Question

At $90^{\circ }C$, pure water has $H_3{O}^{+}$ion concentration of $10^{-6}mol/L^{-1}$. The $K_w$ at $90^{\circ }C$ is :

• A. ${10}^{-6}$
• B. ${10}^{-14}$
• C. ${10}^{-12}$
• D. ${10}^{-8}$

Answer 1

Answer: (C)

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
$=\left(10^{-6}\right)\times \left(10^{-6}\right)$
$\left(\because \left[H^{+}\right]=\left[O{H}^{-}\right]\right)$
$=10^{-12}$