1.

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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