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`K_(w)` of `H_(2)O` at 373 K is `1 xx 10^(-12)`. Identify which of the following is/are correctA. `pK_(w)` of `H_(2)O` is 12B. pH of `H_(2)O` is 6C. `H_(2)O` is neutralD. `H_(2)O` is acidic |
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Answer» Correct Answer - A::B::C `pK_(w) = -logK_(w) - log 1 xx 10^(-12) = 12` `rArr K_(w) = [H^(+)][OH^(-)] = 10^(-12)` `[H^(+)] = [OH^(-)] rArr` `[H^(+)]^(2) = 10^(-12), [H^(+)] = 10^(-6), pH = -log [H^(+)] = -log 10^(-6) = 6` `H_(2)O` is neutral because `[H^(+)] = [OH^(-)]` at 373 K even when pH = 6 (d) Is not correct at 373 K. Water cannot become acidic. |
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