At `80^(@)C` distilled water has `[H_(3)O^(+)]` concentration equal `[OH^(-)] 1xx10^(-6) "mole"//litre`. The value of `K_(w)` at this temperature will be

Options

A. ` 1xx 10^(-6)`
B. ` 1xx 10^(-9)`
C. ` 1 xx 10^(-12)`
D. ` 1xx 10^(-15)`

Answer

C

Explanation

`K_(w)=[H_(3)O^(+)][OH^(-)] `
Concentration of `H_(3)O^(+)` in distilled water = `1 xx 10^(-6)` mol/l
Now `[H_(3)O^(+)]=[OH^(-)] `
`K_(w)= [1 xx 10^(-6)] xx [ 1 xx 10^(-6)] = 1 xx 10^(-12)`