Q1 A solution of sodium hydroxide contained 0.25 mol dm^-3. Using phenolphthalein indicator, titration of 25.0 cm³ of this solution required 22.5 cm³ of a hydrochloric acid solution for complete neutralisation.

(a) write the equation for the titration reaction.

(b) what apparatus would you use to measure out (i) the sodium hydroxide solution? (ii) the hydrochloric acid solution?

(c) what would you rinse your apparatus out with before doing the titration ?

(d) what is the indicator colour change at the end-point?

(e) calculate the moles of sodium hydroxide neutralised.

(f) calculate the moles of hydrochloric acid neutralised.

(g) calculate the concentration of the hydrochloric acid in mol/dm³ (molarity).

Q2 A solution made from pure barium hydroxide contained 2.74 g in exactly 100 cm³ of water. Using phenolphthalein indicator, titration of 20.0 cm³ of this solution required 18.7 cm³ of a hydrochloric acid solution for complete neutralisation. [atomic masses: Ba = 137, O = 16, H = 1)

(a) write the equation for the titration reaction.

(b) calculate the molarity of the barium hydroxide solution.

(c) calculate the moles of barium hydroxide neutralised.

(d) calculate the moles of hydrochloric acid neutralised.

(e) calculate the molarity of the hydrochloric acid.

Q3 4.90g of pure sulphuric acid was dissolved in water, the resulting total volume was 200 cm³. 20.7 cm³ of this solution was found on titration, to completely neutralise 10.0 cm³ of a sodium hydroxide solution. [atomic masses: S = 32, O = 16, H = 1)

(a) write the equation for the titration reaction.

(b) calculate the molarity of the sulphuric acid solution.

(c) calculate the moles of sulphuric acid neutralised.

(d) calculate the moles of sodium hydroxide neutralised.

(e) calculate the concentration of the sodium hydroxide in mol dm^-3 (molarity).

Q4 100 cm³ of a magnesium hydroxide solution required 4.5 cm³ of sulphuric acid (of concentration 0.1 mol dm^-3) for complete neutralisation. [atomic masses: Mg = 24.3, O = 16, H = 1)

(a) give the equation for the neutralisation reaction.

(b) calculate the moles of sulphuric acid neutralised.

(c) calculate the moles of magnesium hydroxide neutralised.

(d) calculate the concentration of the magnesium hydroxide in mol dm^-3 (molarity).

(e) calculate the concentration of the magnesium hydroxide in g cm^-3.

Q5 Magnesium oxide is not very soluble in water, and is difficult to titrate directly. Its purity can be determined by use of a 'back titration' method. 4.06 g of impure magnesium oxide was completely dissolved in 100 cm³ of hydrochloric acid, of concentration 2.0 mol dm^-3 (in excess). The excess acid required 19.7 cm³ of sodium hydroxide (0.20 mol dm^-3) for neutralisation. This 2nd titration is called a 'back-titration', and is used to determine the unreacted acid. [atomic masses: Mg = 24.3, O = 16)

(a) write equations for the two neutralisation reactions.

(b) calculate the moles of hydrochloric acid added to the magnesium oxide.

(c) calculate the moles of excess hydrochloric acid titrated.

(d) calculate the moles of hydrochloric acid reacting with the magnesium oxide.

(e) calculate the moles and mass of magnesium oxide that reacted with the initial hydrochloric acid.

(f) hence the % purity of the magnesium oxide.

(g) what compounds could be present in the magnesium oxide that could lead to a false value of its purity ? explain.

Q6 2 dm³ of concentrated hydrochloric acid (10 M) was spilt onto a laboratory floor. It can be neutralised with limestone powder. [atomic masses: Ca = 40, C = 12, O = 16)

(a) give the equation for the reaction between limestone and hydrochloric acid.

(b) how many moles of hydrochloric acid was spilt?

(c) how many moles of calcium carbonate will neutralise the acid?

(d) what minimum mass of limestone powder is needed to neutralise the acid?

(e) 1000 dm³ of sulphuric acid, of concentration 2 mol dm^-3, leaked from a tank.

Calculate the minimum mass of magnesium oxide required to neutralise it.

Q7 A 50.0 cm3 sample of sulphuric acid was diluted to 1.00 dm³. A sample of the diluted sulphuric acid was analysed by titrating with aqueous sodium hydroxide. In the titration, 25.00 cm³ of 1.00 mol dm^-3 aqueous sodium hydroxide required 20.0 cm³ of the diluted sulphuric acid for neutralisation.

