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InterviewSolution
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The solubility product of `Al(OH)_(3) ` is `2.7xx10^(-11)`. Calculate its solubility in g `L^(-1) ` and also find out pH of this solution . (Atomic mass of Al = 27 u ). |
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Answer» Suppose the solubility is S mol `L^(-1)`. Then `{:(Al(OH)_(3),hArr,Al^(3+),+,3OH^(-),,),(,,S,,3 S,,):}` `K_(sp)= S xx (3 s)^(3) = 27 S^(4)` `:. 27S^(4)=2.7xx10^(-11) or S^(4) = 10^(-12) or S=10^(-3) ` mol `L^(-1)` Molar mass of `Al(OH)_(3) ` in g `L^(-1) = 10^(-3)xx78=7.8xx10^(-2) g L^(-1)` `[OH^(-)]=3 S = 3 xx 10^(-3) ` mol `L^(-1)` `:. pOH = - log (3xx10^(-3))=3-0.4771=3.5229` pH = 14 - 2.5229 = 11.4771. |
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