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?

A solution containing NH_(4)Cl and NH_(4)OH has [OH]=10^(-6) mol L^(-1

1.	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?
Answer» `Mg(OH)_(2) (K_(SP)=3xx10^(-11))` `Fe(OH)_(2) (K_(sp)=8xx10^(-16))` `Cd(OH)_(2) (K_(sp)=8xx10^(-6))` `AgOH(K_(sp)=5xx10^(-3))` Solution :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