The hydronium ion concentration in pure water is `1xx10^(-7) ` mol `L^(-1)`. The degree of dissociation of water is

Options

A. `1.8 xx 10^(-9)`
B. `0.8 xx 10^(-8)`
C. `9.8 xx 10^(-6)`
D. `3xx10^(-9)`

Answer

A

Explanation

If `alpha` is degree of dissociation
`H_(2)OhArr H^(+) +OH^(-)`
Initial conc. `C mol L^(-1)`
Equi. Conc. `C(a-alpha)" "Calpha" "Calpha`
`[H^(+)]=Calpha` or `alpha=([H^(+)])/(C )=(1xx10^(-7))/(55.5)`
`=1.8 xx 10^(-9)`