A solution containing `NH_(4)Cl` and `NH_(4)OH` has `[OH]=10^(-6) mol L^(-1)`, which of the following hydroxides would be precipitated when this solution in added in equal volume to a solution containing `0.1 M` 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)Cl+NH_(4)OH)` and metal ions are mixed (volume becomes double and concentration is halved)
`:. [OH]=(10^(6))/(2),[M^(+n)]=(0.1)/(2)`
`Q_(sp)` ( or `IP`) of metal hydroxides of `M(OH)_(2)` type
`=[M^(+n)][OH]^(2)`
`=((0.1)/(2))((10^(-6))/(2))^(2)=(1)/(8)xx10^(-13)`
`=0.125xx10^(-13)=12.5xx10^(-11)`
`:. Q_(sp) lt K_(sp)` of `Mg(OH)_(2) (1.25xx10^(-11)gt3xx10^(-11))`
and `Q_(sp)gtK_(sp)` of `Fe(OH)_(2) (1.25xx10^(-11)gt8xx10^(-16))`
So both can be precipitated. Since the `K_(sp)` of `Fe(OH)_(2)` is less than `K_(sp)` of `Mg(OH)_(2)`, so `Fe(OH)_(2)` will be precipitated first. Similarly, `Q_(sp)` (or `IP`) of metal hydroxides of `M(OH)` type
`=[M^(+1)][OH]=((0.1)/(2))((10^(-6))/(2))=0.25xx10^(-7)`
`= Q_(sp)ltK_(sp)` of `AgOH(0.25xx10^(-7)lt5xx10^(-3))`
It cannot be precipitated out. Hence `Fe(OH)_(2)` will be precipitated