Chapter U4.91, Problem 6E

Interpretation Introduction

Interpretation:

The milliliters of sodium chloride required to remove silver ions from the solution should be predicted.

Concept Introduction:

Concentration can be expressed in terms of molarity. Molarity is defined as the number of moles of solute dissolved in 1 L of the volume of the solution.

The mathematical expression for molarity is represented as follows:

  $\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$

Answer to Problem 6E

200 mL of sodium chloride would be needed to remove the chloride ions.

Explanation of Solution

The chemical equation for the reaction of sodium chloride and silver nitrate is represented as follows:

  ${\text{AgNO}}_3+\text{NaCl}\to {\text{NaNO}}_{\text{3}}+\text{AgCl}$

Using law of equivalence to calculated the volume required as follows:

  $\begin{array}{c}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\left({\text{AgNO}}_{\text{3}}\right)={\text{M}}_{\text{2}}{\text{V}}_{\text{2}}\left(\text{NaCl}\right)\\ {\text{V}}_{\text{2}}\left(\text{NaCl}\right)=\frac{{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\left({\text{AgNO}}_{\text{3}}\right)}{{\text{M}}_{\text{2}}\left(\text{NaCl}\right)}\\ =\frac{0.20\text{M}\times \text{500}\text{mL}}{0.50\text{M}}\\ {\text{V}}_{\text{2}}\left(\text{NaCl}\right)=200\text{mL}\end{array}$

Thus, 200 mL of sodium chloride would be needed to remove the chloride ions.