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Prove that pH+pOH=pK_(w) at 298 K. |
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Answer» Solution :The ionic PRODUCT of water is given by `[H^(+)][OH^(-)]=K_(w)=1.0xx10^(-14)` at 298 K TAKING LOG on both sides, `log_(10)[H^(+)]+log_(10)[OH^(-)]=log_(10)K_(w)=log_(10)1.0xx10^(-14)` Multipling by -ve SIGN, `-log_(10)[H^(+)]-log_(10)[OH^(-)]=log_(10)K_(w)=-14.0000` wkt by definition, `pH+pOH=pK_(w)=14`. |
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