At `100^@C` the `K_w` of water is 55 times its value at `25^@C` . What will be the pH of neutral solution ?
(log 55=1.74 )

Options

A. `7.00`
B. `7.87`
C. `5.13`
D. `6.13`

Answer

D

Explanation

At `25^(@)C , K_(w) = 1 xx 10^(-14)`
at `100^(@) C , K_(w) = 55 xx 10^(-14)`
`[H^(+)] = [OH^(-)] = 55 xx 10^(-4)`
`[H^(+)] = [OH^(-)] ` for neutral solution so ,
`[H^(+)]^(2) = 55 xx 10^(-14)`
`[H^(+)] = sqrt(55) xx 10^(-7)` ,
`pH = - log [H^(+)] = - log (55^(1//2) xx 10^(-7))`
`pH = -1/2 log 55 - log 10^(-7) = -1/2 xx 1.74 + 7 = 6.13 `