Chapter 20, Problem 73A

Interpretation Introduction

Interpretation: The amount of copper needed in grams to completely reduce silver ions in $85mL$ of $0.15M$ $AgN{O}_3$ solution is to be determined.

Concept introduction: The reaction that involves the reduction of $A{g}^{+}$ in $AgN{O}_3$ is, $Cu\left(s\right)+2Ag\left(N{O}_3\right)+\left(aq\right)\to Cu{\left(N{O}_3\right)}_2+2Ag\left(s\right)$ is a redox reaction.

The formula to calculate the molarity of the solution is given as

  $\text{Molarity}=\frac{\text{Moles solute}}{\text{Volume }\left(\text{Litre}\right)\text{ solution}}$   ...... (1)

Answer to Problem 73A

The mass of copper required to reduce silver ions is $0.4g$ .

Explanation of Solution

The given amount of silver ions is $85mL=0.085L$ .

The molarity of $AgN{O}_3$ solution is 0.15M. The number of moles can be calculated from the molarity equation.

Substitute, $0.15$ for molarity of silver ion and $0.085\text{L}$ for the volume of silver in equation (1).

  $\begin{array}{c}\text{moles}=\text{0}\text{.085}\times 0.15\\ =0.01275\text{ mol}\end{array}$

From the chemical equation, $Cu\left(s\right)+2Ag\left(N{O}_3\right)+\left(aq\right)\to Cu{\left(N{O}_3\right)}_2+2Ag\left(s\right)$ , the stoichiometric ratio of copper and silver ions is,

  $\frac{n\left(AgN{O}_3\right)}{n\left(Cu\right)}=\frac{2}{1}$

Substitute, $0.01275\text{ mol}$ for moles of silver ion in the above equation.

  $\begin{array}{c}\frac{n\left(AgN{O}_3\right)}{n\left(Cu\right)}=\frac{2}{1}\\ 0.01275\text{ mol}=2\left(Cu\right)\\ n\left(Cu\right)=\frac{0.01275\text{ mol}}{2}\\ =0.00637mol\end{array}$

So, the number of moles of copper is $0.00637mol$ .

Now, the mass of copper ions can be calculated. The mass of copper is the product of the number of moles and molecular mass. Hence,

  $\begin{array}{c}Massofcopper=0.00637mol\times 63\\ =0.4g\end{array}$

So, the mass of copper required is $0.4g$ .