The concentration of hydronium ({H_3}{O^{+}}) | vedclass

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(D) In pure water,the auto-ionization reaction is: $2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)$.At $298 \ K$,the ionic product of water is

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$1 	imes 10^{-7} \ g \ ion/litre$

Vedclass · Answer

(D) In pure water,the auto-ionization reaction is: $2H_2O(l) ightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)$.At $298 \ K$,the ionic product of water is $K_w = [H_3O^{+}][OH^{-}] = 1 	imes 10^{-14}$.Since $[H_3O^{+}] = [OH^{-}]$ in pure water,let $[H_3O^{+}] = x$.Therefore,$x^2 = 1 	imes 10^{-14}$,which gives $x = 1 	imes 10^{-7} \ g \ ion/litre$.Thus,the concentration of hydronium ions is $1 	imes 10^{-7} \ g \ ion/litre$.

The concentration of hydronium $({H_3}{O^{+}})$ ion in pure water at $298 \ K$ is:

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