Back Titration Analysis
Complex Titration Worksheet
Challenge Questions
1. A sample of limestone was analysed to determine the percentage of calcium carbonate in the rock. 4.00g of the limestone was reacted against 200cm3 of 0.1moldm-3 H2SO4. The resulting mixture was topped up to 250cm3 in a volumetric flask. 25.0cm3 samples were taken and titrated against 0.01moldm-3 NaOH, where it took an average of 23.4cm3 to neutralise. What is the percentage of CaCO3 in the limestone rock?
Working Out
Step 1: Calculate Excess Acid in 25cm3 Sample
Titration Reaction: H2SO4 + 2NaOH → Na2SO4 + 2H2O
n(NaOH) = 0.01 × 0.0234 = 0.000234 mol
n(H2SO4 sample) = 0.000234 / 2 = 0.000117 mol
n(NaOH) = 0.01 × 0.0234 = 0.000234 mol
n(H2SO4 sample) = 0.000234 / 2 = 0.000117 mol
Step 2: Calculate Total Excess Acid in 250cm3
Scale Factor = 250 / 25 = 10
n(Total Excess) = 0.000117 × 10 = 0.00117 mol
n(Total Excess) = 0.000117 × 10 = 0.00117 mol
Step 3: Calculate Reacted Acid
n(Initial H2SO4) = 0.1 × 0.200 = 0.020 mol
n(Reacted) = Initial – Excess = 0.020 – 0.00117 = 0.01883 mol
n(Reacted) = Initial – Excess = 0.020 – 0.00117 = 0.01883 mol
Step 4: Calculate Mass and Percentage
Reaction: CaCO3 + H2SO4 → CaSO4 + H2O + CO2 (Ratio 1:1)
n(CaCO3) = 0.01883 mol
Mass = 0.01883 × 100.1 = 1.885 g
% = (1.885 / 4.00) × 100 = 47.12%
Answer = 47.1%
n(CaCO3) = 0.01883 mol
Mass = 0.01883 × 100.1 = 1.885 g
% = (1.885 / 4.00) × 100 = 47.12%
Answer = 47.1%
2. An indigestion tablet containing magnesium hydroxide was analysed to determine the percentage mass of magnesium hydroxide in the tablet. A 4.00g tablet was crushed and reacted with 100cm3 of 1.5moldm-3 hydrochloric acid. 15cm3 samples were taken from the excess hydrochloric acid and were titrated against 0.2moldm-3 sodium hydroxide. An average of 12.6cm3 of sodium hydroxide was required to neutralise the excess hydrochloric acid. What percentage of the tablet was magnesium hydroxide?
Working Out
Step 1: Calculate Excess Acid in 15cm3 Sample
Reaction: HCl + NaOH → NaCl + H2O
n(NaOH) = 0.2 × 0.0126 = 0.00252 mol
n(HCl sample) = 0.00252 mol
n(NaOH) = 0.2 × 0.0126 = 0.00252 mol
n(HCl sample) = 0.00252 mol
Step 2: Calculate Total Excess Acid in 100cm3
Scale Factor = 100 / 15 = 6.667
n(Total Excess) = 0.00252 × (100/15) = 0.0168 mol
n(Total Excess) = 0.00252 × (100/15) = 0.0168 mol
Step 3: Calculate Reacted Acid
n(Initial HCl) = 1.5 × 0.100 = 0.15 mol
n(Reacted) = 0.15 – 0.0168 = 0.1332 mol
n(Reacted) = 0.15 – 0.0168 = 0.1332 mol
Step 4: Calculate Mass and Percentage
Reaction: Mg(OH)2 + 2HCl → MgCl2 + 2H2O (Ratio 1:2)
n(Mg(OH)2) = 0.1332 / 2 = 0.0666 mol
Mr(Mg(OH)2) = 58.3
Mass = 0.0666 × 58.3 = 3.883 g
% = (3.883 / 4.00) × 100 = 97.07%
Answer = 97.1%
n(Mg(OH)2) = 0.1332 / 2 = 0.0666 mol
Mr(Mg(OH)2) = 58.3
Mass = 0.0666 × 58.3 = 3.883 g
% = (3.883 / 4.00) × 100 = 97.07%
Answer = 97.1%
3. A 6.00g sample of solid hydrated ethanedioic acid was analysed to determine the number of moles of water in the solid crystal. The solid ethanedioic acid was reacted with 250cm3 0.5moldm-3 potassium hydroxide. The excess potassium hydroxide was reacted against hydrochloric acid in 25cm3 samples. It took an average of 29.7cm3 of 0.1moldm-3 hydrochloric acid to react. What was the formula of the hydrated ethanedioic acid?
