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Chemistry Question on Equilibrium

At 90∘C90^{\circ} C, pure water has H3O+H _{3} O ^{+}ion concentration of 10−6mol/L−110^{-6} mol / L ^{-1}. The KwK_{w} at 90∘C90^{\circ} C is :

A

10−6{{10}^{-6}}

B

10−14{{10}^{-14}}

C

10−12{{10}^{-12}}

D

10−8{{10}^{-8}}

Answer

10−12{{10}^{-12}}

Explanation

Solution

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