By Sunil Bhardwaj

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The product of ionic concentration of water at a given temp. is always constant. This constant is known as ionic product of water. It is written as \({ K }_{ W }\) at 20°C for pure water. The value of \( { K }_{ W } \) is \(1.0\times { 10 }^{ -14 }\) $$ { H }_{ 2 }O \longrightarrow { H }^{ + } + { OH }^{ - } $$ $${ K }_{ W } = { a }_{ { H }^{ + } } \times { a }_{ { OH }^{ - } } $$ also $$ { a }_{ { H }^{ + } } = \frac { { K }_{ W } }{ { a }_{ { OH }^{ - } } } \qquad ...(1) $$The \({ K }_{ W }\) can be determined with the help of cell. The cell must have one electrode $$ Pt, { H }_{ 2 }(g) | { OH }^{ - }.$$ When this electrode is coupled to some suitable electrode for eg. Hydrogen gas electrode or any other suitable electrode, we get the cell. We can find out emf of the cell. This cell will be represented as follows: $$ Pt, \underset { 1 atm }{ { H }_{ 2 }(g) } | \underset { a=unknown }{ { OH }^{ - } } || \underset { a=unknown }{ { H }^{ + } } | \underset { 1 atm }{ { H }_{ 2 }(g) } , Pt $$ at LHE $$ \frac { 1 }{ 2 } { H }_{ 2 } \longrightarrow { H }^{ + }(base) + { e }^{ - }\qquad (Ox) $$ at RHE $$ { H }^{ + }(acid) + { e }^{ - } \longrightarrow \frac { 1 }{ 2 } { H }_{ 2 }\qquad (Red) $$ Net cell reaction is $$ { H }^{ + }(acid) \longrightarrow { H }^{ + }(base) $$ $$ \therefore K = \frac { { a }_{ { H }^{ + } }(base) }{ { a }_{ { H }^{ + } }(acid) } $$ We can substitue the value of \({ a }_{ { H }^{ + } }\)(base) from equation (1) $$ \therefore K = \frac { \left( \frac { { K }_{ W } }{ { a }_{ { OH }^{ - } } } \right) }{ { a }_{ { H }^{ + } }(acid) } = \frac { { K }_{ W } }{ { a }_{ { H }^{ + } } \times { a }_{ { OH }^{ - } } } $$ on applying the Nernst equation $$ { E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { 0.059 }{ 1 } \log { K } $$ $$ { E }_{ cell } = { E }_{ cell }^{ 0 } - 0.059 \log { \left( \frac { { K }_{ W } }{ { a }_{ { H }^{ + } } \times { a }_{ { OH }^{ - } } } \right) } $$ $$ { E }_{ cell } = 0 + 0.059 \log { \left( \frac { { a }_{ { H }^{ + } } \times { a }_{ { OH }^{ - } } }{ { K }_{ W } } \right) } $$ $$ \frac { { E }_{ cell } }{ 0.059 } = \log { \left( \frac { { a }_{ { H }^{ + } } \times { a }_{ { OH }^{ - } } }{ { K }_{ W } } \right) } $$ $$ \frac { { E }_{ cell } }{ 0.059 } = \log { { a }_{ { H }^{ + } } } + \log { { a }_{ { OH }^{ - } } } - \log { { K }_{ W } } $$ $$ \therefore \log { { K }_{ W } } = \log { { a }_{ { H }^{ + } } } + \log { { a }_{ { OH }^{ - } } } - \frac { { E }_{ cell } }{ 0.059 } $$ By knowing \({ a }_{ { H }^{ + } }\) and \({ a }_{ { OH }^{ - } }\) and finding out the emf of the cell we can find out the \({ K }_{ W }\). The value is \({ 10 }^{ -14 }\).

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