In `0.1M` solution, `K_(a)` for dissociation of `H_(2)S` is `4.0xx10^(-3)`. Select the correct statements.

Options

A. Concentration of `H^(+)` is `0.018M`
B. The degree of dissociation of `H_(2)S` is `18%`
C. pH of solution is `1.7447`
D. Concentration of `[H^(+)]is 0.18M`

Answer

A, B, C

Explanation

`{:(H_(2)ShArr,H^(+)+,HS^(-)),(1,0,0),(C(1-alpha),Calpha,Calpha):}`
`K_(a_(1))= (Calpha^(2))/((1-alpha)) , (:' C = 0.1 M)`
`4.0xx10^(-3)= (0.1xxalpha^(2))/((1-alpha))`
`alpha = 0.18= 18%`
Now, `[H^(+)]= C.alpha=0.1xx0.18=0.018M`
`pH= -log [H^(+)]`
`pH= -log 1.8xx10^(-2)= 1.7447`