Question

How many grams of silver chloride can be prepared by the reaction$100.0ml$$of0.2M$ silver nitrate with $100.0mlof0.15M$calcium chloride?. Calculate the concentrations of each ion remaining in solution after precipitation is complete.

Short answer

The concentrations of each ion remaining in solution after the completion of precipitation

$$\begin{gathered} m\left(AgCl\right)=2.87g, \\ c\left(C{a}^{2+}\right)=0.075M, \\ c\left(C{l}^{-}\right)=0.05M, \\ c\left(N{O}_3\right)=0.10M \end{gathered}$$

Step-by-step solution

Definition of concentration of ions

The ionic strength of a solution is ameasure of the concentration of ions in that solution is defined as concentration of ions

Calculation of the concentrations of each ion remaining in solution after precipitation

$$\begin{gathered} 2AgN{O}_3+CaC{l}_3\to 2AgCl+Ca{\left(N{O}_3\right)}_{2}n\left(AgN{O}_3\right)=c.V \\ =0.20M.0.100L \\ =0.02mole \\ n\left(CaC{l}_2\right)=c.V \\ =0.15M.0.100L \\ =0.015mole \\ \frac{n\left(AgCl\right)}{n\left(CaC{l}_2\right)}=\frac{2}{1}\to n\left(AgCl\right) \\ =2\times 0.015 \\ =0.03mole \\ n\left(AgCl\right)=n\left(AgN{O}_3\right) \\ =0.02LIMITINGREACTANT \\ n\left(AgCl\right)=n\times Mr \\ =0.02mole\times 143.32g/mole \\ =2.87g \\ c\left(C{a}^{2+}\right)=\frac{n\left(CaC{l}_2\right)}{V\left(total\right)} \\ =\frac{0.015mole}{0.200L} \\ =0.075M \\ c\left(C{l}^{-}\right)=\frac{2.n\left(CaC{l}_2\right)-n\left(AgCl\right)}{v\left(total\right)} \\ =\frac{2.0.015mole-0.02mole}{0.200L} \\ =0.05M \\ c\left(N{O}_3\right)=\frac{n\left(AgN{O}_3\right)}{V\left(totals\right)} \\ =\frac{0.02mole}{0.200L} \\ =0.10M \end{gathered}$$

The concentrations of each ion remaining in solution after precipitation completed is

$$\begin{gathered} m\left(AgCl\right)=2.87g, \\ c\left(C{a}^{2+}\right)=0.075M, \\ c\left(C{l}^{-}\right)=0.05M, \\ c\left(N{O}_3\right)=0.10M \end{gathered}$$