1.

Derive the relation between pH and pOH.

Answer»

Solution :`pH=-log_10[H_3O^+]`……….(i)
`pOH=-log_10[OH^-]`…….(ii) ADDING equations (1) and (2)
`pH+pOH=(-log_10[H_3O^+])+(-log_10[OH^-])`
`=-[(-log_10[H_3O^+])+(log_10[OH^-])`
`pH+pOH=-log_10[H_3O^+][OH^-]`
`[H_3O^+][OH^-]=K_w`
`therefore pH+pOH=-log_10 K_w`
`pH+pOH=pK_w[ because pK_w=-log_10^(K_w)]`
At `25^@C` the IONIC product of water `K_w=1 times 10^-14`
`pK_w=-log_10 10^-14=14 LOG _10=10=14`
`pK_w=14`
`therefore pH+pOH=14` at `25^@C`


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