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Given that the dissociation constant for `H_(2)O` is `K_(w) = 1 xx 10^(-14) "mole"^(2)//"litre"^(2)`. What is the pH of a 0.001 molar KOH solutionA. `10^(-11)`B. 3C. 14D. 11 |
Answer» Correct Answer - D 0.001 M KOH solution `[OH^(-)] = 0.001 M = 1 xx 10^(-3) M` `[H^(+)] xx [OH^(-)] = 1 xx 10^(-14)` `[H^(+)] = (1 xx 10^(-14))/([OH^(-)])` `[H^(+)] = (1 xx 10^(-14))/(1 xx 10^(-3)) = 1 xx 10^(-14) xx 10^(+3)` `[H^(+)] = 10^(-11) M` `pH = 11`. |
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