The calcium content of an aqueous sample can be determined by the following procedure: Step 1 A few — Physical Chemistry — Kinetics Chemistry Question
Determination of Calcium Ion by Precipitation Followed by Redox Titration
The calcium content of an aqueous sample can be determined by the following procedure:
Step 1 A few drops of methyl red are added to the acidified aqueous sample, followed by thorough mixing with Na2C2O4 solution.
Step 2 Urea ((NH2)2CO) is added and the solution gently boil until the indicator turns yellow (this typically takes 15 min). CaC2O4 precipitates out.
Step 3 The hot solution is filtered and the solid CaC2O4 is washed with ice-cold water to remove excess C2O4 2– ions.
Step 4 The insoluble CaC2O4 is dissolved in hot H2SO4 (c = 0.1 mol dm –3) to give Ca 2+ ions and H2C2O4. The dissolved H2C2O4 is titrated with standardized KMnO4 solution until the purple end point is observed.
Relevant reactions and equilibrium constants:
CaC2O4(s) Ca 2+ (aq) + C2O4 2– (aq) Ksp = 1.30×10 –8
Ca(OH)2(s) Ca 2+ (aq) + 2 OH – (aq) Ksp = 6.50×10 –6
H2C2O4(aq) HC2O4 – (aq) + H + (aq) Ka1 = 5.60×10 –2
HC2O4 – (aq) C2O4 2– (aq) + H + (aq) Ka2 = 5.42×10 –5
H2O H + (aq) + OH – (aq) Kw = 1.00×10 –14
Write a balanced equation for the reaction that takes place upon the addition of urea (Step 2).
Model Answer
(NH2)2CO + H2O CO2 + 2 NH3
The calcium content of a 25.00 cm3 aqueous sample was determined using the above procedure and found to require 27.41 cm3 of a KMnO4 solution (c = 2.50×10–3 mol dm–3) in the final step. Find the concentration of Ca 2+ ions in the sample.
Model Answer
Titration reaction: 5 H2C2O4 + 2 MnO4– + 6 H+ 10 CO2 + 2 Mn2+ + 8 H2O
c(Ca2+) = 2.5×10–3 mol dm–3 × (0.02741 dm3 × (5/2) / 0.025 dm3) = 6.85×10–3 mol dm–3
Calculate the solubility of CaC2O4 in an aqueous solution buffered at pH 4.0. (Neglect activity coefficients.)
Model Answer
Mass–balance:
[Ca 2+] = [C2O4 2–] + [HC2O4 –] + [H2C2O4] = [C2O4 2–] (1 + [H +] / K2 + [H +] 2 / K1K2)
[C2O4 2–] = [Ca 2+] / (1 + [H +] / K2 + [H +] 2 / K1K2) (1)
Substituting (1) into [Ca 2+] [C2O4 2–] = Ksp
[Ca 2+] = 1.92×10 –4
In the above analysis a possible source of error was neglected. The precipitation of CaC2O4 in Step 1 will be incomplete if an excess of C2O4 2– ions is added, due to the following reactions:
Ca 2+(aq) + C2O4 2–(aq) CaC2O4(s) Kf1 = 1.0×10 3
CaC2O4(aq) + C2O4 2–(aq) Ca(C2O4)2 2–(aq) Kf2 = 10
Calculate the equilibrium concentrations of Ca 2+ and C2O4 2– ions in solution after optimal precipitation of CaC2O4 is reached.
Model Answer
cCa = [Ca 2+] + [CaC2O4(aq)] + [Ca(C2O4)2 2–]
= Ksp (1 / [C2O4 2–] + Kf1 + Kf1 Kf2 [C2O4 2–])
dcCa / d[C2O4 2–] = 0 = – Ksp / [C2O4 2–] 2 + Ksp Kf1 Kf2
[C2O4 2–] = 1.0×10 –2
[Ca 2+] = Ksp / [C2O4 2–] = 1.3×10 –6
Calculate the concentrations of H + and Ca 2+ in a saturated solution of CaC2O4. (Neglect activity coefficients. Any assumptions made during calculation must be clearly stated.)
Model Answer
Charge balance: 2 [Ca 2+] + [H +] = 2 [C2O4 2–] + [HC2O4 –] + [OH –] (1)
Mass balance: [Ca 2+] = [C2O4 2–] + [HC2O4 –] + [H2C2O4] (2)
Because Kb2 is too small, [H2C2O4] can be neglected.
Comparing (1), (2), [HC2O4 –] = Kw / [H +] – [H +] (3)
[C2O4 2–] = (K2 Kw) / [H +] 2 – K2 (4)
[Ca 2+] = Ksp / [C2O4 2–] = Ksp [H +] 2 / (K2Kw – K2[H +] 2 ) (5)
Substituting (3), (4), (5) into (2)
K2 [H +] 5 + (K2 2 – Ksp) [H +] 4 – 2 K2 Kw [H +] 3 – 2 K2 2 Kw [H +] 2 + K2 Kw 2 [H +] + K2 2 Kw 2 = 0
Solving [H +], [H +] = 5.5×10 –8 (or pH = 7.26)
Substituting [H +] into (5), [Ca 2+] = 1.04×10 –4