First slide

Ionic product of water

Question

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

Easy

A

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

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

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

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

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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