InterviewSolution
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InterviewSolution
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The concentration of hydroxyl ion in a solution left after mixing 100 mL of `0.1M MgCl_(2)` and 100 mL of `0.2M NaOH` is: `(K_(SP)of Mg(OH)_(2)=1.2xx10^(-11))`A. `2.8xx10^(-3)`B. `2.8xx10^(-2)`C. `2.8xx10^(-4)`D. `2.8xx10^(-5)` |
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Answer» Correct Answer - C `{:(,MgCI_(2)+,2NaOHrarr,Mg(OH)_(2)+,2NaCI),("mm before", 10,20,0,0),("reaction",0,0,10,20):}` Thus, 10 milli-mole of `Mg(OH)_(2)` are formed The product of `[Mg^(2+)][OH^(-)]^(2)` is therefore `[(10)/(200)]xx[(20)/(200)]^(2)=5xx10^(-4)` which is more than `K_(SP)of Mg(OH)_(2)`. Now solubiltiy (S) of `Mg(OH)_(2)` can be derived by `K_(SP)=4S^(3)` `:. S= 3sqrt((K_(SP))/(4))=3sqrt([(1.2xx10^(-11))/(4)])` `=1.4xx10^(-4)` `:. [OH^(-)]= 2S =2.8xx10^(-4)` |
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