   # A 50.0-mL sample of 0.0413 M AgNO 3 ( aq ) is added to 50.0 mL of 0.100 M NaIO 3 ( aq ). Calculate the [Ag + ] at equilibrium in the resulting solution. [ K sp for AgIO 3 ( s ) = 3.17 × 10 −8 .] ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 97CWP
Textbook Problem
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## 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. [Ksp for AgIO3(s) = 3.17 × 10−8.]

Interpretation Introduction

Interpretation: The concentration of [Ag+] at equilibrium in the solution prepared by adding 50mL sample of 0.0413MAgNO3 to 50mL of 0.100MNaIO3 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 of Solution

Explanation

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

The molarity of NaNO3 is 0.02935M_ .

Given

The Ksp for AgIO3 is 3.17×108 .

The volume of 0.0413MAgNO3 solution is 50mL .

The volume of 0.100MNaIO3 solution is 50mL .

The mmol of AgNO3 solution is calculated by the formula,

mmolofAgNO3=Volume(mL)×Molarity

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

mmolofAgNO3=50mL×0.0413M=2.065mmol

The mmol of NaIO3 solution is calculated by the formula,

mmolof0.100M=Volume(mL)×Molarity

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

mmolofNaIO3=50mL×0.100M=5mmol

The equilibrium reaction in the resultant solution is given below.

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

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

Therefore, the amount of NaIO3 solution that remains unreacted is 5mmol2.065mmol=2.935mmol .

Molarity of NaNO3 is calculated by the formula,

Molarity=mmolofNaNO3Totalvolume

The value of mmol of NaNO3 is 2.935mmol .

Total volume =50+50=100mL

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

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