Question

Suppose 3.10g of sodium iodide is dissolved in 250.mL of a 73.0mM 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 2 significant digits.

Answer 1

When NaI reacts with AgNO₃, the reaction is

NaI (aq) + AgNO₃ (aq.) -------------> NaNO₃ (aq) + AgI(s)

Given molarity of AgNO₃ = 75 mM

Volume = 250 mL or 0.25 L

Moles of AgNO₃ = 75 x 10^-3 x 0.25 = 18.75 x 10^-3 or 0.01875 moles

Moles of Ag⁺ = 0.01875

Mass of NaI = 3.1 g

Molar mass of NaI = 150 g

moles of NaI = 3.1/150 = 0.021

As 1 mole of Iodide reacts with one mole of Ag⁺ to form AgI

So, moles of AgI = 0.01875

Left over Iodide = 0.021 - 0.01875 = 0.00225 moles

Molarity of I^- = 0.00225/0.25 = 0.009 M

In two significant figures, it becomes 9.0 x 10^-3 M or 9.0 mM

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