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Derive relation of pH and pOH. |
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Answer» Solution :The ionic EQUILIBRIUM of dissociation of WATER at 298 K temperature is as under. `H_2O_((l)) + H_2O_((l)) hArr H_3O_((aq))^(+) +OH_((aq))^(-)` `K_W=[H_3O^+][OH^-]=1.0xx10^(-14)` Take negative logarithm at both side, -log `K_W=-{log {[H_3O^+][OH^-]}=-log (1xx10^(-14))}` `therefore -log K_W= - log [H_3O^+] -log [OH^-]=-(-14)` but -log `[H_3O^+]=pH` -log `[OH^-]`=pOH So, -log `K_W=pK_W` So, `pK_W=pH+pOH`=14.0 `pK_W`is a very important QUANTITY for aqueous solutions and CONTROLS the relative concen trations of HYDROGEN and hydroxyl ions as their product is a constant. |
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