Question

At ${80}^{\circ }C$, distilled water $\left(H_3{O}^{+}\right)$ has concentration equal to $1\times {10}^{-6}mol/litre$. The value of $K_w$ at this temperature will be:

• A. $1\times {10}^{-6}$
• B. $1\times {10}^{-12}$
• C. $1\times {10}^{-9}$
• D. $1\times {10}^{-15}$

Answer 1

Answer: (B)

$1\times {10}^{-12}$

$\left[H_3{O}^{+}\right]=\left[H^{+}\right]={10}^{-6}$

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$

In pure water, $\left[H^{+}\right]=\left[O{H}^{-}\right]$

So, $\left[O{H}^{-}\right]=\left[H^{+}\right]=\left[H_3{O}^{+}\right]={10}^{-6}$

$K_w=\left[{10}^{-6}\right]\left[{10}^{-6}\right]={10}^{-12}$.

$K_w$ increases with an increase in temperature.