Question

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

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

Answer 1

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

$\left[H_3{O}^{+}\right]=\left[H^{+}\right]=1\times {10}^{-6}$ mole/litre

Now, $K_W=\left[H^{+}\right]\left[O{H}^{-}\right]$

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

$\therefore K_W=[H^{+}{]}^2$

$=(1\times {10}^{-6}{)}^2$

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