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`K_(w)` of `H_(2)O` at `373` K is `1xx10^(-12)` Identify which of the following is/are correct?A. `pH+pOH=12` for every aqueous solutions.B. `pH` of `H_(2)O` is `6`.C. `alpha_(H_(2)O)` has increased from its value at `298`KD. `H_(2)O` is acidic. |
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Answer» `pK_(w)=-logK_(w)=-log1xx10^(-12)=12` `K_(w)=[H^(+)][OH^(-)]=10^(-12)` `[H^(+)]=[OH^(-)]` `rArr [H^(+)]^(2)=10^(-12),[H^(+)]=10^(-6),pH=-log[H^(+)]=-log10^(-6)=6` `H_(2)O` is neutral because `[H^(+)]=[OH^(-)]` at `373` K even when `pH=6` `(D)` is not correct at `373K` .Water cannot become acidic. |
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