Try this worked example of a Chemistry Olympiad question on the use of titration to determine phosphate levels
In this question from the 2007 UK Chemistry Olympiad round one paper, students read about the effects of excess phosphate levels in our water supply and in the blood. They then learn how ammonium molybdate can be used to set up a titration to measure the concentration of phosphate ions in solution, before the question probes this topic in greater depth. The accompanying video walkthrough is designed to help students if they get stuck or want to check their working.
The question is of high difficulty, and covers the following chemistry topics:
- Neutralisation reactions
- Titration calculations
- pH calculations
- Balancing equations
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Excess phosphate is undesirable in our water supply since it promotes the growth of algae. High phosphate levels in the blood (hyperphosphataemia), caused by an inability of the kidneys to filter out excess phosphate from food, cause a range of bone diseases. A safe level of phosphate in blood serum is <1.8 mmol dm–3.
To determine the concentration of phosphate ions in solution, ammonium molybdophosphate (formed by the addition of ammonium molybdate, (NH4)2MoO4, to phosphate ions) is titrated with NaOH using a phenolphthalein indicator. One mole of ammonium molybdophosphate contains one phosphate unit and requires 23 moles of NaOH in the titration before the end point is reached.
Which two of the following statements explain why so many moles of hydroxide are needed per mole of ammonium molybdophosphate?
- A: Sodium hydroxide is a weak alkali.
- B: Ammonium molybdophosphate has a very high molecular mass.
- C: The molybdenum-containing species is acidic.
- D: Phosphate ions are acidic.
- E: The molybdophosphate ion contains many molybdenums.
10.0 cm3 of treated blood serum was reacted with excess ammonium molybdate and the precipitated ammonium molybdophosphate required 17.25 cm3 of 0.0100 mol dm–3 NaOH. Calculate the concentration of phosphate in the blood serum in mol dm–3.
The degree to which a phosphate ion is protonated in aqueous solution depends on the pH. Use the data below to calculate which form of phosphate is prevalent in blood serum at pH 7.4.
- H3PO4(aq) ⇋ H+(aq) + H2PO4–(aq)
Ka1 = 7.9 × 10–3 mol dm–3
- H2PO4–(aq) ⇋ H+(aq) + HPO42–(aq)
Ka2 = 6.2 × 10–8 mol dm–3
- HPO42–(aq) ⇋ H+(aq) + PO43–(aq)
Ka3 = 4.4 × 10–13 mol dm–3
Lanthanum carbonate (marketed as FosrenolTM) is currently being used to treat hyperphosphataemia; aluminium sulfate has been used to control phosphate levels in ponds. Both LaPO4 and AlPO4 are essentially insoluble in water and are therefore removed as precipitates.
Write the equation for the reaction between lanthanum carbonate and the hydrochloric acid present in the stomach.
The saturated concentrations of these phosphates can be worked out from their solubility products (Ksp):
Ksp(LaPO4) = [La3+(aq)]sat[PO43–(aq)]sat = 7.08 × 10–27 mol2 dm–6
Ksp(AlPO4) = [Al3+(aq)]sat[PO43–(aq)]sat = 9.84 × 10–21 mol2 dm–6
Calculate the concentration of a saturated solution of lanthanum phosphate.
If an equimolar solution of aluminium and lanthanum ions is added slowly to a solution containing phosphate ions, which phosphate would precipitate first?
Statements C and E.
Concentration of phosphate: 7.5 x 10–4 mol dm–3.
Form of phosphate: HPO42–.
6HCl + La2(CO3)3 → 2LaCl3 + 3CO2 + 3H2O
8.41 x 10–14 mol dm–3
This question is about phosphate levels in our environment.
The question begins by providing some background information into the problems that can be caused by excess phosphate levels in both water supplies and blood. The introduction then moves on to explain how the concentration of phosphate ions in solution can be determined by addition of ammonium molybdate to the solution of phosphate ions, forming ammonium molybdophosphate, with each mole of ammonium molybdophosphate containing one phosphate unit. The concentration of ammonium molybdophosphate formed can be determined by titration with sodium hydroxide using phenolphthalein indicator. One mole of ammonium molydophosphate requires 23 moles of sodium hydroxide for neutralisation. Parts (a) and (b) of the question look at the chemistry of this titration.
Part (a) of the question asks you to choose two statements out of a choice of five which explain why so many moles of sodium hydroxide are needed to neutralise each mole of ammonium molybdophosphate. If we take each of these statements in turn we can see which ones we think are correct.
