Question

Calculate the pH of each of the following solutions of a strong acid in water.

a.$\text{0.10M}\text{HCl}$

b.$\text{5.0}\text{M}{\text{HClO}}_4$

c.$\text{1.0}\times {\text{10}}^{-11}\text{M}\text{HI}$

Short answer

For a solution at ${25}^{\circ }$

$\text{pH + pOH}=14$

Acidic solution cannot have pH more than 7, so if concentration of acid is very low$\left[{\text{H}}^{\text{+}}\right]$, concentration of data-custom-editor="chemistry" $\left[{\text{H}}^{\text{+}}\right]$ from water is also taken in consideration.

pH of an aqueous HCl solutiondata-custom-editor="chemistry" $\left(0.1\text{M}\right)=1$

pH of an aqueous HClO₄ solution data-custom-editor="chemistry" $\left(\text{5 M}\right)\text{= -}\text{0.6989}$

pH of an aqueous HI solution data-custom-editor="chemistry" $\left(1\times {\text{10}}^{\text{- 11}}\text{M}\right)\text{= 6.99}$

Step-by-step solution

Subpart (a) pH of an aqueous HCl solution with concentration 0.1 M

pH of a solution is a measure of $\left[{\text{H}}^{\text{+}}\right]$ion concentration. It can be calculated using the formula,

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

As HCl is a strong acid hence it completely ionises in solution, so

$\left[{\text{H}}^{\text{+}}\right]=0.1\text{M}$

$\begin{array}{rcl}\text{pH}& =& -\log \left[H^{+}\right]\\ & =& -log[0.1]\\ & =& 1\\ & & \end{array}$

pH of an aqueous  HClO4 solution with concentration 5M

HClO₄ is also a strong acid hence it also completely ionises in solution, so

$\left[{\text{H}}^{\text{+}}\right]=5\text{M}$

$\begin{array}{rcl}\text{pH}& =& -\log \left[H^{+}\right]\\ & =& -log\left[5\right]\\ & =& -0.6989\\ & & \end{array}$

pH of an aqueous HI solution with concentration  1×10-11 M

HI is also a strong acid hence it also completely ionises in solution, so

$\left[{\text{H}}^{\text{+}}\right]=1\times {10}^{-11}\text{M}$

As the concentration of $\left[{\text{H}}^{\text{+}}\right]$of very low due to acid so$\left[{\text{H}}^{\text{+}}\right]$ from water cannot be ignored.

$\begin{array}{rcl}\text{Total}{\text{[H}}^{+}\text{]}& =& 1\times {10}^{-11}+1\times {10}^{-7}\\ & =& 1.0001\times {10}^{-7}\\ & & \end{array}$

role="math" localid="1657516576230" $\begin{array}{rcl}\text{pH}& =& -\log \left[H^{+}\right]\\ & =& -log[1.0001\times {10}^{-7}]\\ & =& 6.99\\ & & \end{array}$