# Answer to Question #120338 in Inorganic Chemistry for courtnie

Question #120338
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
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2020-06-05T07:28:02-0400

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-) = CM(KMnO4) × V(KMnO4)

CM(MnO4-) = CM(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:

CM(Sn2+) = n(Sn2+) / V(Sn2+)

CM(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.

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