The concentration in M of OH in `0.0014 MH_(2) SO_(4)` is

Options

A. `1xx10^(-13)`
B. `0.5 xx10^(-12)`
C. `5xx10^(-12)`
D. `0.5 xx10 ^(-13)`

Answer

C

Explanation

Given,
`0.001 M H _(2) SO_(4) i.e., H _(2) SO_(4) hArr 2H^(+) + SO _(4) ^(2-)`
Then,
Concentration of `H ^(+) , [H^(+)] = 2xx 0.001`
`= 0.002M.`
Since,
`pH + pOH = 14`
`- log [H^(+)] + - log [OH^(-)] =14`
`-log (0.002) -log [OH^(-)] = 14`
` - (-2.699 )- log [H^(-)] =14`
`log [OH^(-)] = 2.699 -14`
`log [OH^(-)] =- 11.3010`
`[OH^(-)] = 10 ^(-11.3010)`
`therefore [OH^(-)] = 5xx 10 ^(-12)`