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Define the ionic product of water. Give its value at room temperature. |
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Answer» SOLUTION :Ionic product of water (kw) is given by the product of concentration of HYDRONIUM `(H_3 O^+)` and hydroxide `(OH^-) "ion"^-.` `2H_2O hArr H_3O^++OH^-` `k_(aq)=([H_3O^+][OH^-])/([H_2O]^2)` Since water is a solvent and is TAKEN in excess, change in concentration due to dissociation is negligible. `therefore "Keg"=[H_2O]^2=[H_3O^+][OH^-]=kW` At `298 k (25^@ C) kw=1xx10^(-14) mol^2 dm^(-6)`. |
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