The ionic product of water is 1.0 × 10−14 at 25o C. Assuming the density of water independent from c

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The ionic product of water is 1.0 × 10−14 at 25o C. Assuming the density of water independent from change in temperature, the ionic product of water at 50o C will be

Accepted Answer

$\ln\frac{K_{2}}{K_{1}} = \frac{\Delta H}{R}\left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$

$\text{Or, ln}\frac{K_{2}}{10^{- 14}} = \frac{13.7 \times 10^{3}}{2}\left( \frac{1}{298} - \frac{1}{323} \right) \Rightarrow K_{2} = 5.9 \times 10^{- 14}$

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