(a) give the equation for the full neutralisation of sulphuric acid by sodium hydroxide.

(b) calculate how many moles of sodium hydroxide were used in the titration?

(c) calculate the concentration of the diluted acid.

(d) calculate the concentration of the original concentrated sulphuric acid solution.

Q8 A sample of sodium hydrogencarbonate was tested for purity using the following method. 0.40g of the solid was dissolved in 100 cm³ of water and titrated with 0.20 mol dm^-3 hydrochloric acid using methyl orange indicator.

23.75 cm3 of acid was required for complete neutralisation. [Ar's: Na = 23, H = 1, C = 12, O = 16]

(a) Write the equation for the titration reaction.

(b) Calculate the moles of acid used in the titration and the moles of sodium hydrogencarbonate titrated.

(c) Calculate the mass of sodium hydrogen carbonate titrated and hence the purity of the sample.

Q9 This question involves theoretical calculations to do with 'how much to weigh out' for titrations and a common requirement to show development in coursework projects. They involve reagents such as pure anhydrous sodium carbonate, standardised hydrochloric acid and EDTA titrations (theory).

Atomic masses: O = 16, H = 1, Na = 23, C = 12, Ca = 40, P = 31.0

9(a)(i) Write out the equation, complete with state symbols for the reaction between hydrochloric acid and sodium carbonate.

(ii) A pipetted 25cm³ aliquot of a solution of sodium carbonate is to be titrated with an approximately 1.0 mol dm^-3 hydrochloric acid to be standardised.

What mass of dried anhydrous sodium carbonate must be dissolved in 250 cm³ of deionised water, so that a 25cm³ aliquot of the carbonate solution will give a 20.0 cm³ titration with the hydrochloric acid?

What is the molarity of the sodium carbonate solution, assuming 100% purity of

9(b)(i) The simplified molecular structure of 2-ethanoylhydroxybenzoic acid ('Aspirin') is CH₃COOC₆H₄COOH.

Give the equation of its reaction with sodium hydroxide.

(ii) A sample of aspirin was to be analysed for purity by titrating it with standardised 0.10 mol dm^-3 sodium hydroxide using phenolphthalein indicator. Assuming 100% purity and access to a 4 decimal place electronic balance, calculate the mass of Aspirin that should be weighed out to give a titration of 23.0 cm³ of the alkali.

(iii) The main contaminant is likely to be unreacted 2-hydroxybenzoic acid. Why is this likely to be an impurity? and how will this affect the % purity you calculate i.e. why and how will the % purity be in error?

9(c) Pure calcium carbonate can be used to make a standard calcium ion solution to practice a complexometric titration of calcium ions with EDTA or determine the molarity of the EDTA reagent.

(i) Give a simple equation to show the chelation reaction between hydrated calcium ions and the EDTA anion at pH10 and what sort of reaction is it?

(ii) To make a standard calcium ion solution 0.25 g of A.R. calcium carbonate was dissolved in a little dilute hydrochloric acid and made up to 250 cm³ in a calibrated volumetric flask. Calculate the molarity of the calcium ion in this solution.

(iii) Approximately 1.0g of the solid disodium dihydrate salt of EDTA was dissolved in 250 cm³ of water in a volumetric flask. 25 cm³ of this was pipetted into a conical flask and 1 cm³ of a conc. ammonia/ammonium chloride pH10 buffer was added. After adding a few drops of Eriochrome Black T indicator, the solution was titrated with the EDTA solution until the reddish tinge turns to blue at the endpoint. If 25.70 cm³ of the EDTA solution was required to reach the equivalence point, what was the molarity of the EDTA?

(iv) In human teeth, approximately 96% of the outer enamel and 70% of the inner dentine are composed of the apatite mineral, calcium hydroxy phosphate.

The simplest empirical formula is Ca₅(PO₄)₃OH. Calculate the % calcium in the apatite mineral.

(v) A dried 1.40g human tooth was dissolved in a small quantity of hot conc. nitric acid. A drop of methyl orange indicator was added followed by drops of 6M sodium hydroxide until the indicator turned orange to neutralise the solution. The solution was then made up to 250 cm³ in a volumetric flask. 10.0 cm³ of this solution was pipetted into a conical flask and 1 cm³ of a conc. ammonia/ammonium chloride pH10 buffer was added. The solution was titrated with 0.02 mol dm^-3 EDTA using Eriochrome Black T indicator. The indicator turned blue after 22.5 cm³ of EDTA was added. Calculate the average % by mass of calcium throughout the tooth.

Q10 25 cm³ of seawater was diluted to 250 cm³ in a graduated volumetric flask. A 25 cm³ aliquot of the diluted seawater was pipetted into a conical flask and a few drops of potassium chromate(VI) indicator solution was added.

 On titration with 0.1 mol dm^-3 silver nitrate solution, 13.8 cm³ was required to precipitate all the chloride ion. [Atomic masses: Na = 23, Cl = 35.5]

Q11 0.12 g of rock salt was dissolved in water and titrated with 0.1 mol dm^-3 silver nitrate until the first permanent brown precipitate of silver chromate is seen.

19.7 cm³ was required to titrate all the chloride ion. [Atomic masses: Na = 23, Cl = 35.5]

(a) How many moles of chloride ion was titrated?

(b) What mass of sodium chloride was titrated?

(b) What was the % purity of the rock salt in terms of sodium chloride?

Q12 5.0 g of a solid mixture of anhydrous calcium chloride(CaCl₂) and sodium nitrate (NaNO₃) was dissolved in 250 cm³ of deionised water in a graduated volumetric flask. A 25 cm³ aliquot of the solution was pipetted into a conical flask and a few drops of potassium chromate(VI) indicator solution was added.

On titration with 0.1 mol dm^-3 silver nitrate solution, 21.2 cm³ was required to precipitate all the chloride ion. [Atomic masses: Ca = 40, Cl = 35.5]

(a) Calculate the moles of chloride ion titrated.

(b) Calculate the equivalent moles of calcium chloride titrated.

(c) Calculate the equivalent mass of calcium chloride titrated.

(d) Calculate the total mass of calcium chloride in the original 5.0 g of the mixture.

(e) The % of calcium chloride and sodium nitrate in the original mixture.

Q13 A bulk solution of hydrochloric acid was standardised using pure anhydrous sodium carbonate (Na₂CO₃, a primary standard). 13.25 g of sodium carbonate was dissolved in about 150 cm³ of deionised water in a beaker. The solution was then transferred, with appropriate washings, into a graduated flask, and the volume of water made up to 250 cm³, and thoroughly shaken (with stopper on!) to ensure complete mixing.

25 cm³ of the sodium carbonate solution was pipetted into a conical flask and screened methyl orange indicator added. The aliquot required 24.65 cm³ of a hydrochloric acid solution, of unknown molarity, to completely neutralise it. [atomic masses: Na = 23, C = 12, O = 16]

(a) Calculate the molarity of the prepared sodium carbonate solution.

(b) Write out the equation between sodium carbonate and hydrochloric acid, including state symbols.

(c) How many moles of sodium carbonate were titrated?

(d) How many moles of hydrochloric acid were used in the titration?

(e) What is the molarity of the hydrochloric acid?

Q14 For this question relevant formula mass and equation are in the answers to Q13.

Q18 Using the method described in Q15(a), sodium hydroxide solution can be standardised. 0.25g of very pure benzoic acid (C₆H₅COOH) was titrated with a solution of sodium hydroxide of unknown molarity. If 22.50 cm³ of the alkali was required for neutralisation, calculate ...

EDTA is an acronym abbreviation for the old name EthyleneDiamineTetraAcetic acid and is used in equations.

It is a hexadentate ligand i.e. it can donate 6 electron pairs to form 6 dative-covalent bonds and binds strongly with many metal cations Mn+ where n is usually 2 or 3.

Multi-dentate ligands are called chelating* agents, because the two ligand bonds become part of a five membered ring system. (*The chelate is from the Greek word meaning a crab's claw)

The solid EDTA 'off the shelf' used in analysis is usually the disodium dihydrate salt, which has the structure

(Na^+-OOCCH₂)(HOOCCH₂)NCH₂CH₂N(CH₂COOH)(CH₂COO^-Na⁺).2H₂O (M_r = 372.2)

The full unionised structure is (HOOCCH₂)₂NCH₂CH₂N(CH₂COOH)₂ which we could abbreviate to H₄EDTA since theoretically four hydrogens from the four carboxylic acid groups are ionisable.

of which H₂EDTA^2- is the most prominent chelating species in solutions of pH10 in titrating calcium ions though the complex is actually formed by the combination of a metal ion and the EDTA^4- ion.

The theory behind the titration of calcium ions with EDTA reagent is a bit complicated and the titration should be carried out in the presence of magnesium ions, usually included in the EDTA volumetric reagent, but if not, they must be in the mixture being titrated. This may seem to prelude an incorrect titration for calcium since magnesium ions reacting with EDTA, but it doesn't (see explanation later).

