Advanced A/AS Level Quantitative Chemistry: Answers to non-redox volumetric titration questions part 2 Doc Brown's GCE AS A2 Level Chemistry

Advanced Level Chemistry Revision on Titrations

GCE A Level AS-A2 IB Acid-base and other non-redox volumetric titration quantitative calculation ANSWERS to PART 2 Questions 21 to onwards

PART 2 also includes some A Level gas volume and gravimetric questions as well as more acid-alkali and acid-carbonate titrations and standardising hydrochloric acid calculations. [SEARCH BOX]

If you find these useful or spot a silly error please EMAIL query?comment

ILLUSTRATIONS OF ACID-ALKALI TITRATIONS and SIMPLE STARTER CALCULATIONS

The non-redox titration question Answers

I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT! Most of the answers have been rounded up or rounded down to three significant figures (3sf)

Q21 ANSWERS

(a) mol HCl = 0.100 x 10.5/1000 = 0.00105

from the neutralisation equation Ca(OH)2(aq) + 2HCl(aq) ===> CaCl2(aq) + 2H2O(l)

2 mol HCl needed to neutralise 1 mol Ca(OH)2

therefore mol Ca(OH)2 = 0.00105/2 = 0.000525 (in 25 cm3 or 25/1000 = 0.025 dm3)

Concentration of calcium hydroxide = 0.000525/0.025 = 0.0210 mol dm-3

(b) (i) Mr[Ca(OH)2] = 74, mass = mol x formula mass

mass Ca(OH)2 in 1 dm3 = 0.021 x 74 = 1.554 g

Therefore concentration of Ca(OH)2 = 1.55 g dm-3

(ii) Since 100 cm3 is 1/10th of 1000 cm3 or 1 dm3

The concentration of Ca(OH)2 = 0.155 g/100 cm-3

Q22 ANSWERS

NaOH + HCl ===> NaCl + H2O

from equation mol NaOH = mol HCl = 0.100 x 20.55/1000 = 0.002055

25.0 cm3 = 25/1000 = 0.025 dm3

so, molarity NaOH = 0.002055/0.025 = 0.0822 mol dm-3

Q23 ANSWERS

mol NaOH in titration = 0.1025 x 17.65/1000 = 0.001809125

from the equation 2NaOH + H2SO4 ==> Na2SO4 + 2H2O

mol H2SO4 = mol NaOH/2

mol H2SO4 = 0.001809125/2 = 0.0009046 in 25.0/1000 = 0.0250 dm3

molarity H2SO4 = 0.0009046/0.025 = 0.0362 mol dm-3 (3sf)

Q24 ANSWERS

In the titration: mol NaOH = 0.1 x 22.5/1000 = 2.25 x 10-3

Each mol of citric acid requires 3 mol of NaOH for complete neutralisation,

therefore, mol citric acid in 25.0 cm3 of the diluted solution = 2.25 x 10-3/3 = 7.5 x 10-4 mol

Since only 1/10th of the diluted solution was used in the titration ...

... the total moles of citric acid in the original cordial is 10 x 7.5 x 10-4 = 7.5 x 10-3 mol

Mr(citric acid, C6H8O7) = 192, mass = mol x formula mass,

so mass of citric acid = 192 x 7.5 x 10-3 = 1.44 g in 25.0/1000 = 0.0250 dm3

So the concentration of citric acid in the original cordial is 1.44/0.025 = 57.6 g/dm3

Q25 ANSWERS

Mass hydrated salt = 4.28, mass anhydrous salt = 1.89, mass water driven off = 4.28 - 1.89 = 2.39

Mr(H2O) = 18, Mr(Na2SO4) = 142

mol H2O = 2.39/18 = 0.1328, mol Na2SO4 = 1.89/142 = 0.01331

molar ratio H2O/Na2SO4 = 0.1328/0.01331 = 9.98 i.e. ~10, therefore x = 10,

so, in this hydrated form of sodium sulphate, the formula is Na2SO4.10H2O

Although the ratio is not precisely 10.0 (actually 0.02 x 100/9.98, about 0.2% off!), its pretty close given that the masses were only quoted to three significant figures and would be highly unlikely to be any other integer!