Working Out
Step 1: Calculate Excess KOH in Sample
n(HCl) = 0.1 × 0.0297 = 0.00297 mol
n(KOH sample) = 0.00297 mol (1:1 ratio)
n(KOH sample) = 0.00297 mol (1:1 ratio)
Step 2: Calculate Total Excess KOH
Scale Factor = 250 / 25 = 10
n(Total Excess) = 0.00297 × 10 = 0.0297 mol
n(Total Excess) = 0.00297 × 10 = 0.0297 mol
Step 3: Calculate Reacted KOH
n(Initial KOH) = 0.5 × 0.250 = 0.125 mol
n(Reacted) = 0.125 – 0.0297 = 0.0953 mol
n(Reacted) = 0.125 – 0.0297 = 0.0953 mol
Step 4: Moles of Acid
Reaction: H2C2O4 + 2KOH → K2C2O4 + 2H2O (Ratio 1:2)
n(Acid) = 0.0953 / 2 = 0.04765 mol
n(Acid) = 0.0953 / 2 = 0.04765 mol
Step 5: Formula Mass & Water
Mr(Hydrated) = 6.00 / 0.04765 = 125.9 g mol-1
Mr(Anhydrous H2C2O4) = 90.0
Mass of Water = 125.9 – 90.0 = 35.9
Moles of Water = 35.9 / 18 = 1.99 ≈ 2
Formula = H2C2O4 · 2H2O
Mr(Anhydrous H2C2O4) = 90.0
Mass of Water = 125.9 – 90.0 = 35.9
Moles of Water = 35.9 / 18 = 1.99 ≈ 2
Formula = H2C2O4 · 2H2O
4. Fertiliser was analysed to determine the amount of magnesium nitrate in it. 2.0g of fertiliser was warmed with a 100cm3 solution of 0.5moldm-3 sodium hydroxide. The solid magnesium hydroxide formed in the reaction was filtered off. Then, the excess sodium hydroxide was made up to 500cm3 and 25cm3 samples were taken to be titrated against 0.1moldm-3 nitric acid. It took an average of 12.4cm3 of nitric acid. What percentage of the fertiliser is magnesium nitrate?
Working Out
Step 1: Calculate Excess NaOH in Sample
n(HNO3) = 0.1 × 0.0124 = 0.00124 mol
n(NaOH sample) = 0.00124 mol
n(NaOH sample) = 0.00124 mol
Step 2: Calculate Total Excess NaOH (500cm3)
Scale Factor = 500 / 25 = 20
n(Total Excess) = 0.00124 × 20 = 0.0248 mol
n(Total Excess) = 0.00124 × 20 = 0.0248 mol
Step 3: Calculate Reacted NaOH
n(Initial NaOH) = 0.5 × 0.100 = 0.050 mol
n(Reacted) = 0.050 – 0.0248 = 0.0252 mol
n(Reacted) = 0.050 – 0.0248 = 0.0252 mol
Step 4: Percentage Calculation
Reaction: Mg(NO3)2 + 2NaOH → Mg(OH)2 + 2NaNO3 (Ratio 1:2)
n(Mg(NO3)2) = 0.0252 / 2 = 0.0126 mol
Mr = 148.3
Mass = 0.0126 × 148.3 = 1.869 g
% = (1.869 / 2.0) × 100 = 93.4%
Answer = 93.4%
n(Mg(NO3)2) = 0.0252 / 2 = 0.0126 mol
Mr = 148.3
Mass = 0.0126 × 148.3 = 1.869 g
% = (1.869 / 2.0) × 100 = 93.4%
Answer = 93.4%
5. Phosphorous pentachloride is commonly used in the production of lithium-ion batteries. A sample of phosphorous pentachloride was thought to be contaminated. To determine whether this was the case, a 15.0g sample was reacted with water to form phosphoric acid and hydrochloric acid. The mixture of acids was added to 300cm3 of 2.4 moldm‑3 sodium hydroxide. The excess NaOH was then topped up to 500cm3 and 25cm3 samples were taken to be titrated. The samples were titrated against 0.5moldm-3 HCl where they were found to require 15.1cm3 to neutralise. Was the sample of phosphorous pentachloride contaminated?
