Answer to Question #348047 in General Chemistry for Andrea

Question

Suppose 47.6g of sodium iodide is dissolved in 350.mL of a 0.80 M aqueous solution of silver nitrate.

Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Round your answer to 1 significant digit.

Answer 1

n = C_M * V

n(AgNO₃) = 0,8 * 0,35 = 0,28 mol

n = m/M

M(NaI) = 150 g/mol

n(NaI) = 47,6 / 150 = 0,31 mol

1mol of AgNO3 and 1mol of NaI. that is, 0.28 mol of AgNO3 reacted with 0.28 mol of NaI and the resulting AgI precipitated. The excess NaI remained in solution. 1mol NaI forms 1mol I-anion in solution. Given the above, we find the mole of the I-anion in solution.

NaI + AgNO₃ = AgI + NaNO₃

1mol NaI — 1mol AgNO₃

X — 0,28mol AgNO₃

X = 1* 0,28 / 1 = 0,28 mol NaI reacted

we find the amount of excess NaI:

0,31 - 0,28 = 0,03 mol

NaI = Na⁺ + I^-

0,03mol NaI —> 0,03mol I^-

C_M(I^-) = n / V =0,03mol /0,35 l = 0,0857 ≈ 0,086 mol/l

ANSWER: 0,086 mol/l