The concentration in M of `OH^-` in 0.001 M `H_2SO_4` is

Options

A. `1 xx 10^(-13) `
B. `0.5 xx 10^(-12)`
C. `5 xx 10^(-12)`
D. `0.5 xx 10^(-13)`

Answer

C

Explanation

`H_2SO_4 (0.001M) iff 2H^(+) + SO_(4)^(2-)`
Concentration of `H^(+) = 2 xx 0.001M = 0.002 M`
`pH = - log [H^(+)] = - log (0.002)`
`pH = 3 - 0.3010 = 2.6990`
From , `pH + pOH = 14`
` pOH = 14 - 26990 = 11.3010`
`pOH = - log [OH^(-) ]`
or ` 11.3010 = - log [OH^(-) ]`
or ` [OH^(-) ] = 5 xx 10^(-12) `