2. Complexation titration/calculation (15 marks) ntnal oation constant Kay of th
ID: 1034434 • Letter: 2
Question
2. Complexation titration/calculation (15 marks) ntnal oation constant Kay of the complex Ba-EDTA at a p? of 50, 90 and to, i) Calcium ion forms a weak Ll complex with nitrate ion with a formation constant of 2.0. What would be the equilibrium concentrations of Ca and Ca(NO,) in a solution prepared by adding 10.0 mL each of 0.010 M CaClh and 2.0 M NaNO,? Neglect diverse ion effects. i) What is the pCu if 2.217 g Cu(NO,)26H O (MW-295.644 g/mol) is dissolved in 500 mL of 100 M aqueous ammonia? Given the formation constant (k) of [Cu(NH) 21E13.Explanation / Answer
i) Kf = Kf x alpha[Y4-]
alpha[Y4-] for EDTA at pH 5.0 = 3.7 x 10^-7
alpha[Y4-] for EDTA at pH 9.0 = 5.4 x 10^-2
alpha[Y4-] for EDTA at pH 11.0 = 0.85
logKf for CuEDTA = 18.80
Kf = 6.31 x 10^18
So conditional formation constant,
Kf' at pH 5.0 = 6.31 x 10^18 x 3.7 x 10^-7 = 2.34 x 10^12
Kf' at pH 5.0 = 6.31 x 10^18 x 5.4 x 10^-2 = 3.41 x 10^17
Kf' at pH 5.0 = 6.31 x 10^18 x 0.85 = 5.36 x 10^18
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ii)
For the reaction,
Ca2+ + NO3- <==> Ca(NO3)+
Kf = [Ca(NO3)+]/[Ca2+][NO3-] = 2.0
[Ca2+] = 0.01 M
[NO3-] = 2.0 M
let x be the change at equilibrium
Ca2+ + NO3- <==> Ca(NO3)+
I 0.01 2.0 -
C -x -x +x
E 0.01-x 2-x x
So,
2.0 = (x)/(0.01-x)(2-x)
2x^2 - 5.02x + 0.04 = 0
x = 0.008 M
therefore,
Equilibrium concentration of [Cu2+] = 0.010 - 0.008 = 0.002 M
Equilibrium concentration of [Cu(NO3)+] = 0.008 M
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iii) [Cu2+] = 2.217 g/295.644 g/mol x 0.5 L = 0.015 M
[NH3] = 1.0 M
Kf = [Cu(NH3)4]2+/[Cu2+][NH3]^4 = 2.1 x 10^13
Cu2+ + 4NH3 <====> Cu(NH3)4^2+
I 0.015 1.0 -
C -x -4x +x
E 0.015-x 1-4x x
So,
2.1 x 10^13 = (x)/(0.015-x)(1-4x)^4
let x be a small change then,
2.1 x 10^13 = (x)/(0.015-x)(1)^4
3.15 x 10^11 = 2.1 x 10^13x
x = 0.015 M
pCu = -log[0.015] = 1.83
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