Question

25cm3 of 0.018 of potassium permanganate is pipetted into a conical flask,some dilute sulphuric acid and excess potassium iodide are then added and the liberated iodine is titrated in sodium thiosulphate of unknown concentration ,

Average titration figure is 15.6 cm3

a)write the equation for the two reactions hence find the mole ratio

b)calculate the concentration of sodium thiosulphate in moles per litre

Answer 1

Solution:

a)

(1): 2KMnO₄ + 10KI + 8H₂SO₄ → 2MnSO₄ + 6K₂SO₄ + 5I₂ + 8H₂O

2MnO₄^- + 16H⁺ + 10I^- → 2Mn²⁺ + 5I₂ + 8H₂O

According to the equation (1): n(KMnO₄)/2 = n(I₂)/5

Mole ratio: n(KMnO₄) : n(I₂) = 2 : 5

(2): 2Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI

2S₂O₃^2- + I₂ → S₄O₆^2- + 2I^-

According to the equation (2): n(Na₂S₂O₃)/2 = n(I₂)

Mole ratio: n(Na₂S₂O₃) : n(I₂) = 2 : 1

n(KMnO₄) : n(I₂) = 2 : 5

n(Na₂S₂O₃) : n(I₂) = 2 : 1 = 10 : 5

So, the mole ratio: n(KMnO₄) : n(I₂) : n(Na₂S₂O₃) = 2 : 5 : 10

b)

n(KMnO₄) : n(Na₂S₂O₃) = 2 : 10 = 1 : 5

5×n(KMnO₄) = n(Na₂S₂O₃)

5 × C_M(KMnO₄) × V(KMnO₄) = C_M(Na₂S₂O₃) × V(Na₂S₂O₃)

5 × (0.018 M) × (25 cm³) = C_M(Na₂S₂O₃) × (15.6 cm³)

C_M(Na₂S₂O₃) = (5 × 0.018 M × 25 cm³) / (15.6 cm³) = 0.144 M = 0.144 mol/L

C_M(Na₂S₂O₃) = 0.144 mol/L

Answers:

a)

2KMnO₄ + 10KI + 8H₂SO₄ → 2MnSO₄ + 6K₂SO₄ + 5I₂ + 8H₂O

2Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI

n(KMnO₄) : n(I₂) : n(Na₂S₂O₃) = 2 : 5 : 10

b)

The concentration of sodium thiosulphate (Na₂S₂O₃) is 0.144 mol/L