Question

At $80^{\circ} C$, distilled water has concentration equal to $1 \times 10^{-6}$ mole/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)

$\left[ H _{3} O ^{+}\right]=1 \times 10^{-6} \,mol\, L ^{-1}$

or $\left[ H ^{+}\right]=1 \times 10^{-6}\, mol \,L ^{-1}$

As for distilled water $\left[ H ^{+}\right]=\left[ OH ^{-}\right]$

$\left[ H ^{+}\right]=\left[ OH ^{-}\right]=1 \times 10^{-6}\, mol\, L ^{-1}$

$K _{ w }=\left[ H ^{+}\right]\left[ OH ^{-}\right] $

$=10^{-12}\, mol ^{2}\, L ^{-2}$