Calculate the dissociation constant of water at room temperature

1.	Calculate the dissociation constant of water at room temperature
Answer» Solution :DISSOCIATION of water is given as, `H_2 O hArr H^(+)_OH^(-)` Dissociation constant `K_a` is given as `K_a ([H^(+)][OH^(-)])/([H_2 O]) = (k_w )/( [H_2O]) =(1XX 10^(-14))/( 55.5 )` Dissociation constant of acid `= 1.8 xx10^(-16)mol L^(-1)`

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