Frequently Asked Question

The provided information describes a titration experiment. Here's an analysis and answer to the questions:
b. Indicator:
The appropriate indicator for this titration would be potassium permanganate (KMnO4) itself. KMnO4 acts as its own indicator due to its intense purple color. The endpoint of the titration is reached when the solution turns a faint pink color, indicating that the permanganate ions are in slight excess.
c. Ionic Equation:
The reaction involves the oxidation of chloride ions (Cl-) by permanganate ions (MnO4-) in an acidic medium. The balanced ionic equation is:
2MnO4- (aq) + 16H+ (aq) + 10Cl- (aq) → 2Mn2+ (aq) + 8H2O (l) + 5Cl2 (g)
d. Calculations:
i. Concentration of B (KCl) in mol/dm³:
This requires the volume of A (KMnO4) used to reach the endpoint. You'll need to provide the volume of KMnO4 used in the titration for the calculation. Let's assume it is 15.0 cm³:


Step 1: Calculate the moles of KMnO4 used:
Moles of KMnO4 = Concentration × Volume = 0.02 mol/dm³ × (15.0 cm³/1000 cm³/dm³) = 3.0 × 10⁻⁴ mol

Step 2: Use the stoichiometry of the balanced equation to determine moles of KCl:
From the equation: 2 moles MnO4- react with 10 moles Cl-.
Therefore, 3.0 × 10⁻⁴ mol MnO4- react with (10/2) × 3.0 × 10⁻⁴ mol Cl- = 1.5 × 10⁻³ mol Cl-

Step 3: Calculate the concentration of KCl:
Concentration of KCl = Moles/Volume = 1.5 × 10⁻³ mol / (25 cm³/1000 cm³/dm³) = 0.06 mol/dm³
ii. Concentration of A (KMnO4) in g/dm³:


Step 1: Calculate the molar mass of KMnO4 (39.1 + 54.9 + 4 × 16.0 = 158 g/mol)

Step 2: Convert the concentration from mol/dm³ to g/dm³:
Concentration of KMnO4 (g/dm³) = Concentration (mol/dm³) × Molar mass (g/mol)
= 0.02 mol/dm³ × 158 g/mol = 3.16 g/dm³
iii. Concentration of B (KCl) in g/dm³:


Step 1: Calculate the molar mass of KCl (39.1 + 35.5 = 74.6 g/mol)

Step 2: Convert the concentration from mol/dm³ to g/dm³:
Concentration of KCl (g/dm³) = Concentration (mol/dm³) × Molar mass (g/mol)
= 0.06 mol/dm³ × 74.6 g/mol = 4.48 g/dm³