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The dissociation constant of water at 10⁰C is 5.31 × 10^–17. The ionic product of water is

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$large large 0.295, times ,{10^{ - 14}},mol{e^2}li{t^{ - 2}}$

$0.6, times ,{10^{ - 14}},mol{e^2}li{t^{ - 2}}$

$0.54, times ,{10^{ - 14}},mol{e^2}li{t^{ - 2}}$

$0.82, times ,{10^{ - 14}},mol{e^2}li{t^{ - 2}}$

answer is A.

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Detailed Solution

$large {K_i} = frac{{{K_w}}}{{55.55}} Rightarrow {K_w} = {K_i} times 55.55$

$large Rightarrow {K_w} = 5.31 times {10^{ - 17}} times 55.55 = 2.95 times {10^{ - 15}},mol{e^2}li{t^{ - 2}}$

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