Q26 ANSWERS

the equation will be M + 2HCl ==> MCl2 + H2, from equation molar ratio M : H2 of 1 : 1

therefore mol M = mol H2 = 75/24000 = 3.125 x 10-3

mol M = mass M/Ar(M), therefore Ar = mass M/mol M = 0.428/3.125 x 10-3 = 136.96

Ar is ~137 which corresponds to barium Ba

Q27 ANSWERS

Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2

Mr(Na2CO3) = 106, so mol Na2CO3 titrated = 0.132/106 = 0.001245

from the molar equation, mol HCl = 2 x mol Na2CO3 = 0.00249

molarity HCl = mol/volume in dm3 = 0.00249/(24.8/1000) = 0.0996 mol dm-3 (0.0996M)

Q28 ANSWERS

(a) Mr(Na2CO3) = 106, mol Na2CO3 = 1.30/106 = 0.01236 mol

volume = 250 cm3 = 25.0/1000 = 0.025 dm3

therefore molarity of Na2CO3 solution = 0.01236/0.025 = 0.04904 mol dm-3 (4sf)

(b) mol Na2CO3 in each titration = 0.04904 x 25.0/1000 = 0.001226

from the molar equation Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2

each mole of Na2CO3 needs two moles of HCl for complete neutralisation

therefore mol HCl = 0.001226 x 2 = 0.002452 in 24.35 cm3 (or 24.35/1000 = 0.02435 dm3)

therefore molarity HCl = 0.002452/0.02435 = 0.1007 mol dm-3 (0.101 M 3sf)

Q29 ANSWERS

From equation mol Na2CO3 = mol HCl/2, Mr(Na2CO3) = 106, mass = mol x Mr

Analysis (i)

(a) mol HCl = 1.00 x 20.95/1000 = 0.02095, (b) mol Na2CO3 = 0.02095/2 = 0.010475

(c) mass titrated based on HCl titre = 0.010475 x 106 = 1.11035g, (d) % purity = 100 x 1.11035/1.113 = 99.76%

Analysis (ii)

(a) mol HCl = 1.00 x 20.55/1000 = 0.02055, (b) mol Na2CO3 = 0.02055/2 = 0.010275

(c) mass based on HCl titre = 0.010275 x 106 = 1.089g

(d) % purity = 100 x 1.08915/1.092 = 99.74%

Analysis (iii)

(a) mol HCl = 1.00 x 21.90/1000 = 0.0219

(b) mol Na2CO3 = 0.0219/2 = 0.01095

(c) mass based on HCl titre = 0.01095 x 106 = 1.1607g

(d) % purity = 100 x 1.1607/1.166 = 99.54%

Most likely analysis result

All three values are reasonably close together, so the best estimate would be to average them

(99.76 + 99.73 + 99.55)/3 = 99.68%, 99.7% (1dp, 3sf) the analysis method is not really accurate enough for 4sf

In this method the main error is from the titration value, where an error of ≥0.05 cm3 is likely.

For a titration of 21 cm3, this equates to an error of 100 x 0.05/21 = 0.24%, so quoting beyond 3 significant figures or 1 decimal place is inappropriate i.e. the analytical result is best quoted as 99.7+/-0.2%

Q30 ANSWERS

Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2, 1 mol hydrochloric acid relates to 0.5 moles of sodium carbonate

mol HCl used = 0.100 x 24.65/1000 = 0.002465, mol Na2CO3 titrated = 0.002465/2 = 0.0012325

Mr(Na2CO3) = 106, so mass Na2CO3`titrated = 106 x 0.0012325 = 0.130645

Therefore mass of H2O in sample= 0.352 - 0.130645 = 0.221355 g

mole ratio Na2CO3 : H2O is therefore 0.0012325 : 0.221355/18, giving 0.0012325 : 0.0122975

diving through by 0.0012325 gives a ratio of 1 : 9.98 (only a 0.2% error if x = 10)

Therefore the value of x can be reliably deduced as 10, since it would be expected to be an integer.

i.e. the formula of hydrated sodium carbonate crystals ('washing soda') is Na2CO3.10H2O

Q31 ANSWERS

(a) mass CuSO4.xH2O = 3.33, mass  anhydrous CuSO4 = 2.13g

therefore mass of water driven off = 3.33 - 2.12 = 1.21 g

% water of crystallisation = 100 x 1.21/3.33 = 36.3% (63.7% is CuSO4)

(b) Mr(H2O) = 18,  Mr(CuSO4) = 159.5

using the % composition from (a) you can calculate a mole ration based on 100g of the hydrated salt.

mol CuSO4 = 63.7/159.5 = 0.399, mol H2O = 36.3/18 = 2.016

the mole ratio H2O/CuSO4  = 2.016/0.399 = 5.05

To within a 1% error x = 5, so the formula of hydrated copper(II) sulfate is CuSO4.5H2O

(Note: assuming 5 is the answer, the error would be 100 x  (5.05 - 5.0)/5 = 1.0%, which would be quite acceptable.

