Redox Titrations
Redox Worksheet
Practice Questions
1.
A student carried out an experiment to find the mass of FeSO4.7H2O in a brand of iron deficiency tablets. The tablet was dissolved in 250cm3 of sulfuric acid. 25.0 cm3 of this solution reacted with 13.4 cm3 of a 0.0250 moldm–3 solution of K2Cr2O7. Calculate the mass of hydrated iron sulfate in the tablet.
6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
Step 1: Moles of Dichromate
0.0250 × (13.4/1000) = 0.000335 mol
Step 2: Moles of Iron (II) in 25cm3
Ratio Fe2+ : Cr2O72- is 6:1
0.000335 × 6 = 0.00201 mol
0.000335 × 6 = 0.00201 mol
Step 3: Moles of Iron (II) in 250cm3 (Tablet)
0.00201 × 10 = 0.0201 mol
Step 4: Mass of FeSO4.7H2O (Mr = 277.9)
0.0201 × 277.9 = 5.59 g (3 s.f.)
2.
An industrial bleach contains sodium chlorate (V) (NaClO3). To determine the concentration, a 25.0cm3 sample is diluted to 250cm3. A 10.0cm3 sample was added to 25.0cm3 of 0.500moldm-3 methanoic acid. The remaining acid required 22.0cm3 of 0.500moldm-3 NaOH to neutralise. What is the concentration of the bleach?
ClO3– + 3HCOOH → Cl– + 3CO2 + 3H2O
Step 1: Total Moles of Methanoic Acid Added
0.500 × (25.0/1000) = 0.0125 mol
Step 2: Moles of Excess Acid (Reacted with NaOH)
Ratio HCOOH : NaOH is 1:1
0.500 × (22.0/1000) = 0.0110 mol
0.500 × (22.0/1000) = 0.0110 mol
Step 3: Moles of Acid Reacted with Chlorate
0.0125 – 0.0110 = 0.0015 mol
Step 4: Moles of NaClO3 in 10cm3
Ratio NaClO3 : HCOOH is 1:3
0.0015 ÷ 3 = 0.0005 mol
0.0015 ÷ 3 = 0.0005 mol
Step 5: Concentration of Original Bleach
Moles in 250cm3 diluted = 0.0005 × 25 = 0.0125 mol
Conc = 0.0125 / (25.0/1000) = 0.500 mol dm-3
Conc = 0.0125 / (25.0/1000) = 0.500 mol dm-3
3.
A 6.00g tablet containing iron (II) ethanedioate (FeC2O4) was dissolved in acid and made up to 250cm3. A 25.0cm3 quotient was titrated against 0.100moldm-3 KMnO4, requiring 12.1cm3. What percentage of the tablet is iron (II) ethanedioate? (Both cation and anion are oxidised).
Total electrons lost per FeC2O4 = 3 (1 from Fe2+, 2 from C2O42-)
Ratio FeC2O4 : MnO4– = 5 : 3
Ratio FeC2O4 : MnO4– = 5 : 3
Step 1: Moles of Manganate(VII)
0.100 × (12.1/1000) = 0.00121 mol
Step 2: Moles of FeC2O4 in 25cm3
0.00121 × (5/3) = 0.002017 mol
Step 3: Mass in Tablet (250cm3)
Moles in total = 0.002017 × 10 = 0.02017 mol
Mr(FeC2O4) = 143.8
Mass = 0.02017 × 143.8 = 2.900 g
Mr(FeC2O4) = 143.8
Mass = 0.02017 × 143.8 = 2.900 g
Step 4: Percentage Mass
(2.900 / 6.00) × 100 = 48.3%
4.
A water company reacts 100cm3 of chlorinated water with excess KI. The resulting I2 is titrated against 0.200moldm-3 S2O32-, taking 12.3cm3. What is the concentration of chlorine in the water?
Ratio Cl2 : I2 is 1:1
Ratio I2 : S2O32- is 1:2
Therefore, Ratio Cl2 : S2O32- is 1:2
Ratio I2 : S2O32- is 1:2
Therefore, Ratio Cl2 : S2O32- is 1:2
Step 1: Moles of Thiosulfate
0.200 × (12.3/1000) = 0.00246 mol
Step 2: Moles of Chlorine
0.00246 ÷ 2 = 0.00123 mol
Step 3: Concentration of Chlorine
0.00123 / (100/1000) = 0.0123 mol dm-3
5.
A 3.80g sample of hydrated ethanedioic acid, H2C2O4.xH2O, was dissolved in 250cm3. A 25.0cm3 quotient required 24.2cm3 of 0.0500moldm-3 KMnO4. What is the value of X?
2MnO4– + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2 + 8H2O
Ratio MnO4– : Acid = 2:5
Ratio MnO4– : Acid = 2:5
Step 1: Moles of Manganate
0.0500 × (24.2/1000) = 0.00121 mol
Step 2: Moles of Acid in 25cm3
0.00121 × (5/2) = 0.003025 mol
Step 3: Moles in 250cm3 (Total Sample)
0.003025 × 10 = 0.03025 mol
Step 4: Molar Mass of Hydrated Acid
Mass / Moles = 3.80 / 0.03025 = 125.6 g mol-1
Step 5: Calculate X
Mr(H2C2O4) = 90.0
Mass of water = 125.6 – 90.0 = 35.6
X = 35.6 / 18.0 ≈ 2
Mass of water = 125.6 – 90.0 = 35.6
X = 35.6 / 18.0 ≈ 2