Question

Use the solubility-product constant for Cr (OH) 3 ( Ksp = 6.7×10?31) and the formation constant for Cr (OH) ?4 from the following table to determine the concentration of Cr (OH) ?4 in a solution that is buffered at pH= 11 and is in equilibrium with solid Cr (OH) 3.
Table Formation Constants for Some Metal Complex Ions in Water at 25 ?C

Complex Ion	Kf	Equilibrium Equation
Ag(NH3)+2	1.7×107	Ag+(aq)+2NH3(aq)?Ag(NH3)+2(aq)
Ag(CN)?2	1×1021	Ag+(aq)+2CN?(aq)?Ag(CN)?2(aq)
Ag(S2O3)3?2	2.9×1013	Ag+(aq)+2S2O2?3(aq)?Ag(S2O3)3?2(aq)
CdBr2?4	5×103	Cd2+(aq)+4Br?(aq)?CdBr2?4(aq)
Cr(OH)?4	8×1029	Cr3+(aq)+4OH?(aq)?Cr(OH)?4(aq)
Co(SCN)2?4	1×103	Co2+(aq)+4SCN?(aq)?Co(SCN)2?4(aq)
Cu(NH3)2+4	5×1012	Cu2+(aq)+4NH3(aq)?Cu(NH3)2+4(aq)
Cu(CN)2?4	1×1025	Cu2+(aq)+4CN?(aq)?Cu(CN)2?4(aq)
Ni(NH3)2+6	1.2×109	Ni2+(aq)+6NH3(aq)?Ni(NH3)2+6(aq)
Fe(CN)4?6	1×1035	Fe2+(aq)+6CN?(aq)?Fe(CN3)4?6(aq)
Fe(CN)3?6	1×1042	Fe3+(aq)+6CN?(aq)?Fe(CN)3?6(aq)

Answer 1

The formation of complex ion [ Cr(OH)4]^- involves two reaction steps:

Cr(OH)3(s) <---->Cr³⁺(aq) +3OH^-(aq) ,Ksp[Cr³⁺][OH^-]6.7*10^-31

Cr³⁺(aq) +4OH^-(aq) <----> [ Cr(OH)4]^- ,Kf [ Cr(OH)4]-/[OH-]8.0*10^29

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net rxn:Cr(OH)3(s) +OH^-(aq) <----> [ Cr(OH)4]^- ,K[ Cr(OH)4]^- /[OH-]

KKsp*Kf(6.7*10^-31)(8.0*10^-29)0.536

The solubility of Cr(OH)3 increases on adding excess OH- due to the formation of the complex ion

given: pH11.0-log [H+]

[H+]1*10^-11M

[H+][OH-]kw1*10^-14Mionic product of water

[OH-]1*10^-14M/[H+](1*10^-14)/(1*10^-11)0.001M (initial concentration)

Consider the net equation , the ICE table: solids are not included in the equilibrium constant equation,as the concentration of solids are constant and equal to unity

	Cr(OH)3(s)	[OH-]	[Cr(OH)4]-
initial	NA	0.001M	0
change	NA	-x	+x
equilibrium	NA	0.001-x	x

K0.536[ Cr(OH)4]^- /[OH-]x/(0.001-x)

or, 0.536x/(0.001-x)

or, 5.36*10^-4-0.536xx

or, x3.489*10^-4M[ Cr(OH)4]^- (concentration of complex ion)