Chapter 19, Problem 109A

Interpretation Introduction

Interpretation : The pH of the solution is to be calculated.

Concept Introduction : The negative logarithm of the concentration of hydrogen ions is the pH of a solution. The formula to calculate pH is given as

  $\text{pH=}14-\text{pOH}$

pOH stands for the potential of hydroxide ions. It measures the acidity or basicity of any solution. The pOH range lies between 0-14. pOH is calculated by the formula:

  $pOH=-log\left[O{H}^{-}\right]$

Answer to Problem 109A

The pH of the solution is $10.66$ .

Explanation of Solution

The concentration of sodium cyanide is $0.01\text{M}$ .

The base dissociation constant is ${\text{K}}_{\text{b}}=2.1\times {10}^{-5}$ .

The reaction between sodium cyanide and water gives sodium hydroxide and hydrogen cyanide.

  ${\text{NaCN+H}}_{\text{2}}\text{O}\to \text{NaOH+HCN}$

The concentration of $\text{NaOH}$ is $x$ .

The concentration of $\text{HCN}$ is $x$ .

The formula of the base dissociation constant is given as

  $\begin{array}{c}{\text{K}}_{\text{b}}=\frac{\left[\text{NaOH}\right]\left[\text{HCN}\right]}{\left[\text{NaCN}\right]}\\ =\frac{\left[x\right]\left[x\right]}{\left[\text{NaCN}\right]}\end{array}$

Rearrange the above equation to isolate the term $x$ .

  $x^2=\left[\text{NaCN}\right]\times {\text{K}}_{\text{b}}$

To calculate the value of $x$, substitute the values in above equation:

  $\begin{array}{c}{x}^2=\left[\text{NaCN}\right]\times {\text{K}}_{\text{b}}\\ x^2=0.01\times 2.1\times {10}^{-5}{\text{M}}^2\\ x=4.58\times {10}^{-4}\text{M}\end{array}$

So, the concentration of $\text{HCN}$ is $4.58\times {10}^{-4}\text{M}$ and the concentration of $\text{NaOH}$ is $4.58\times {10}^{-4}\text{M}$ .

pOH stands for the potential of hydroxide ions. pOH is calculated by the formula:

  $pOH=-log\left[O{H}^{-}\right]$

To calculate the pOH, substitute the values in above formula:

  $\begin{array}{c}pOH=-log\left[O{H}^{-}\right]\\ =-\log \left[4.58\times {10}^{-4}\right]\\ =3.34\end{array}$

The formula of pH is given as

  $\begin{array}{c}\text{pH=}14-\text{pOH}\\ \text{=14}-3.34\\ =10.66\end{array}$