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 `100^(@)C`
`K_(w) = 55 xx 10^(-14)`
`H^(+) = sqrt(55 xx 10^(-14))`
`= 7.41 xx 10^(-7)`
`pH = -log [H^(+)]`
`= -log[7.41 xx 10^(-7)]`
`= -[log 7.41 + log 10^(-7)]`
`= -[0.86 - 7] = -[-6.13] = 6.13`