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Prove that pH+pOH=pK(w) at 298 K....

Prove that `pH+pOH=pK_(w)` at 298 K.

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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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