The dissociation constant of water is `1 xx 10^(-14) mol ^(-2) litre ^(-2)`. What is the pH of 0.001 M KOH soluiton ?

Options

A. `10^(-11)`
B. 3
C. 14
D. 11

Answer

D

Explanation

0.001 M KOH solution
`[OH^(-)] = 0.001 M = 1 xx 10^(-3) M`
`[H^(+)] xx [OH^(-)] = 1 xx 10^(-14)`
`[H^(+)] = (1 xx 10^(-14))/([OH^(-)])`
`[H^(+)] = (1 xx 10^(-14))/(1 xx 10^(-3)) = 1 xx 10^(-14) xx 10^(+3)`
`[H^(+)] = 10^(-11) M`
`pH = 11`.