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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)`

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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