Q32 ANSWERS in preparation

(a)(i) On adding the acid to the alkaline carbonate mixture solution, what are the colour changes for the end-points of titration (1) and (2)?

In titration (1) phenolphthalein turns from pink to colourless (exactly when it becomes first colourless!)

In titration (2) methyl orange changes from yellow to the first orange colour.

Detailed theory on acid-base titrations, pH curves and use of indicators

(a)(ii) Give the equations for what happens in each titration.

In titration (1) only the Na2CO3 is half-neutralised to NaHCO3

(i) Na2CO3 + HCl ==> NaCl + NaHCO3

In titration (2) all of the carbonates are neutralised in two stages

(i) Na2CO3 + HCl ==> NaCl + NaHCO3

(ii) NaHCO3 + HCl ==> NaCl + H2O + CO2

(b)(i) Calculate the moles of Na2CO3 in the prepared solution and calculate its molarity.

From titration (1) the Na2CO3 is half-neutralised by 9.80 cm3 of HCl.

From the equation (i) 1 mole of HCl reacts with 1 mole of Na2CO3

mol = molarity x volume

Therefore mol HCl = mol Na2CO3 = 0.5000 x 9.80/1000 = 0.0049 mol

Since each aliquot titrated is 1/10th of the prepared solution, total mol Na2CO3 = 10 x 0.0049 = 0.049

250 cm3 = 0.25 dm3, so molarity Na2CO3 = 0.049/0.25 = 0.196 mol/dm3

(b)(ii) Calculate the mass of sodium carbonate in the solution and hence the percentage Na2CO3 in the original solid mixture.

mass Na2CO3 = Mr(Na2CO3) x mol = 106 x 0.049 = 5.194 g

% Na2CO3 = 5.194 x 100 / 7.357 = 70.6%

Extra calculations for further practice and will partly help you to solve Q33

(c)(i) In the titrations, what total volume of the HCl was used to neutralise the Na2CO3?

Since it took 9.80 cm3 HCl to effect Na2CO3 + HCl ==> NaCl + NaHCO3

on an equimolar basis, it takes another 9.8 cm3 HCl to effect NaHCO3 + HCl ==> NaCl + H2O + CO2

Therefore a total volume of 19.60 cm3 HCl would be required to totally neutralise the Na2CO3

(c)(ii) In titration (2) what volume of HCl was used to neutralise the NaHCO3 in the original mixture?

If 19.60 cm3 HCl was used to neutralise the Na2CO3 in the original mixture, then,

the volume of HCl needed to neutralise the original NaHCO3 must = 24.75 - 19.60 = 5.15 cm3 HCl

(c)(iii) Calculate the number of moles of NaHCO3 in the 25.0 cm3 aliquot at the start and the total moles in the prepared solution.

from ? one mole of HCl reacts with 1 mole NaHCO3, therefore

moles HCl = mol NaHCO3 = 0.5000 x 5.15 / 1000 = 0.002575

therefore total moles = 0.002575 x 10 = 0.02575 mol NaHCO3 in the prepared solution.

(c)(iv) Calculate the mass off NaHCO3 in the prepared solution, and hence its % in the original solid mixture.

mass = moles x formula mass, mass = 0.02575 x 84 = 2.163 g

% NaHCO3 in mixture = 2.163 x 100 / 7.357 = 29.4%

If your answer to (b)(ii) + (c)(iv) do not add up to 100%, you have made an error somewhere!

Q33 ANSWERS

(a) Explaining the titrations ...

There are three bases present, in order of strength: (i) OH- > (ii) CO32- > (iii) HCO3-, so this gives the order in which they will be neutralised, shown in equations (i) to (iii) below ...

Titration (1) using phenolphthalein indicator corresponds to titration of NaOH plus Na2CO3 (but only half neutralised to NaHCO3)

The sequence of reaction is as follows as the pH decreases on adding the acid  ...

(i) NaOH + HCl ==> NaCl + H2O (normally a sharp inflection at pH 7 in the pH curve, but presence of Na2CO3 blurs this and the pH is much higher)

actual ionic equation: OH-(aq) + H+(aq) ==> H2O(l)

(ii) Na2CO3 + HCl ==> NaCl + NaHCO3 (complete at ~pH 8-9, clear 2nd inflection in the pH curve and detectable with an appropriate indicator)

actual ionic equation: CO32-(aq) + H+(aq) ==> HCO3-(aq)

The pKind of phenolphthalein (9.3) can detect the end-point at the end of the reaction (ii).

In this titration you will NOT observe effervescence at all, except if you overshoot the end-point by some margin, the colour change is pink to colourless.

