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Q. 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 :

ManipalManipal 2005Equilibrium

Solution:

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