Chapter 18.4, Problem 44LC

Interpretation Introduction

Interpretation: Whether the precipitate formed or not on mixing 500 mL of 0.02 M solution of silver nitrate with 500 mL of a 0.001 M solution of NaCl is to be stated.

Concept Introduction: The insoluble solid so formed by the reaction of two or more salts. The reaction in which precipitate is formed is called a precipitation reaction.

Answer to Problem 44LC

Yes, the precipitate will form.

Explanation of Solution

Given;

Volume of silver nitrate solution = 500 mL

Concentration of silver nitrate solution = 0.02 M

Volume of sodium chloride solution = 500 mL

Concentration of sodium chloride solution = 0.001 M

The balanced chemical equation for the reaction of silver nitrate and sodium chloride is given as;

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

Calculating the moles of silver nitrate and sodium chloride solution as;

  $\begin{array}{c}\text{Moles}\text{of}{\text{AgNO}}_{\text{3}}=\text{Molarity}\times \text{Volume}\\ \text{=0}\text{.02 }\frac{\text{mol}}{\text{L}}\times \text{500}\text{mL}\times \frac{\text{1}\text{L}}{\text{1000}\text{mL}}\\ =0.01\text{}\text{mol}\end{array}$

  $\begin{array}{c}\text{Moles}\text{of}\text{NaCl}=\text{Molarity}\times \text{Volume}\\ \text{=0}\text{.001 }\frac{\text{mol}}{\text{L}}\times \text{500}\text{mL}\times \frac{\text{1}\text{L}}{\text{1000}\text{mL}}\\ =0.0005\text{}\text{mol}\end{array}$

According to the reaction, one mole of silver nitrate reacts with one mole of sodium chloride thus, 0.0005 moles of sodium chloride will react with 0.0005 moles of silver nitrate as sodium chloride is the limiting reagent.

Now, calculate the ionic product for AgCl as;

  $\begin{array}{c}\left[\text{AgCl}\right]=\left[{\text{Ag}}^{+}\right]\left[{\text{Cl}}^{-}\right]\\ =\left[0.0005\right]\left[0.0005\right]\\ =0.00000025\end{array}$

The ionic product of AgCl is greater than the solubility product $\left({\text{K}}_{\text{sp}}=1.5\times {10}^{-10}\right)$ thus, a precipitate will form.