A 50.0-mL sample of 0.0413 M AgNO₃(aq) is added to 50.0 mL of 0.100 M NaIO₃(aq). Calculate the [Ag⁺] at equilibrium in the resulting solution. [K_sp for AgIO₃(s) = 3.17 × 10⁻⁸.]

Interpretation: The concentration of [Ag+] at equilibrium in the solution prepared by adding 50 mL sample of 0.0413 M AgNO3 to 50 mL of 0.100M NaIO3 is to be calculated.

Concept introduction: Solubility product is defined as the mathematical product of the dissolved ion concentration of a substance raised to the power of its stoichiometric coefficients. When sparingly soluble ionic compound releases ions in the solution, it gives relevant solubility product. The solvent is generally water. Precipitate formation takes place in solution if ionic product is greater than the solubility product.

Explanation

To determine: The concentration of [Ag+] at equilibrium in the resultant solution.

The molarity of NaNO3 is 0.02935 M_ .

Given

The Ksp for AgIO3 is 3.17×10−8 .

The volume of 0.0413 M AgNO3 solution is 50 mL .

The volume of 0.100M NaIO3 solution is 50 mL .

The mmol of AgNO3 solution is calculated by the formula,

mmol of AgNO3=Volume(mL)×Molarity

Substitute the value of volume of the AgNO3 solution in mL and molarity in the above formula.

mmol of AgNO3=50 mL×0.0413 M=2.065 mmol

The mmol of NaIO3 solution is calculated by the formula,

mmol of 0.100M=Volume(mL)×Molarity

Substitute the value of volume of the NaIO3 solution in mL and molarity in the above formula.

mmol of NaIO3=50 mL×0.100M=5 mmol

The equilibrium reaction in the resultant solution is given below.

AgNO3(aq)+NaIO3(aq)⇌AgIO3(aq)+NaNO3(aq)

The 2.065 mmol AgNO3 solution will combine with 2.065 mmol NaIO3 solution to give 2.065 mmol AgIO3 .

Therefore, the amount of NaIO3 solution that remains unreacted is 5 mmol−2.065 mmol=2.935 mmol .

Molarity of NaNO3 is calculated by the formula,

Molarity=mmol of NaNO3Total volume

The value of mmol of NaNO3 is 2.935 mmol .

Total volume =50+50=100 mL

Substitute the value of total volume and mmol of NaNO3 in the above formula