Q51 / 180 · AIIMS 2002 · Chemistry
At $80^{\circ} \mathrm{C}$, distilled water has $\mathrm{H}^{+}$ions concentration equal to $1 \times 10^{-6} \mathrm{~mole} /$ litre. The value of $K_w$ at this temperature will be
- A$1 \times 10^{-6}$
- Correct answerB$1 \times 10^{-12}$
- C$1 \times 10^{-9}$
- D$1 \times 10^{-15}$.
Solution
$\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=1 \times 10^{-6} \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1}$
or $\left[\mathrm{H}^{+}\right]=1 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$
As for distilled water $\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]$
$$
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]=1 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}} \\
& \mathrm{~K}_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-12} \mathrm{~mol}^2 \mathrm{~L}^{-2}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.