Titration (2) using methyl orange indicator corresponds to the titration of NaOH + all of Na2CO3

The sequence of reactions is as follows as the pH decreases on adding the acid ...

(i) NaOH + HCl ==> NaCl + H2O (normally a sharp inflection at pH 7 in the pH curve, but presence of Na2CO3 blurs this and the pH is much higher)

actual ionic equation: OH-(aq) + H+(aq) ==> H2O(l)

(ii) Na2CO3 + HCl ==> NaCl + NaHCO3 (complete at ~pH 8-9, clear 2nd inflection in the pH curve and detectable with an appropriate indicator)

actual ionic equation: CO32-(aq) + H+(aq) ==> HCO3-(aq)

(iii) NaHCO3 + HCl ==> NaCl + H2O + CO2 (complete at ~pH 4, clear 3rd inflection in the pH curve and detectable with an appropriate indicator, effectively you have a solution of 'carbonic acid' at the end of the titration)

actual ionic equation: HCO3-(aq) + H+(aq) ==> H2O(l) + CO2(aq/g)

Note that (ii) + (iii) add up to (iv) Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2

AND reaction (iii) is the difference between the two titrations.

The pKind of methyl orange is 3.7 and matches the pH of end-point of the reaction (iii) when everything is completely neutralised, colour change is yellow to first trace or orange.

In this titration you may observe effervescence in the final stages as carbon dioxide is formed.

Note: You can do this titration exercise as a double indicator single titration. The reason is quite simple.

When titration (1) is complete, the phenolphthalein has gone colourless {reactions (i) + (ii) completed}, so, at this point, you can add the methyl orange indicator and continue the titration to give the extra HCl needed to complete the neutralisation {to complete reaction (iii)}.

Detailed theory on acid-base titrations, pH curves and use of indicators

(b) Calculating the molarity of the sodium carbonate formed by solution X on absorbing CO2

The difference between the titrations is the reaction (iii) NaHCO3 + HCl ==> NaCl + H2O + CO2

From this we can calculate the concentration of NaHCO3, hence the original concentration of Na2CO3

titration (2) - (1) = 22.55  - 12.45 = 10.10 cm3 of 1.000 mol/dm3 HCl

therefore: mol HCl reacting with NaHCO3 =  1.000 x 10.10/1000 = 0.0101

From the titration equation: NaHCO3 + HCl ==> NaCl + H2O + CO2

mol NaHCO3 = mole HCl = 0.0101, {and 25 cm3 = 25/1000 = 0.025 dm3, also required in (c)}

molarity = mol / volume in dm3, {and 25 cm3 = 25/1000 = 0.025 dm3, also required in (c)}

molarity NaHCO3 formed = 0.0101/0.025 = 0.404 mol/dm3

but in reaction (ii) each mol of Na2CO3 gives 1 mol of NaHCO3

so, the original molarity of Na2CO3 formed = 0.404 mol/dm3

(c) Calculating the molarity of the unreacted sodium hydroxide

Since 10.10 cm3 of the HCl was used to effect ...

(iii) NaHCO3 + HCl ==> NaCl + H2O + CO2

Prior to this, on an equimolar basis, another 10.10 cm3 of the HCl would be used to effect ...

(ii) Na2CO3 + HCl ==> NaCl + NaHCO3

Therefore a total of 20.20 cm3 of the HCl was used to completely neutralise the Na2CO3

therefore 22.55 - 20.20 = 2.35 cm3 of the HCl was used to neutralise the residual NaOH

mol HCl used = 1.000 x 2.35/1000 = 0.00235

since titration equation is HCl + NaOH ==> NaCl + H2O,

mol HCl = mol NaOH, so mol NaOH = 0.00235

molarity = mol / volume in dm3, 25 cm3 = 0.025 dm3

molarity NaOH = 0.00235/0.025 = 0.094 mol/dm3

(d) Calculating the volume of carbon dioxide gas absorbed by the NaOH (molar gas volume = 24 dm3 at RTP)

from calculation (b) molarity of Na2CO3 = 0.404 mol/dm3

and from the equation for CO2 absorbed: 2NaOH + CO2 ==> Na2CO3 + H2O

mol Na2CO3 = mol CO2 , so in 2 dm3 of the solution there is the equivalent of 2 x 0.404 mol = 0.808 mol CO2

so volume CO2 absorbed = 0.808 x 24 = 19.392 = 19.4 dm3 (3sf, 1dp)

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I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT!

ILLUSTRATIONS OF ACID-ALKALI TITRATIONS and SIMPLE STARTER CALCULATIONS

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