Question

pH of a pure water is ____.

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

pH of Pure Water. Due to autoionization (\(H_2O \rightleftharpoons H^+ + OH^-\)), pure water at 25°C has [H\(^+\)] = [OH\(^-\)] = \(10^{-7\) M. pH = -log[H\(^+\)] = 7. This is the definition of neutrality on the pH scale.

Answer 1

The Correct Option is C

Pure water undergoes autoionization: $$ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- $$ At standard temperature (25°C), the equilibrium constant for this reaction, known as the ionic product of water (\(K_w\)), is \((1)0 \times 10^{-14}\)
In pure water, the concentration of hydrogen ions ([H\(^+\)]) is equal to the concentration of hydroxide ions ([OH\(^-\)])
$$ K_w = [\text{H}^+][\text{OH}^-] = (1)0 \times 10^{-14} $$ Since [H\(^+\)] = [OH\(^-\)], let this concentration be \(x\)
$$ x^2 = (1)0 \times 10^{-14} $$ $$ x = \sqrt{(1)0 \times 10^{-14}} = (1)0 \times 10^{-7} \, \text{M} $$ So, in pure water at 25°C, [H\(^+\)] = \((1)0 \times 10^{-7}\) M
The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration: $$ \text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}((1)0 \times 10^{-7}) = -(-7) = 7 $$ Thus, the pH of pure water at 25°C is 7, which represents neutrality