ionic product of water

Question

At 25°C, the ionic product of water (Kw) is 1.0 × 10⁻¹⁴. The pH of pure water at this temperature is:

RankGuru · Accepted Answer

The **Ionic Product of Water ($K_w$)** is the product of concentrations of $H^+$ and $OH^-$ ions.
$K_w = [H^+][OH^-] = 1.0 	imes 10^{-14}$ at $25^\circ C$.
- In pure water (neutral), $[H^+] = [OH^-]$.
- Let $[H^+] = x$. Then $x^2 = 1.0 	imes 10^{-14}$.
- $x = [H^+] = \sqrt{10^{-14}} = 10^{-7} 	ext{ M}$.
- $pH = -\log[H^+] = -\log(10^{-7}) = 7$.
- This is why a $pH$ of 7 defines a neutral solution at $25^\circ C$.

Rank Guru

ionic product of water

Select the correct option: