Doc Brown's Chemistry Clinic

Advanced Level Chemistry Revision on Titrations

GCE-AS-A2-IB Acid-base and other non-redox volumetric titration calculation ANSWERS

ORIGINAL Q's for the non-redox titration Q answers below * Redox titration Q's

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The non-redox titration question Answers

(b) (i) pipette (ii) burette

(c) everything with distilled water, then pipette with a little of the NaOH(aq) and the burette with a little of the HCl(aq)

(d) pink to colourless, the first drop of excess acid removes the pink alkaline colour of phenolphthalein

(e) moles sodium hydroxide neutralised: 0.25 x 25/1000 = 0.00625 mol NaOH

(remember: moles = molarity x volume in dm³ and its two rearrangements and 1 dm³ = 1000 cm³)

(f) moles HCl = moles NaOH (equation) = 0.00625 mol HCl (in 22.5 cm³)

(g) concentration hydrochloric acid = 0.0062 x 1000 ÷ 22.5 = 0.278 mol dm^-3

(scaling up to 1 dm³ = 1000 cm³ to get the molarity)

(b) moles H₂SO₄ = 4.90 ÷ 98 = 0.05 mol in 200cm³

scaling up to get molarity of the sulphuric acid solution, 0.05 x 1000 ÷ 200 = 0.25 mol dm^-3

(c) moles of sulphuric acid neutralised = 0.25 x 20.7/1000 = 0.005175 mol

(d) moles of sodium hydroxide neutralised = 2 x 0.005175 = 0.01035 mol (2 : 1 in equation)

(e) concentration of the sodium hydroxide = 0.01035 x 1000 ÷ 10 = 1.035 mol dm^-3 (molarity 1.04, 3sf)

(b) moles of sulphuric acid neutralised = 0.1 x 4.5/1000 = 0.00045 mol

(c) moles of magnesium hydroxide neutralised also = 0.00045 (1:1 in equation) in 100 cm³

(d) concentration of the magnesium hydroxide in mol dm^-3 = 0.00045 x 1000 ÷ 100 = 0.0045

(scaling up to 1000cm³=1dm³,  to get molarity)

(e) molar mass of Mg(OH)₂ = 58.3

so concentration of the magnesium hydroxide = 0.0045 x 58.3 = 0.26 g dm^-3 (= g per 1000 cm³),

so concentration = 0.26 ÷ 1000 = 0.00026 g cm^-3

(a)(ii) NaOH_(aq) + HCl_(aq) ==> NaCl_(aq) + H₂O_(l)

(b) moles of hydrochloric acid added to the magnesium oxide = 2 x 100/1000 = 0.20 mol HCl

(c) moles of excess hydrochloric acid titrated = 19.7 ÷ 1000 x 0.2 = 0.00394 mol HCl

{mole ratio NaOH:HCl is 1:1 from equation (ii)}

(d) moles of hydrochloric acid reacting with the magnesium oxide = 0.20 - 0.00394 = 0.196 mol HCl

(e) mole MgO reacted = 0.196 ÷ 2 = 0.098 {1: 2 in equation (i)}

the formula mass of MgO = 40.3

therefore mass of MgO reacting with acid = 0.098 x 40.3 = 3.95 g

(f) % purity = 3.95 ÷ 4.06 x 100 = 97.3% MgO

(g) Mg(OH)₂ from MgO + H₂O, MgCO₃ from the original mineral source, both of these compounds react with acid and would lead to a false titration value.

(b) moles of hydrochloric acid was spilt = 2 x 10 = 20 mol HCl

(c) moles of calcium carbonate to neutralise the acid = 20 ÷ 2 = 10 mol CaCO₃ (1:2 in equation)

(d) formula mass of CaCO₃ = 100,

so mass of limestone powder needed to neutralise the acid = 100 x 10 = 1000g CaCO₃

(e) the neutralisation reaction is MgO + H₂SO₄ ==> MgSO₄ + H₂O,

moles H₂SO₄ = 1000 x 2 = 2000 mol acid, 2000 mol MgO needed (1:1 in equation),

mass MgO needed = 2000 x 40.3 = 80600 g or 80.6 kg

(b) moles of sodium hydroxide used in the titration = 25 x 1/1000 = 0.025 mol NaOH

(c) mol H₂SO₄ = mol NaOH ÷ 2 = 0.0125 mol in 20 cm³,

so scaling up to 1000 cm³ to get molarity of diluted acid = 0.0125 x 1000 ÷ 20 = 0.625 mol dm^-3

(or molarity = 0.0125 mol/0.02 dm³ = 0.625 mol dm^-3)

(d) scaling up from 50 to 1000 cm³, gives the concentration of the original concentrated sulphuric acid solution,

 = 0.625 x 1000 ÷ 50 = 12.5 mol dm^-3

(b) mol = molarity x volume in dm³, mol acid = 0.2 x 23.75/1000 = 4.75 x 10^-3 mol HCl

from equation HCl:NaHCO₃ is 1:1 by ratio

so mol HCl = mol NaHCO₃ = 4.75 x 10^-3

(c) mass = mol x formula mass, f. mass NaHCO₃ = 23 + 1 + 12 + (3 x 16) = 84

mass NaHCO₃ = 4.75 x 10^-3 x 84 =  0.399 g

% purity of NaHCO₃ = 0.399 x 100/0.40 =  99.75% (99.8%, 3sf)

9a(ii) 20 cm³ of 1.0 mol dm^-3 hydrochloric acid contains 1.0 x 20/1000 = 0.02 mol HCl

From the equation, 0.02 mol HCl reacts with 0.01 mol Na₂CO₃, M_r(Na₂CO₃) = 106

therefore mass Na₂CO₃ titrated = 0.01 x 106 = 1.06 g per aliquot,

since 250 cm³ is ¹/₁₀th of the aliquot, 10 x 1.06 = 10.6 g of Na₂CO₃ would be used to make up the solution.

Molarity of Na₂CO_3(aq) = (10.6 g/106 g mol^-1)/0.25 dm³ = 0.40 mol dm^-3

9(b)(i) CH₃COOC₆H₄COOH + NaOH ==> CH₃COOC₆H₄COO^-Na⁺ + H₂O

9b(ii) 23.0 cm³ of 0.1 mol dm^-3 NaOH contains 0.1 x 23/1000 = 0.0023 mol NaOH

From the equation, mol Aspirin = mol NaOH, M_r(CH₃COOC₆H₄COOH) = 180

so need Aspirin mass of 0.0023 x 180 = 0.414 g

9b(iii) The last stage in the synthesis of 2-ethanoylhydroxybenzoic acid ('Aspirin') is made by esterifying 2-hydroxybenzoic acid with ethanoic anhydride.

Mr(HOC₆H₄COOH) = 138, is 42 units less than aspirin. In terms of this particular impurity the % aspirin will be overestimated for the following reason. The 2-hydroxybenzoic acid will also be titrated with the aspirin, and, with its smaller molecular mass, it will need more alkali to neutralise than aspirin per equivalent mass of material. This can result in >100% purity!!!!

9(c)(i) It is a ligand substitution/replacement reaction.

[Ca(H₂O)₆]²⁺_(aq) + EDTA^4-_(aq) ==> [CaEDTA]^2-_(aq) + 6H₂O_(l)

more simply Ca²⁺_(aq) + EDTA^4-_(aq) ==> [CaEDTA]^2-_(aq)

or [Ca(H₂O)₆]²⁺_(aq) + H₂EDTA^2-_(aq) ==> [CaEDTA]^2-_(aq) + 2H⁺_(aq) + 6H₂O_(l)

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

9(c)(ii) M_r(CaCO₃) = 100, mol CaCO₃ = mol Ca²⁺ in solution = 0.25/100 = 0.0025 mol

since 250 cm³ = 0.25 dm³, molarity Ca²⁺ = 0.0025/0.25 = 0.01 mol dm^-3

9(c)(iii) mole CaCO₃ = mol Ca²⁺ = mol EDTA used in titration.

Therefore from c(ii) mol Ca²⁺ = mol EDTA = 0.01 x 25/1000 = 0.00025 mol in 25.70 cm³ (0.0257 dm³) EDTA solution,

so molarity EDTA = 0.00025/0.0257 = 0.00973 mol dm^-3 (3 sf)

9(c)(iv) M_r(apatite) = (5 x 40) + 3 x (31 + 4 x 16) + (16 + 1) = 502

% Ca in apatite = 200 x 100/502 = 39.8%

9(c)(v) In the titration mol Ca²⁺ = mol EDTA,

therefore mol Ca²⁺ = 22.5 x 0.02/1000 = 0.00045,

since ¹⁰/₂₅₀ of the original solution was used in the titration,

the total mol of calcium in the tooth solution = 0.00045 x 250/10 = 0.01125 mol Ca

total mass of Ca in tooth = 0.01125 x 40 =  0.45 g

% by mass Ca in the tooth = 0.45 x 100/1.40 = 32.1 %

Q10 (a) Ag⁺_(aq) + Cl^-_(aq) ==> AgCl_(s) (sodium ions and nitrate ions etc. are spectator ions)

(b) from equation: moles silver nitrate (AgNO₃) = moles chloride ion (Cl^-)

moles = molarity AgNO₃ x volume of AgNO₃ in dm³

= 0.1 x 13.8/1000 = 1.38 x 10^-3 mol Cl^- (in 25 cm³ aliquot)

(c) moles in 1 dm³ of diluted seawater = 1.38 x 10^-3 x 1000/25 = 0.0552 (scaling up to 1000 cm³)

So molarity of chloride in diluted seawater is 0.0552 mol dm^-3

(d) Now in the titration 25 cm3 of the 250 cm³ was used,

so the molarity of chloride ion in the original seawater must be scaled up accordingly.

molarity of chloride ion in seawater = 0.0552 x 250/25 = 0.552 mol dm^-3

(e) M_r(NaCl) = 23 + 35.5 = 58.5

concentration of NaCl in g dm^-3 = molarity x formula mass = 0.552 x 58.5 =  32.3 g dm^-3

[AgNO₃:NaCl or Ag⁺:Cl^- is 1:1, see Q10(a)/(b)] f. mass NaCl = 23 + 35.5 = 58.5

(b) mass = mol x formula mass = 1.97 x 10^-3 x 58.5 = 0.1152 g NaCl

(c) % purity = 0.1152 x 100/0.12 = 96.0 % in terms of NaCl

Q12 (a) mole Cl^- = moles Ag⁺ [=AgNO₃, see Q10(a)/(b)]

mole Cl^- = molarity AgNO₃ x vol AgNO₃ = 0.1 x 21.2/1000 = 2.12 x 10^-3 mol Cl^-

(b) Since calcium chloride is CaCl₂, mol CaCl₂ = mole Cl^-/2  = 2.12 x 10^-3/2 = 1.06 x 10^-3 mol CaCl₂

(c) M_r(CaCl₂) = 40 + (2 x 35.5) = 111

mass = mol x f. mass = 1.06 x 10^-3 x 111 = 0.1177 g CaCl₂

(d) Since ¹/₁₀th of original solution titrated, original mass of CaCl₂ in mixture = 10 x 0.1177 g = 1.177g CaCl₂

(e) Therefore % = 1.177 x 100/5.0 =  23.5% CaCl₂(3 sf)

and % NaNO₃ = 100 - 23.5  = 76.5% (3 sf)

molarity = mol/vol. in dm³, 250 cm³ = 0.25 dm³,

molarity Na₂CO₃ = 0.125/0.25 = 0.50 mol dm^-3

(b) Na₂CO_3(aq) + 2HCl_(aq) ==> 2NaCl_(aq) + H₂O_(l) + CO_2(g)

(c) mol = molarity x volume

mol Na₂CO₃ titrated = 0.5 x 25/1000 = 0.0125 mol Na₂CO₃ (in the 25 cm³ aliquot pipetted)

(d) from equation, mole ratio Na₂CO₃:HCl is 1:2,

so mol HCl = 2 x mol Na₂CO₃  = 2 x 0.0125 = 0.025 mol HCl (in the 24.65 cm³ titre)

(e) molarity = mol/vol. in dm³, dm³ = cm³/1000, 24.65/1000 = 0.02465 dm³

therefore molarity HCl = 0.025/0.02465 = 1.014 mol dm^-3 (1.01 3sf)

(b) reacting mole ratio, Na₂CO₃:HCl is 1:2

so mol Na₂CO₃ titrated = mol HCl/2 = 0.0253/2 = 0.01265 mol

(c) mass = mol x f. mass, so mass Na₂CO₃ = 0.01275 x 106 = 1.34 g

therefore % purity of Na₂CO₃ =  1.34 x 100/1.35 =  99.3% (3sf)

Q15 (a) An appropriate quantity of the acid is weighed out, preferably on a 4 sf electronic balance. It can be weighed into a conical flask by difference i.e. weight acid added to flask = (weight of boat + acid) - (weight of boat).  The acid is dissolved in water, or aqueous-ethanol if not very soluble in water. The solution is titrated with standard sodium hydroxide solution using phenolphthalein indicator until the first permanent pink.

The burette should be rinsed with the sodium hydroxide solution first. The flask should be rinsed around the inside to ensure all the acid and alkali react, and drop-wise addition close to the end-point to get it to the nearest drop.

The pK_ind for phenolphthalein is 9.3, and its effective pH range is 8.3 to 10.0. The pH of a solution of the sodium salt of the acid (from strong base + weak acid) is in this region and so the equivalence point can be detected with this indicator. (page on acid-base equilibria and theory of indicators is in production)

(b) moles = molarity x volume in dm³ (dm³ = cm³/1000)

mol NaOH = 0.10 x 20.5/1000 = 0.00205 mol

(c) mol NaOH = mol RCOOH = 0.00205

because 1:1 mole ratio for a monobasic acid: RCOOH + Na⁺OH^- ==> RCOO^-Na⁺ + H₂O

(d) moles = mass (g)/M_r, so M_r = mass/mol = 0.279/0.00205 = 136.1

(e) The simplest aromatic carboxylic acid is benzoic acid C₆H₅COOH, M_r = 122

136-122 = 14, which suggests an 'extra' CH₂ (i.e. -CH₃ attached to the benzene ring instead of a H), so, since the COOH is attached to the ring, there are three possible positional/chain isomers of CH₃C₆H₄COOH (M_r = 136)

2-, 3- or 4-methylbenzoic acid.

Q16 (a) moles = molarity x volume in dm³ (dm³ = cm³/1000)

mol NaOH = 0.0995 x 19.85/1000 = 0.001975 mol

(b) The titration reaction for complete neutralisation is:

R(COOH)₂ + 2Na⁺OH^- ==> R(COO^-Na⁺)₂ + 2H₂O

this 1:2 reaction mole ratio means that mol dibasic acid = mol NaOH/2 = 0.001975/2 = 0.0009875

(c) moles = mass (g)/M_r, so M_r = mass/mol = 0.103/0.0009875 = 104.3 (approx. 104)

(d) Since a dibasic acid, and 2 x COOH = 2 x 45 = 90 mass units, the remaining 14 units could be CH₂,

and so the structure is likely to be HOOC-CH₂-COOH,

propanedioic acid (malonic acid), M_r = 144

Q17 (a) mol NaOH = 0.1 x 19.25/1000 = 0.001925

(b) mol NaOH = mol acid = 0.001925, mol ratio 1:1, C₆H₅COOH + Na⁺OH^- ==> C₆H₅COO^-Na⁺ + H₂O

M_r (C₆H₅COOH) =122, mol = mass (g)/M_r or mass = mol x M_r

so mass acid = 0.001925 x 122 = 0.2349 g

(c) % purity = actual mass of acid titrated x 100 / mass of original sample

% purity = 0.2349 x 100/0.236 = 99.5% (3 sf)

Q18 (a) M_r (C₆H₅COOH) =122, mol acid = mass (g)/M_r = 0.25/122 = 0.002049 mol

(b) mol alkali = 0.002049 mol, since mol acid, 1:1 mole ratio in reaction (see Q17(b)).

(c) Since 0.002049 mol NaOH in 22.5 cm³, so scaling up

molarity NaOH = 0.002049 x 1000/22.5 =  0.0911 mol dm^-3 (3 sf)

Q19 (a) The ratio of 2M/0.1M is 20, so need to do 20 fold dilution.

So 25 cm³ of 2M diluted to 500 cm³ gives an approximately 0.1 mol dm^-3 solution. You can do this with a measuring cylinder and beaker.

In practice you could make up 12.5 cm³ of the 2M acid up to 250 cm³. You could do this with a burette and a 250 cm³ standard volumetric flask, but standardisation of the acid is still required.

(b) The solubility of calcium hydroxide is low, so it would give a very inaccurate tiny titration value with relatively concentrated acid. For any accurate work e.g. to 3sf, standardisation of reagents is required.

(c) 1 x 50 cm³ pipette or 2 x 25 cm³ pipette would be the most convenient. Phenolphthalein is used for strong base-strong acid titrations.

(d) Ca(OH)_2(aq) + 2HCl_(aq) ==> CaCl_2(aq) + 2H₂O_(l)

(e) from equation, mole ratio Ca(OH)₂ : HCl is 1:2

and since moles solute = molarity x volume in dm³

mol HCl used in titration = 0.1005 x 15.22/1000 = 0.001530 mol HCl

therefore mol Ca(OH)₂ = 0.001558/2 = 0.000765 mol Ca(OH)₂

(f) Scaling up from mol Ca(OH)₂ in 50 cm³ to 1 dm³ (1000 cm³)

molarity Ca(OH)₂ = 0.00765 x 1000/50 = 0.1558 = 0.153 mol dm^-3 (to 3sf)

(g) Mr[Ca(OH)₂] =74, so solubility in g dm^-3 = 0.153 x 74 = 1.13 g dm^-3 (3 sf)

Since density of water is 1.0 g cm^-3, the solubility of Ca(OH)₂ is about 0.113 g/100 cm³ H₂O

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