Chapter 15, Problem 1KSP

The $K_{\text{a}}$ for hydrocyanic acid $\left(\text{HCN}\right)\text{ is 4}\text{.9 }\times {\text{ 10}}^{\text{-10}}$. Determine the concentration of ${\text{H}}^{+}$ in a solution that is 0.25 M in

$\text{HCN}\text{.}$

$\begin{array}{l}\left(\text{a}\right)\hfill \\ \text{1}\text{.2 }\times {\text{ 10}}^{\text{-10}}M\hfill \\ \left(\text{b}\right)\hfill \\ \text{5}\text{.0 }\times {\text{ 10}}^{\text{-4}}M\hfill \\ \left(\text{c}\right)\hfill \\ \text{2}\text{.2 }\times {\text{ 10}}^{\text{-5}}M\hfill \\ \left(\text{d}\right)\hfill \\ \text{2}\text{.5 }\times {\text{ 10}}^{\text{-5}}M\hfill \\ \left(\text{e}\right)\hfill \\ \text{1}\text{.1 }\times {\text{ 10}}^{\text{-5}}M\hfill \end{array}$

Interpretation Introduction

Interpretation:

The concentration of $H^{+}$ ions in the $\text{HCN}$ solution is to be determined.

Concept introduction:

Equilibrium constant is the ratio of concentrations of the reactants and products present in the chemical reaction.

The equilibrium concentration of reactants and products, present in liquid or solid phase, in the reaction is considered to be $1$ in the expression of equilibrium constant.

Answer to Problem 1KSP

Correct answer: Option (e)

Explanation of Solution

Given information:

The equilibrium reaction is given as follows:

$HCN\underset{}{\rightleftharpoons }{H}^{+}+C{N}^{-}$

The concentration of $\text{HCN}$ solution is $0.25\text{M}$.

$K_a$ for hydrocyanic acid is $4.9\times {10}^{-10}$.

Reason for correct option:

The change in the concentration of the species in the given reaction is as follows:

$\begin{array}{cccc}& HCN& H^{+}& C{N}^{-}\\ Initial& 0.25M& 0& 0\\ Change& -x& +x& +x\\ Atequilibrium& \left(0.25-x\right)& +x& +x\end{array}$

The concentration of hydrogen ions in the $\text{HCN}$ solution is calculated by the following expression as:

$K_a=\frac{\left[H^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}$

Here, $K_a$ is the equilibrium constant, $\left[H^{+}\right],\left[C{N}^{-}\right]$, and $\left[HCN\right]$ are the concentrations of $H^{+},C{N}^{-}$, and $HCN$, respectively.

Substitute the values of $K_a,\left[H^{+}\right],\left[C{N}^{-}\right]$, and $\left[HCN\right]$ in the above equation as follows:

$4.9\times {10}^{-10}=\frac{x^2}{\left(0.25-x\right)}$

As the value of $x$ is very small, therefore, the value $\left(0.25-x\right)$ is approximately $0.25$.

$\begin{array}{c}4.9\times {10}^{-10}\times 0.25=x^2\\ x=\sqrt{1.225\times {10}^{-10}}\\ =1.1\times {10}^{-5}M\end{array}$

Therefore, the concentration of $H^{+}$ ion in the $\text{HCN}$ solution is $1.1\times {10}^{-5}M$.

Hence, option (e) is correct.

Reason for incorrect options:

Option (a) is incorrect because according to the above relation, the value $1.2\times {10}^{-10}M$ is not obtained. So, it is the wrong answer.

Option (b) is incorrect because according to the above relation, the value $5.0\times {10}^{-4}M$ is not obtained. So, it is the wrong answer.

Option (c) is incorrect because according to the above relation, the value $2.2\times {10}^{-5}M$ is not obtained. So, it is the wrong answer.

Option (d) is incorrect because according to the above relation, the value $2.5\times {10}^{-5}M$ is not obtained. So, it is the wrong answer.

Hence, options (a), (b), (c), and (d) are incorrect.