Doc Brown's GCE AS A2 A Level Chemistry - Advanced Level Chemistry Revision on Volumetric 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.

All my advanced A level organic chemistry notes

PART 1 Questions * PART 1 Question Answers

PART 2 Questions * Redox Titration Q's * Qualitative Analysis

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

ILLUSTRATIONS OF ACID-ALKALI TITRATIONS and SIMPLE STARTER CALCULATIONS

(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

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

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)

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

M_r(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

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

M_r(H₂O) = 18, M_r(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!

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

A_r is ~137 which corresponds to barium Ba

Na₂CO₃ + 2HCl ==> 2NaCl + H₂O + CO₂

M_r(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)

(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)

 

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

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%

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

M_r(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

(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) M_r(H₂O) = 18,  M_r(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.

(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 = M_r(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) CO₃^2- > (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: CO₃^2-_(aq) + H⁺_(aq) ==> HCO₃^-_(aq)

The pK_ind 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: CO₃^2-_(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) + CO_2(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 pK_ind 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)

[SEARCH BOX]

Website content © Dr Phil Brown 2000+. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial.