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Chemistry Question on Equilibrium

At 80∘C80^{\circ} C, distilled water has concentration equal to 1×10−61 \times 10^{-6} mole/litre. The value of KwK_{w} at this temperature will be

A

1×10−61 \times 10^{-6}

B

1×10−121 \times 10^{-12}

C

1×10−91 \times 10^{-9}

D

1×10−151 \times 10^{-15}

Answer

1×10−121 \times 10^{-12}

Explanation

Solution

The correct option is(A): 1x10-12.

[H3O+]=1×10−6 mol L−1\left[ H _{3} O ^{+}\right]=1 \times 10^{-6} \,mol\, L ^{-1}

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

As for distilled water [H+]=[OH−]\left[ H ^{+}\right]=\left[ OH ^{-}\right]
[H+]=[OH−]=1×10−6 mol L−1\left[ H ^{+}\right]=\left[ OH ^{-}\right]=1 \times 10^{-6}\, mol\, L ^{-1}
Kw=[H+][OH−]K _{ w }=\left[ H ^{+}\right]\left[ OH ^{-}\right]
=10−12 mol2 L−2=10^{-12}\, mol ^{2}\, L ^{-2}.