Question

The $\text{pH}$ of pure water at $25^{\circ}\text{C}$ is:

• A. 0
• B. 5
• C. 7
• D. 14

Hint:

Neutrality means $[\text{H}^+] = [\text{OH}^-]$. This condition occurs when $\text{pH} = \text{pOH} = 7$ at $25^{\circ}\text{C}$. Remember that $\text{pH}$ is temperature-dependent because $K_w$ is temperature-dependent.

Answer 1

The Correct Option is C

At $25^{\circ}\text{C}$, the ion product of water $K_w$ is $1.0 \times 10^{-14}$.
In pure water, the concentrations of $\text{H}^+$ and $\text{OH}^-$ are equal.
$[\text{H}^+] = \sqrt{K_w} = \sqrt{1.0 \times 10^{-14}}$.
$[\text{H}^+] = 1.0 \times 10^{-7} \text{ M}$.
The $\text{pH}$ is calculated as:
$\text{pH} = -\log(10^{-7}) = 7$.
Pure water is neutral, and at $25^{\circ}\text{C}$, its $\text{pH}$ is 7.