Question

What is the concentration of lead ions in a solution if they are completely precipitated from 32.7 mL of solution by 24.9 mL of 0.33M potassium iodide.

Answer 1

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Given reaction can be represented as:

Pb2+ + 2KI → Pb12(s) + 2K+

From the reaction, we have, 1 mol of Pb solution requires 2 mol of Potassium iodide solution.

1 mol Pb2+ = 2 mol KI (1)

No.of moles of lead ions in the solution,

ni = M1 x V1 = M1 X 32.7 ml

No.of moles of potassium iodide in the solution,

n2 = M, XV2 = 0.33 M x 24.9 mL = 8.217 m.mol

From (1):

ni = 2n

:: M1 X 32.7 mL = 2 x 8.217 m.mol

:: M 2 x 8.217 m.mol = - -= 0.503 M 32.7 ml

Therefore no.of moles of Pb ions initially in the solution:

ni = M1 x V1 = 0.503 M X 32.7 mL = 16.434 m.mol

Total volume of the final solution:

V = 32.7 mL + 24.9 mL = 57.6 mL

Therefore, concentration of precipitated lead in the final solution:

16.434 m/mol 57,6 mL

::C= 0.285 M

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