Question

A CH₃COOH / NaCH₃COO buffer system is to be constructed that is also 0.044 M in Cr³⁺.

What is the minimum [CH₃COOH] / [CH₃COO^-] that must be maintained to keep a precipitate of chromium(III) hydroxide from forming?

minimum	[CH3COOH]	=	---------
	[CH3COO-]		1

Answer 1

Get the minimum OH- required for precipitaiton

Ksp = [Cr+3][OH-]^3

Cr(OH)3 3×10–29

3*10^-29 = (0.044)(OH-)^3

[OH-] = ((3*10^-29)/0.044)^(1/3)

[OH-] = 8.801*10^-10

then,

[H3O+] = (10^-14)/(8.801*10^-10) = 1.136*10^-5 M

pH = -log(H3O+) = -log(1.136*10^-5) = 4.944

apply buffer equation

pH = pKa + log(NaCH3COO / CH3COOH)

4.944= 4.75 + log(NaCH3COO / CH3COOH)

NaCH3COO / CH3COOH = 10^(4.944- 4.75 ) = 1.5631

we need the inverse, so

(NaCH3COO / CH3COOH)^1 = 1.5631^-1

CH3COOH/NaCH3COO = 0.6397