Answer to Question #120338 in Inorganic Chemistry for courtnie

Question

To determine the concentration of a Sn2+(aq) solution, a student titrated a 50.00 mL sample of acidified Sn2+(aq) with 1.44 mmol/L KMnO4(aq). The titration required 24.83 mL of KMnO4(aq) to reach the endpoint

Answer 1

Solution:

(MnO₄^- + 8H⁺ + 5e --> Mn²⁺ + 4H₂O) × 2

(Sn²⁺ --> Sn⁴⁺ + 2e) × 5

The balanced net ionic equation for this titration reaction is:

2MnO₄^-(aq) + 16H⁺(aq) + 5Sn²⁺(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 5Sn⁴⁺(aq)

According to the equation: n(MnO₄^-)/2 = n(Sn²⁺)/5

For every two mole of MnO₄^-, there five mole of Sn²⁺

n(MnO₄^-) = C_M(KMnO₄) × V(KMnO₄)

C_M(MnO₄^-) = C_M(KMnO₄) = 1.44 mmol/L = 0.00144 mol/L

n(MnO₄^-) = (0.00144 mol/L) × (0.02483 L) = 3.576×10^-5 mol

So, the moles of Sn²⁺ is:

n(Sn²⁺) = 5 × n(MnO₄^-) / 2 = 5 × (3.576×10^-5 mol) / 2 = 8.94×10^-5 mol

Finally, the concentration of a Sn²⁺(aq) solution is:

C_M(Sn²⁺) = n(Sn²⁺) / V(Sn²⁺)

C_M(Sn²⁺) = (8.94×10^-5 mol) / (0.05 L) = 0.001788 mol/L = 1.79 mmol/L

Therefore, the concentration of a Sn²⁺(aq) solution is 1.79 mmol/L.

Answer: The concentration of a Sn²⁺(aq) solution is 1.79 mmol/L.