is the equilibrium constant for the acidic reaction.
• $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ is the concentration of hydronium ion.
• $\left[{\text{HClO}}_{\text{2}}\right]$ is the concentration of ${\text{HClO}}_{\text{2}}$ .
• $\left[{\text{ClO}}_2^{-}\right]$ is the concentration of ${\text{ClO}}_2^{-}$ ion.

Substitute the value of $K_{\text{a}}$ for ${\text{HClO}}_{\text{2}}$ and equilibrium concentrations from the above table in the above expression.

$\begin{array}{c}0.011=\frac{\left(x\right)\left(x\right)}{0.155-x}\\ 0.011=\frac{x^2}{0.155-x}\end{array}$

The value of $x$ is very small, so it is neglected in the above expression.

$\begin{array}{c}\sqrt{0.011}=\sqrt{\frac{x^2}{0.155}}\left(\text{M}\right)\\ \sqrt{\left(0.011\right)\left(0.155\right)}=x\\ 41.2\times {10}^{-3}\text{M}=x\end{array}$

Therefore, the concentration of ${\text{H}}_3{\text{O}}^{+}$ in ${\text{HClO}}_{\text{2}}\left(aq\right)$ is $41.2\times {10}^{-3}\text{M}$ .

The $\text{pH}$ is calculated by the formula,

$\text{pH=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$

Substitute the value of concentration of ${\text{H}}_3{\text{O}}^{+}$ ion in the above expression.

$\begin{array}{c}\text{pH}=-\log \left(41.2\times {10}^{-3}\right)\\ =-\left(-1.39\right)\\ =1.39\end{array}$

Therefore, $\text{pH}$ value of the given ${\text{HClO}}_{\text{2}}\left(aq\right)$ solution of $\text{0}\text{.155}\text{M}$ concentration is $1.39$ .