Statement A is that sodium hydroxide is a weak alkali. You should know that sodium hydroxide is probably the best known example of a strong alkali and so we can disregard this statement. Also since the indicator for the titration is phenolphthalein we know that sodium hydroxide cannot be a weak alkali, as the phenolphthalein is not a suitable indicator for titrations with weak alkalis, owing to the colour change taking place at a pH above the equivalence point. This supports our decision that Statement A is incorrect and hence can be disregarded.
Statement B is that ammonium molybdophosphate has a very high molecular mass. Although we do not know the formula of the compound we can imagine that this is probably true. However the mass of the compound would not explain why so many moles of base were needed as with titrations it is the number of moles of the compounds involved which is important.
Statement C is that the molybdenum-containing species is acidic. This must be true for neutralisation to be possible by the addition of base. Although it doesn’t help us to explain why so many moles of hydroxide are needed it does help us to explain why hydroxide is needed and so we can tick this as one of our two statements.
Statement D states that phosphate ions are acidic. By definition, a Brønsted-Lowry acid is a substance that can donate a proton. Looking at the formula of a phosphate ion, PO43- it clearly has no protons to donate and hence we can rule out statement D.
Finally then, statement E is that the molybdophosphate ion contains many molybdenums. Although we do not fully know the structure of the ammonium molybdophosphate complex, since the phosphate is not acidic then it must be the ammonium molybdo bit which is. Ammonium ions are weakly acidic, but since one mole of hydroxide reacts with one mole of acid then we know that the ammonium molydo bit must contain many moles of acid and so we can say that this statement, statement E does help to explain why so many moles of hydroxide are needed per mole of ammonium molydophosphate. Hence our answer is C and E.
Part (b) of the question then asks you to use the information given about the titration to calculate the concentration of phosphate in 10 cm3 of treated blood serum, given that when the blood serum was reacted with an excess of ammonium molybdate, the precipitated ammonium molybdophosphate required 17.25 cm3 of a 0.0100 mol dm–3 solution of sodium hydroxide for neutralisation. We can calculate the number of moles of sodium hydroxide required for the neutralisation using the equation, moles = concentration × volume. Applying the concentration of hydroxide of 0.0100 mol dm–3 and the volume of 17.25 cm3 or (17.25 ¸ 1000) dm3 we can calculate the number of moles of sodium hydroxide needed to be 1.725 × 10–4 mol.
From our initial equation, written using the information in the introduction, we know that 1 mole of ammonium molybdophosphate requires 23 moles of sodium hydroxide before the end point is reached. Therefore if we know that the number of moles of sodium hydroxide involved in the reaction is 1.725 × 10–4 mol, we can calculate that the number of moles of ammonium molydophosphate formed by the reaction of the phosphate in the 10 cm3 of blood serum with the ammonium molybdate, is equal to 1.725 × 10–4 mol ¸ 23 = 7.5 × 10–6 mol. In the introduction, we are told that one mole of ammonium molydophosphate contains one phosphate unit. Therefore we know that this 7.5 × 10–6 mol is also equal to the number of moles of phosphate ion present in the 10 cm3 of blood serum. Finally then, if there are 7.5 × 10–6 mol of phosphate present in 10 cm3 of the blood serum, we can calculate the concentration of phosphate in the blood serum to be 7.5 × 10–6 mol ÷ 0.010 dm3 = 7.5 × 10–4 mol dm–3.
Part (c) of the question introduces you to the fact that the degree to which a phosphate ion is protonated in aqueous solution, depends on the pH of the solution. The question provides the acid dissociation constants for each of the three dissociation steps of phosphoric acid, and asks you to calculate which form of phosphate is prevalent in blood serum at pH 7.4. If we start off by looking at the dissociation of a weak acid HA into H+ and A–, we know that the acid dissociation constant Ka = [H+][A–] / [HA].
In order to determine which form of phosphate is prevalent in blood serum, we need to determine the ratio of the deprotonated to the protonated form for each of the dissociation steps at the pH given. This can represented by [A–] / [HA]. Rearranging our equation for Ka given above, this ratio can therefore be calculated as Ka / [H+]. Since pH = –log[H+] we can calculate the [H+] at pH 7.4 to be 10–7.4 or 3.98 × 10–8 mol dm–3. With this value to hand, we can now run through each of the dissociation steps and calculate the ratio of [A–] / [HA] by dividing the acid dissociation constant by the concentration of [H+]. So, for the first dissociation, dividing Ka1 by [H+], 3.98 × 10–8 mol dm–3 we calculate that [H2PO4–] / [H3PO4] = 198439, ie the predominant species is H2PO4– and the equilibrium lies to the right.
