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Question: How will you calculate the pH of this solution: \([{H^ + }] = 8.3 \times {10^{ - 10}}M\)....

How will you calculate the pH of this solution: [H+]=8.3×10−10M[{H^ + }] = 8.3 \times {10^{ - 10}}M.

Explanation

Solution

By the definition of pH which is the measure of the hydrogen ion concentration or the acidity of the solution, the pH of the solution is calculated as the negative log of hydrogen ion concentration.

Complete step by step answer:
It is given that the hydrogen ion concentration of the solution is [H+]=8.3×10−10M[{H^ + }] = 8.3 \times {10^{ - 10}}M.
The pH of the solution is the measure of the hydrogen ion concentration which in turns is the measure of its acidity.
The pure water dissociates into equal concentrations of hydrogen ion and hydroxyl ion. It is shown below.
H2O⇌H++OH−{H_2}O \rightleftharpoons {H^ + } + O{H^ - }
When the hydrogen ion is in excess, the solution is considered as acidic and when the hydrogen ion is limited (hydroxyl ion) is in excess than the solution is basic in nature.
For a neutral solution the hydrogen ion concentration is 10−7{10^{ - 7}} or the pH is 7, for an acidic solution the pH will be less than 7 and for the basic solution the pH is greater than 7.
The pH of the solution is defined as the negative logarithm of hydrogen ion concentration.
The equation used to calculate the pH is shown below.
pH=−log⁡[H+]pH = - \log [{H^ + }]
To calculate the pH of the solution, substitute the value of hydrogen ion concentration is the equation.
⇒pH=−log⁡[8.3×10−10]\Rightarrow pH = - \log [8.3 \times {10^{ - 10}}]
⇒pH=9.08\Rightarrow pH = 9.08

Therefore, the pH of the solution is 9.08.

Note: Here the calculated value of pH obtained is 9.08 which shows that the solution is slightly basic in nature as the pH is above the neutral range. Also, the equilibrium constant Kw{K_w} is the product of hydrogen ion concentration and hydroxyl ion concentration.
[H+][OH−]=Kw=10−14[{H^ + }][O{H^ - }] = {K_w} = {10^{ - 14}}