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The maximum pH of a solution which is `0.10 M` is `Mg^(2+)` from which `Mg(OH)_(2)` is not precipitated is [`K_(sp)` of `Mg(OH)_(2) = 1.2 xx 10^(-11) M^(3)`]A. `4.96`B. `6.96`C. `7.04`D. `9.04` |
Answer» Correct Answer - D For `Mg(OH)_(2)` not to be precipitated `[OH^(-)] lt [(K_(sp)(Mg(OH)_(2)))/([Mg^(2+)])]^(1//2)` `[OH^(-)] lt [(1.2x10^(-11)M^(3))/(0.10M)]^(1//2) lt 1.035x10^(-5)M` `pOH lt 4.36,pH gt 14-4.36 = 9.04` |
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