Similarly, for the second dissociation step, dividing the acid dissociation constant, Ka2 by the concentration of [H+] we calculate that [HPO42–] / [H2PO4–] = 1.55. In this case the equilibrium is much more balanced with HPO42– slightly dominating. Finally then, for the final dissociation, [PO43–] / [ HPO42–] = Ka3 / 3.98 × 10–8 mol dm–3 = 1.11 × 10–5. In this case the equilibrium lies far to the left and HPO42– dominates. So what do these calculations tell us about the phosphate species present at pH 7.4? The first dissociation lies to the right and H2PO4– dominates. The second dissociation step however tells us that this H2PO4– dissociates further into HPO42–. However, the HPO42– doesn’t dissociate any further because the third dissociation step lies to the left at this pH and hence we can say from the data, that the form of phosphate which is prevalent at pH 7.4 is HPO42–.
The final part of the question moves on to look at some treatments for high phosphate levels in the blood, known as hyperphosphataemia, and for high phosphate levels in ponds. In both cases the excess phosphate ions are removed as precipitates, in blood as a precipitate of lanthanum phosphate formed by the addition of lanthanum carbonate which is marketed as the drug Fosrenol or in pond water as a precipitate of aluminium sulphate formed by the addition aluminium sulfate.
Part (d) of the question asks you to write an equation for the reaction between lanthanum carbonate and the hydrochloric acid present in your stomach. This first thing to do in this question is to work out the correct formula for the lanthanum carbonate. Since we are told in the introduction that lanthanum phosphate has a formula of LaPO4, then we know that a lanthanum ion must carry a 3+ charge. Since a carbonate ion has a 2– charge and knowing that the overall charge on an ionic compound must be equal to zero, then we know that the two ions must combine in a ratio of 2:3 giving the formula of lanthanum carbonate as La2(CO3)3.
Now, when a carbonate reacts with an acid, the products of the reaction are a salt, carbon dioxide and water. Using this equation then, we can write that when lanthanum carbonate reacts with hydrochloric acid the products of the reaction are a salt, lanthanum chloride (which must have a formula LaCl3), carbon dioxide and water. Finally then, balancing the equation, looking at the lanthanum ions first, each equivalent of lanthanum carbonate contains two lanthanum ions and hence we must form two equivalents of lanthanum chloride. Each equivalent of lanthanum chloride contains three chloride ions and so we know the reaction must involve six equivalents of hydrochloric acid. Looking now at the carbonate ions, each equivalent of lanthanum carbonate contains three carbonate ions and hence three equivalents of carbon dioxide must be formed. Finally then balancing the hydrogen atoms we can see that we must also form three equivalents of water.
The final part of the question then moves on to look at the solubility of the phosphate salts formed in the precipitation reactions. The question introduces something the solubility product of a salt, Ksp and tells us that it is equal to the product of the concentrations of each of the ions in the salt at the point of saturation. So for lanthanum phosphate, the product from the precipitation reaction between lanthanum carbonate and excess phosphate ions, Ksp = [La3+]sat [PO43–]sat.
Part (e) of the question asks you to use these values to calculate the concentration of a saturated solution of lanthanum phosphate. Since the formula for lanthanum phosphate is LaPO4 then each mole of lanthanum phosphate contains an equal number of lanthanum and phosphate ions. Therefore at saturation, we know that [La3+] = [PO43–]. Knowing this then, we can say that 7.08 × 10–27 mol2 dm-6 = either [La3+]2 or [PO43–]2 and hence at saturation the concentration of each of these ions is 8.41 × 10–14 mol dm3. Since each mole of lanthanum phosphate contains one mole of each ion, then the concentration of each ion at saturation is equal to the concentration of the salt at saturation and hence the concentration of lanthanum phosphate at saturation is 8.41 × 10–14 mol dm–3.
Finally in part (f) of the question you are asked, if an equimolar solution of aluminium and lanthanum ions is added slowly to a solution of phosphate ions, which of the two possible phosphates, aluminium phosphate or lanthanum phosphate would precipitate first. A salt precipitates out when its concentration goes above the concentration of the saturated solution; the solution can ‘hold’ no more salt. So, from the calculation in part(e) we know that lanthanum phosphate would precipitate out at a concentration of 8.41 × 10–14 mol dm–3. If we do an identical calculation for AlPO4 since it too contains an equal number of Al3+ and PO43– ions, we can say that [Al3+] = [PO43–] = [AlPO4]sat = square route of Ksp(AlPO4) and hence calculate that the aluminium phosphate would precipitate out at a concentration of 9.91 × 10–11 mol dm–3. Finally then, from our calculations we can see that the concentration of lanthanum phosphate at saturation is lower than the concentration of aluminium phosphate at saturation and hence lanthanum phosphate would precipitate first.
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