Reactions of complex formation are frequently used in titrimetric methods of determination of variou — Organic Chemistry Chemistry Question
Complexation reactions in the determination of inorganic ions
Reactions of complex formation are frequently used in titrimetric methods of determination of various inorganic ions. For example, fluoride forms a stable complex with aluminum(III):
6 F– + Al3+ = AlF6 3–
In water the complex gives a neutral solution. This process can be used for the direct titration of fluoride and indirect determinations of other species.
In the first experiment, a sample solution containing fluoride was neutralized with methyl red, solid NaCl was added to saturation, and the solution was heated to 70 – 80°C. The titration was performed with AlCl3 solution (c = 0.15 mol dm-3) until yellow color of the indicator turned pink.
What process occurred at the endpoint?
Model Answer
After the endpoint, the excessive Al3+ ions undergo hydrolysis, which makes the medium acidic, and the indicator turns red:
[Al(H2O)6 ] 3+ + H2O ⇆ [Al(OH)(H2O)5] 2+ + H3O +
Why heating increased the endpoint sharpness?
Model Answer
On heating, the hydrolysis equilibrium shifts rightwards.
What is the purpose of adding sodium chloride?
Model Answer
Cryolite Na3AlF6 being formed upon the titration is only slightly soluble in water. Hence, NaCl was added to further decrease its solubility and shift the equilibrium of complex formation rightwards.
In the second experiment, the content of calcium was determined in the following way. An excess of NaCl together with 0.500 g NaF were added to the sample, and the resulting solution was titrated with a standard solution of AlCl3 (0.1000 mol dm-3) in the presence of methyl red. The endpoint was attained with 10.25 cm3 of the titrant.
What operation (absolutely necessary to make the determination correct!) is missing from the description of the procedure? Compare with the first experiment described above.
Model Answer
Neutralization of the sample solution before titration is missing. This operation is mandatory if an acid–base indicator is used to observe the endpoint and the sample is suspected to contain acids. Heating makes the endpoint sharper but is not as critical.
Write down the reactions taking place in this procedure.
Model Answer
In this case a reverse titration was applied. Fluoride precipitates calcium:
Ca2+ + 2 F– = CaF2↓,
and the excess of fluoride is titrated with AlCl3:
6 F– + Al3+ = AlF6 3–
Calculate the amount of calcium in the sample.
Model Answer
10.25 cm3 of 0.1000 mol dm-3 AlCl3 gives 1.025 mmol of Al3+, corresponding to 6.15 mmol of F–. The initial amount of NaF was 0.500 g, or 11.91 mmol, i.e. 5.76 mmol of F– was spent for the precipitation of calcium. The amount of calcium is 2.88 · 10–3 mol.
Similar principles are used in determination of silicic acid. To the neutralized colloidal solution of the sample, 0.5 g of KF was added, which was followed by introduction of HCl (10.00 cm3, c = 0.0994 mol dm-3) up to a definite excess. The resulting mixture was then titrated with a standard solution of alkali in the presence of phenyl red (5.50 cm3 of NaOH solution with a concentration of 0.1000 mol dm-3 were spent).
What chemical reaction(s) is the determination based on? Write silicic acid as Si(OH)4.
Model Answer
Si(OH)4 + 6 KF + 4 HCl → K2SiF6 + 4 KCl + 2 H2O
As can be seen from the equation, HCl is spent in this process, and its excess is titrated with NaOH in the presence of an acid-base indicator. (To be more precise, the excess of HCl reacts with KF yielding a weak acid HF, which is then titrated with NaOH.)
What indicator should be used when neutralizing the sample of silicic acid before the titration? The pKa values of indicators: methyl red, 5.1; phenol red, 8.0; thymolphthalein, 9.9.
Model Answer
The solution of free silicic acid (a weak acid with pKa of about 10) will be slightly acidic; hence, the indicator used in the neutralization of the sample should change its color in a weakly acidic medium (methyl red, pKa ≈ 5). In weakly alkaline media (color change range of two other indicators), a considerable part of the silicic acid will be present in the form of a silicate ion, the buffer solution of which will consume a certain amount of the reacting HCl.
Calculate the amount of silicic acid in the sample solution.
Model Answer
The amount of NaOH and the excess of HCl are the same and equal to 0.550 mmol. Hence, the amount of HCl spent for the reaction with silicic acid is 0.994 – 0.550 = 0.444 mmol, and the amount of silicic acid is 0.111 mmol.