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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 :A. `11.78`B. `10.78`C. `2.5`D. `2.22` |
Answer» Correct Answer - A 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` |
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