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))`

Options

A. `2.8xx10^(-3)`
B. `2.8xx10^(-2)`
C. `2.8xx10^(-4)`
D. `2.8xx10^(-5)`

Answer

C

Explanation

`{:(,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)`