Question

A solution is tested for pH and conductivity as pictured below:

The solution contains one of the following substances: HCl, NaOH, NH₄Cl, HCN, NH₃, HF, or NaCN. If the solute concentration is about 1.0 M, what is the identity of the solute?

Short answer

The Solute is HCN.

Step-by-step solution

Ruling out the Wrong Options

As the pH of the given solution is 4.6 so it cannot be NaOH as it is strong base (pH=14), cannot be HCl(strong acid).It cannot be NaCN and NH₃ also as both of these are basic in nature and must be having pH more than 7. Thus,the Solution may contain NH₄Cl, HCN and HF.

Calculating Ka

$\begin{array}{l}{\text{[H}}^{+}\text{]}={10}^{-\text{pH}}\\ {\text{[H}}^{+}\text{]}={10}^{-4.6}\\ {\text{[H}}^{+}\text{]}=2.5\times {10}^{-5}\end{array}$

Acid dissociates in water as

$\begin{array}{l}{\text{HA+H}}_{\text{2}}\text{O\hspace{1em}}\rightleftarrows {\text{\hspace{1em}H}}_3{\text{O}}^{\text{+}}{\text{+\hspace{1em}A}}^{\text{-}}\\ 100\\ 1-\text{x\hspace{1em}\hspace{0.17em}\hspace{0.17em}\hspace{1em}\hspace{1em}\hspace{1em}}2.5\times {10}^{-5}\text{\hspace{0.17em}\hspace{0.17em}x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a}}=\frac{{\text{x}}^2}{1-\text{x}}\\ {\text{K}}_{\text{a}}=\frac{{(2.5\times {10}^{-5})}^2}{1-(2.5\times {10}^{-5})}\\ {\text{K}}_{\text{a}}=6.25\times {10}^{-10}\end{array}$

As the K_a is matching with K_a value for HCN so the given solution is HCN.