Question

Distilled water at ${80}^{\circ }C$ has hydronium ion $\left(H_3{O}^{+}\right)$ concentration equal to $1\times {10}^{-6}mol/L$.
The value of $K_w$ at this temperature would be:

• A. $1\times {10}^{-6}$
• B. $1\times {10}^{-9}$
• C. $1\times {10}^{-14}$
• D. $1\times {10}^{-12}$

Answer 1

The correct option is D $1\times {10}^{-12}$

The distilled water is neutral with equal concentrations of hydrogen ion and hydroxide ion. Whatever the temperature may be the water will always dissociate to give equal conc. of $H^{+}$ ion and $O{H}^{-}$ ion, that is it will always be neutral.
Hence, $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
=$1\times {10}^{-6}\times 1\times {10}^{-6}$
=$1\times {10}^{-12}$