Citric acid (H_3Cit) can be used as a household cleaning agent to dissolve rust
ID: 1024544 • Letter: C
Question
Citric acid (H_3Cit) can be used as a household cleaning agent to dissolve rust stains. The rust (i.e., iron(III) hydroxide) dissolves because the citrate ion forms a soluble complex with iron(III). Using the fact that K_f = 6.3 times 10^11 for iron(III) citrate, K_sp = 2.6 times 10^-39 for iron(III) hydroxide, K_w = 1.0 times 10^-14, and pK_a1 = 3.15, pK_a2 = 4.77 and pK_a3 = 6.39 for citric acid, answer the following: What is the equilibrium constant for the reaction of iron(III) hydroxide with citric acid to form iron(III) citrate and water? What is the molar solubility of iron(|III) hydroxide in a 0.500 M solution of citric acid?Explanation / Answer
5.
(a) For reaction of Fe(OH)3 with citric acid
Fe(OH)3 + citric acid ---> Fe citrate + 3H2O
with,
pKa = -logKa
and given values for,
Fe(OH)3 <==> Fe3+ + 3OH- .... Ksp = 2.6 x 10^-39
Fe3+ + citrate^3- <==> Fe citrate ... Kf = 6.3 x 10^11
H3Citrate <==> H+ + H2Citrate^- ... Ka1 = 7.08 x 10^-4
H2citrate^- <==< H+ + Hcitrate^2- ... Ka2 = 1.70 x 10^-5
Hcitrate^2- <==< H+ + citrate^3- ... Ka3 = 4.07 x 10^-7
3 (H+ + OH- <==> H2O ....... 1/Kw^3 = 1 x 10^42
So,
equilibrium constant = 2.6 x 10^-39 x 6.3 x 10^11 x 7.08 x 10^-4 x 1.70 x 10^-5 x 4.07 x 10^-7 x 1 x 10^42
= 8.024
(b) Molar solubility of Fe(OH)3(s) in 0.5 M citric acid
Fe(OH)3 + citric acid <==> Fe citrate + 3H2O
I 0.5 -
C -x +x
E 0.5-x +x
So,
8.024 = x/0.5-x
molar solubility = x = 0.44 M