Question

Calculate the pH of the following solutions:
(i) \[1.0{\text{ }} \times {10^{ - 8}}{\text{ M HCl}}\]
(ii) \[1.0{\text{ }} \times {10^{ - 8}}{\text{ M NaOH}}\]

Answer 1

Hint:
(i) You can calculate the pH of the solution by using the following formula
\[{\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]\]
Here, the hydronium ion concentration is the sum of the hydronium ion concentrations from the ionization of hydrochloric acid and the autoionization of water.
(ii) You can calculate the pOH of the solution by using the following formula
\[{\text{pOH}} = - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]\]
Here, the hydroxide ion concentration is the sum of the hydroxide ion concentrations from the ionization of sodium hydroxide and the autoionization of water.
From pOH, you can calculate the pH using the formula \[{\text{pOH = 14}} - {\text{pH}}\] .

Complete answer:
(i) From \[1.0{\text{ }} \times {10^{ - 8}}{\text{ M HCl}}\] solution,
\[\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = 1.0{\text{ }} \times {10^{ - 8}}{\text{ M}}\]
From autoionization of water
\[\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = 1.0{\text{ }} \times {10^{ - 7}}{\text{ M}}\]
Calculate the total hydronium ion concentration
\[\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = 1.0{\text{ }} \times {10^{ - 8}}{\text{ M}} + 1.0{\text{ }} \times {10^{ - 8}}{\text{ M = }}1.1{\text{ }} \times {10^{ - 7}}{\text{ M}}\]
Calculate the pH of the solution
\[{\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]{\text{ = }} - {\log _{10}}1.1{\text{ }} \times {10^{ - 7}}{\text{ M = 6}}{\text{.96}}\]
Hence, the pH of \[1.0{\text{ }} \times {10^{ - 8}}{\text{ M HCl}}\] solution is 6.96.

(ii) From \[1.0{\text{ }} \times {10^{ - 8}}{\text{ M NaOH}}\] solution,
\[\left[ {{\text{O}}{{\text{H}}^ - }} \right] = 1.0{\text{ }} \times {10^{ - 8}}{\text{ M}}\]
From autoionization of water
\[\left[ {{\text{O}}{{\text{H}}^ - }} \right] = 1.0{\text{ }} \times {10^{ - 7}}{\text{ M}}\]
Calculate the total hydroxide ion concentration
\[\left[ {{\text{O}}{{\text{H}}^ - }} \right] = 1.0{\text{ }} \times {10^{ - 8}}{\text{ M}} + 1.0{\text{ }} \times {10^{ - 8}}{\text{ M = }}1.1{\text{ }} \times {10^{ - 7}}{\text{ M}}\]
Calculate the pOH of the solution
\[{\text{pOH}} = - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ = }} - {\log _{10}}1.1{\text{ }} \times {10^{ - 7}}{\text{ M = 6}}{\text{.96}}\]
Calculate the pH of the solution
\[{\text{pH = 14}} - {\text{pOH}} = 14 - {\text{6}}{\text{.96 = 7}}{\text{.04}}\]

Hence, the pH of \[1.0{\text{ }} \times {10^{ - 8}}{\text{ M NaOH}}\] solution is 7.04.

Note:
If you do not consider the autoionization of water, then you will get the wrong answer. For example, for \[1.0{\text{ }} \times {10^{ - 8}}{\text{ M HCl}}\] the pH value will be
 \[{\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]{\text{ = }} - {\log _{10}}1.0{\text{ }} \times {10^{ - 8}}{\text{ M = 8}}\]
An acidic solution cannot have pH greater than 7.
Again, for \[1.0{\text{ }} \times {10^{ - 8}}{\text{ M NaOH}}\] the pOH value will be
 \[{\text{pOH}} = - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ = }} - {\log _{10}}1.0{\text{ }} \times {10^{ - 8}}{\text{ M = 8}}\]
A basic solution cannot have pOH greater than 7.