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

Calculate the pH of the following solutions:
(i) 1.0 ×108 M HCl1.0{\text{ }} \times {10^{ - 8}}{\text{ M HCl}}
(ii) 1.0 ×108 M NaOH1.0{\text{ }} \times {10^{ - 8}}{\text{ M NaOH}}

Explanation

Solution

(i) You can calculate the pH of the solution by using the following formula
pH=log10[H3O+]{\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
pOH=log10[OH]{\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 pOH = 14pH{\text{pOH = 14}} - {\text{pH}} .

Complete answer:
(i) From 1.0 ×108 M HCl1.0{\text{ }} \times {10^{ - 8}}{\text{ M HCl}} solution,
[H3O+]=1.0 ×108 M\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = 1.0{\text{ }} \times {10^{ - 8}}{\text{ M}}
From autoionization of water
[H3O+]=1.0 ×107 M\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = 1.0{\text{ }} \times {10^{ - 7}}{\text{ M}}
Calculate the total hydronium ion concentration
[H3O+]=1.0 ×108 M+1.0 ×108 M = 1.1 ×107 M\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
pH=log10[H3O+] = log101.1 ×107 M = 6.96{\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 ×108 M HCl1.0{\text{ }} \times {10^{ - 8}}{\text{ M HCl}} solution is 6.96.

(ii) From 1.0 ×108 M NaOH1.0{\text{ }} \times {10^{ - 8}}{\text{ M NaOH}} solution,
[OH]=1.0 ×108 M\left[ {{\text{O}}{{\text{H}}^ - }} \right] = 1.0{\text{ }} \times {10^{ - 8}}{\text{ M}}
From autoionization of water
[OH]=1.0 ×107 M\left[ {{\text{O}}{{\text{H}}^ - }} \right] = 1.0{\text{ }} \times {10^{ - 7}}{\text{ M}}
Calculate the total hydroxide ion concentration
[OH]=1.0 ×108 M+1.0 ×108 M = 1.1 ×107 M\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
pOH=log10[OH] = log101.1 ×107 M = 6.96{\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
pH = 14pOH=146.96 = 7.04{\text{pH = 14}} - {\text{pOH}} = 14 - {\text{6}}{\text{.96 = 7}}{\text{.04}}

**Hence, the pH of 1.0 ×108 M NaOH1.0{\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 ×108 M HCl1.0{\text{ }} \times {10^{ - 8}}{\text{ M HCl}} the pH value will be
pH=log10[H3O+] = log101.0 ×108 M = 8{\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 ×108 M NaOH1.0{\text{ }} \times {10^{ - 8}}{\text{ M NaOH}} the pOH value will be
pOH=log10[OH] = log101.0 ×108 M = 8{\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.