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Chemistry Question on Ionic Equilibrium In Solution

The ionic product of water at 25∘C25^{\circ} C is 10−1410^{-14}. Its ionic product at 90∘C90^{\circ} C will be

A

1×10−141 \times 10^{-14}

B

1×10−121 \times 10^{-12}

C

1×10−201 \times 10^{-20}

D

1×10−161 \times 10^{-16}

Answer

1×10−121 \times 10^{-12}

Explanation

Solution

The product of H+H^{+} and OHOH - ions in water at a particular temperature is known as Ionic product of water. It is denoted as KwK_{w}
Kw=[H+][OH−]K_{ w }=\left[ H ^{+}\right]\left[ OH ^{-}\right]

The value of KwK_{w} increase with the increase of temperature i.e. concentration of H+H ^{+} and OH−OH ^{-} ions increase with increase of temperature, but it still exists in the range of 10−1410^{-14}

e.g. 25∘C  1.00×10−1425^{\circ} C\,\, 1.00 \times 10^{-14}
100∘C  7.50×10−14100^{\circ} C \,\, 7.50 \times 10^{-14}