A solution containing `NH_(4)CI` and `NH_(4)OH` has `[overset(Theta)OH] = 10^(-6) molL^(-1)`, which of the following hydroxides would be precipitated when this solution in added in equal volume to a solution containing `0.1M` of metal ions?

Options

A. `Mg(OH)_(2),(K_(sp)=3xx10^(-11))`
B. `Fe(OH)_(2)(K_(sp)=8xx10^(-16))`
C. `Cd(OH)_(2)(K_(sp)=8xx10^(-6))`
D. `AgOH(K_(sp)=5xx10^(-3))`

Answer

B

Explanation

When equal volumes of `(NH_(4)CI + NH_(4)OH)` and metal ions are mixed (volume becomes double and concentration is halved)
`:. [overset(Theta)OH] = (10^(-6))/(2), [M^(+n)] = (0.1)/(2)`
`Q_(sp) (or IP)` of metal hydroxides of `M(OH)_(2)`type
`=[M^(+n)] [overset(Theta)OH]^(2)`
`= ((0.1)/(2)) ((10^(-6))/(2))^(2) = (1)/(8) xx 10^(-13)`
`= 0.125 xx 10^(-13) = 12.5 xx 10^(-11)`
`:. Q_(sp) gt K_(sp) of Mg(OH)_(2) (12.5 xx 10^(-11) gt3 xx 10^(-11))` and `Q_(sp) gt K_(sp)` of `Fe(OH)_(2) (12.5 xx 10^(-11) gt 8 xx 10^(-16))`
So both can be precipitated. Since the `k_(sp)` of `Fe(OH)_(2)` is less than `Ksp` of `Mg(OH)_(2)`, so `Fe(OH)_(2)` will be precipiated first.
Similarly, `Q_(sp)` (or IP) of metal hydroxides of `M(OH)` type
`=[M^(+1)] [overset(Theta)OH] = ((0.1)/(2)) ((10^(-6))/(2))`
`= 0.25 xx 10^(-7)`
`:.Q_(sp) lt K_(sp) of AgOH (0.25 xx 10^(-7) lt 5 xx10^(-3))`
IT can not be precipitated out. Hence `Fe(OH)_(2)` will be precipitated.