Hydrated/hexa-aqua metal ions like aqueous calcium and magnesium ions (M²⁺) give the following reaction with EDTA reagent,

(eq 1) [M(H₂O)₆]²⁺_(aq) + H₂EDTA^2-_(aq) ==> [MEDTA]^2-_{(aq, colourless)} + 2H⁺_(aq) + 6H₂O_(l)

which, for theoretical explanation of the titration and subsequent calculations, is best simplified to

Both calcium and magnesium EDTA complexes are strongly formed i.e. virtually 100% to the right BUT the K_stab for the formation of the EDTA-calcium ion complex is greater than that for the EDTA-magnesium ion complex i.e. the calcium ion complex is more stable and calcium ions will displace magnesium ions from their EDTA complex.

The indicators used e.g. Eriochrome Black T (represented in a 'free' anionic form as HIn^2-) weakly complexes with ions such as the magnesium ion.

(eq 4) Mg²⁺_{(aq, colourless)} + HIn^2-_{(aq, blue)} [MgIn]^-_{(aq, red)} + H⁺_(aq) (M = Mg, Ca)

Both metal ions form a weak complex with the indicator at the start of the titration, but the indicator is displaced by the stronger binding EDTA (ligand displacement reactions), but much more slowly from the calcium complex than the magnesium complex i.e.

(eq 5) [MgIn]^-_{(aq, red)} + H₂EDTA^2-_(aq) = fast => [MgEDTA]^2-_{(aq, colourless)} + HIn^2-_{(aq, blue)} + H⁺_(aq)

(eq 6) [CaIn]^-_{(aq, red)} + H₂EDTA^2-_(aq) = slow => [CaEDTA]^2-_{(aq, colourless)} + HIn^2-_{(aq, blue)} + H⁺_(aq)

and this means without the presence of magnesium ions, the end-point is sluggish giving an inaccurate with just the calcium ions present, because (eq 6) is too slow.

Therefore the order of complex ion stability is [CaEDTA]^2-_(aq) > [MgEDTA]^2-_(aq) > [MgIn]^-_(aq) and this order of stability is crucial to the success of the titration as the ensuing argument will show.

In the EDTA solution are the Mg-EDTA complex ion plus excess uncomplexed EDTA ions. As the EDTA reagent is run into the calcium ion solution the calcium ion-EDTA complex is formed by reactions (eq 7) or (eq 8).

(eq 7) Ca²⁺_(aq) + H₂EDTA^2-_(aq) ==> [CaEDTA]^2-_(aq) + 2H⁺_(aq)

(eq 8) Ca²⁺_(aq) + [MgEDTA]^2-_(aq) ==> [CaEDTA]^2-_(aq) + Mg²⁺_(aq)

In (eq 8)  the magnesium ion is displaced from its EDTA complex on a 1 : 1 molar basis by the calcium ion and then the free magnesium ions form a red complex with the blue indicator (eq 4) below. This continues as long as there are still Ca²⁺ ions to titrate and magnesium ions to be displaced i.e. no blue colour is seen yet.

(eq 4) Mg²⁺_{(aq, colourless)} + HIn^2-_{(aq, blue)} [MgIn]^-_{(aq, red)} + 2H⁺_(aq)

Continued addition of EDTA eventually converts all the 'free' calcium ions into their EDTA complex ion via (eq 7 or 8), BUT, the 1st drop of excess EDTA after the calcium ions are all complexed then releases the blue form of the indicator via the fast reaction (eq 5),

(eq 5) [MgIn]^-_{(aq, red)} + H₂EDTA^2-_(aq) ==> [MgEDTA]^2-_{(aq, colourless)} + HIn^2-_{(aq, blue)} + H⁺_(aq)

so giving the sharp end point from red to blue at pH10.

In the case of analysing a mixture of calcium and magnesium ions in the same mixture, one method is to analyse for Ca²⁺ as above (V_Ca). Obviously, you do NOT add magnesium ions to the EDTA or the mixture being titrated if you wish to estimate the total (Mg²⁺ + Ca²⁺). You add a known excess of standardised EDTA solution (V_excess) and then back titrate with another standardized M²⁺ ion solution (V_back) and the end point is blue to red.

Therefore: (V_excess) - (V_back) = (V_Ca+Mg) and (V_Mg) = (V_Ca+Mg) - (V_Ca)

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