Question

Suppose 35.7g of potassium iodide is dissolved in 350.mL of a 0.50 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 potassium iodide is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

KI(aq) + AgNO₃(aq) → AgI(s) + KNO₃(aq)
In reaction, mole ratio KI : AgNO₃ = 1 : 1

KI(aq) + AgNO₃(aq) → AgI(s) + KNO₃(aq)

Molar mass of KI = 166 g/mol
Initial no. of moles of KI = (35.7 g) / (166g) = 0.215 mol
Initial no. of moles of AgNO₃ = (0.50 mol/L) × (350/1000 L) = 0.175 mol < 0.215 mol
Hence, KI is in excess, and AgNO₃ completely reacts.

No. of moles of AgNO₃ reacted = 0.175 mol
No. of moles of KI reacted = 0.175 mol
No. of moles of KI left unreacted = (0.215 - 0.175) mol = 0.040 mol

1 mole of KI contains 1 mole of I⁻ anion.
No. of moles of I⁻ left unreacted = 0.040 mol
Volume of the final solution = 350 mL = 0.350 L
Molarity of I⁻ anion left unreacted in the final solution = (0.040 mol) / (350 mL) = 0.114 M

The final molarity of iodide anion in the solution = 0.114M