Working Out
Step 1: Reaction Stoichiometry
PCl5 + 4H2O → H3PO4 + 5HCl
H3PO4 reacts with 3 NaOH.
5 HCl reacts with 5 NaOH.
Total Ratio: 1 mol PCl5 : 8 mol NaOH
H3PO4 reacts with 3 NaOH.
5 HCl reacts with 5 NaOH.
Total Ratio: 1 mol PCl5 : 8 mol NaOH
Step 2: Back Titration (Excess NaOH)
n(HCl titrant) = 0.5 × 0.0151 = 0.00755 mol
n(NaOH excess in sample) = 0.00755 mol
n(Total Excess in 500cm3) = 0.00755 × 20 = 0.151 mol
n(NaOH excess in sample) = 0.00755 mol
n(Total Excess in 500cm3) = 0.00755 × 20 = 0.151 mol
Step 3: Reacted NaOH
n(Initial) = 2.4 × 0.300 = 0.720 mol
n(Reacted) = 0.720 – 0.151 = 0.569 mol
n(Reacted) = 0.720 – 0.151 = 0.569 mol
Step 4: Purity
n(PCl5) = 0.569 / 8 = 0.0711 mol
Mass = 0.0711 × 208.5 = 14.83 g
Purity = (14.83 / 15.0) × 100 = 98.9%
Answer: Yes, slightly contaminated (98.9% pure)
Mass = 0.0711 × 208.5 = 14.83 g
Purity = (14.83 / 15.0) × 100 = 98.9%
Answer: Yes, slightly contaminated (98.9% pure)
6. A group 2 carbonate was left in a sample bottle without a label. A 10g sample of the metal carbonate was mixed with 250cm2 of 1.5moldm-3 HCl. The excess HCl had its volume increased to 500cm3 with distilled water and then portioned into 25cm3 portions and they were titrated against 0.2moldm-3 NaOH, where it took 21.0cm3 of NaOH to reach neutralisation. Identify the metal in the compound.
Working Out
Step 1: Calculate Excess Acid
n(NaOH) = 0.2 × 0.021 = 0.0042 mol
n(Excess HCl in sample) = 0.0042 mol
n(Total Excess in 500cm3) = 0.0042 × 20 = 0.084 mol
n(Excess HCl in sample) = 0.0042 mol
n(Total Excess in 500cm3) = 0.0042 × 20 = 0.084 mol
Step 2: Calculate Reacted Acid
n(Initial HCl) = 1.5 × 0.250 = 0.375 mol
n(Reacted) = 0.375 – 0.084 = 0.291 mol
n(Reacted) = 0.375 – 0.084 = 0.291 mol
Step 3: Identify Metal
Ratio MCO3 : HCl is 1:2
n(MCO3) = 0.291 / 2 = 0.1455 mol
Mr(MCO3) = 10 / 0.1455 = 68.7
Ar(M) = 68.7 – 60.0 = 8.7
Closest Group 2 metal is Beryllium (9.0).
Identity: Beryllium Carbonate (BeCO3)
n(MCO3) = 0.291 / 2 = 0.1455 mol
Mr(MCO3) = 10 / 0.1455 = 68.7
Ar(M) = 68.7 – 60.0 = 8.7
Closest Group 2 metal is Beryllium (9.0).
Identity: Beryllium Carbonate (BeCO3)