If first dissociation of `X(OH)_(32)` is `100%` where as second dissociation is `50%` and third dissciation is negligible then the pH of `4 xx 10^(-3) M, X(OH)` is :

Options

A. `11.78`
B. `10.78`
C. `2.5`
D. `2.22`

Answer

A

Explanation

First dissociation
`X(OH)_(3) rarr X(OH)_(2)^(+) + OH^(-)`
Second dissociation :
`X(OH)_(2)^(+) rarr X(OH)^(2+) + OH^(-)`
Total `[OH^(-)] = 4 xx 10^(-3) + 2 xx 10^(-3) = 6 xx 10^(-3)`
`pOH = 3 log 6 = 2.22`
`:. pH = 11.78`