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The solubility of `Mg (OH)_(2)` in pure water is `9.57 xx 10^(-3) gL^(-1)`. Calculate its solubility (in `gL^(-1))` in `0.02M Mg(NO_(3))_(2)` solution.A. `1.5 xx 10^(-4)`B. `8.69 xx 10^(-4)`C. `0.5 xx 10^(-3)`D. `0.5 xx 10^(-5)` |
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Answer» Correct Answer - B Solubility of `Mg(OH)_(2)` `= (9.57 xx 10^(-3))/(58) = 1.65 xx 10^(-4) "mol lits"^(-1)` `K_(sp) = 4s^(3)` `= 4xx (1.65 xx 10^(-4))^(3)` `= 17.96 xx 10^(-12)` But in presence of `Mg(NO_(3))_(2)` `K_(s) = (x+c)(2s)^(2)` `K_(s) = 17.96 xx 10^(-12) = (s+0.02)(2s)^(2)` `s = 14.98 xx 10^(-6) "mollts"^(-1)` `s = 14.98 xx 10^(-6) xx 58 gr"lt"^(-1)` `8.69 xx 10^(-4) gr "lits"^(-1)` |
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