Chapter 12, Problem 12.131QP

Interpretation Introduction

Interpretation:

A solution of – $0.375molN{a}_{2}C{O}_3,\text{ 0}{\text{.125 mol Ca(NO}}_3{\text{)}}_2$ and $\text{0}{\text{.200 mol AgNO}}_3$ are prepared in $2.00\text{ L}$ of water.  Occurrence of precipitation reaction has to be confirmed by writing balanced equation.  Molarity of each ion in its solution has to be calculated.

Concept Introduction:

Molarity is a term used to express concentration of a solution.  It is expressed as,

$\text{Molarity = }\frac{number\text{ of }molesof\text{ solute}}{volume\text{ of solution in L}}$

Answer to Problem 12.131QP

Molarity of $N{a}^{+},C{O}_3{}^{2-}andN{O}_3{}^{-}$ ions are calculated as $0.375M,0.0750M,0.225M$ respectively.

Explanation of Solution

Determine the number of moles of each ion present in the solution. Totally five types ions are present in the solution as follows – $N{a}^{+},{\text{ CO}}_3{}^{2-},{\text{ Ca}}^{2+},{\text{ Ag}}^{+}{\text{ and NO}}_3{}^{-}$.

$\begin{array}{l}No.{\text{ of moles of 2 Na}}^{+}\text{ ions}\text{= 2}\times \text{0}\text{.375 mol}\\ \text{= 0}\text{.750 mol}\\ \\ \text{No}{\text{. of moles of CO}}_3{}^{2-}\text{ ions}\text{= 0}\text{.375 mol}\\ \\ \text{no}{\text{. of moles of Ca}}^{2+}\text{ ions}\text{= 0}\text{.125 mol}\\ \\ \text{no}{\text{.of moles of Ag}}^{+}\text{ ions}\text{= 0}\text{.200 mol}\\ \\ \text{no}{\text{.of moles of NO}}^{3-}\text{ ions}\text{= 0}\text{.200 mol + 2(0}\text{.125 mol)}\\ \text{= 0}\text{.450 mol}\end{array}$

Two compounds are precipitated.  The precipitation is represented in equation as follows–

$\begin{array}{l}C{a}^{2+}(aq){\text{ + CO}}_3{}^{2-}(aq)\stackrel{}{\to }{\text{ CaCO}}_3(s)\\ 2A{g}^{+}(aq){\text{ + CO}}_3{}^{2-}(aq)\stackrel{}{\to }{\text{ Ag}}_2{\text{CO}}_3(s)\end{array}$

$N{a}^{+}$ ions are not precipitated by ${\text{CO}}_3{}^{2-}$ ions.  All amount of ${\text{CO}}_3{}^{2-}$ ions don’t involve in precipitation reaction.  The remaining moles of ${\text{CO}}_3{}^{2-}$ are –

$0.375\text{ mol}-0.125\text{ mol}-\frac{0.200}{2}\text{ mol = 0}\text{.150 mol}$

Volume of the solution is $2.00\text{ L}$.  As ${\text{Ca}}^{2+},{\text{ Ag}}^{+}$ ions are precipitated, only $N{a}^{+},{\text{CO}}_3{}^{2-}{\text{and NO}}_3{}^{-}$ ions are left over in the solution.  Molarity of each ion in the solution is calculated as,

$\begin{array}{l}{\text{Molarity of Na}}^{+}\text{ ions}\text{= }\frac{number\text{ of }molesof{\text{ Na}}^{+}\text{ ions}}{volume\text{ of solution in L}}\\ \text{= }\frac{0.750{\text{ mol Na}}^{+}}{\text{2}\text{.000 L}}\text{ = 0}\text{.375 M}\\ {\text{Molarity of CO}}_3{}^{2-}\text{ ions}\text{= }\frac{number\text{ of }molesofC{O}_3{}^{2-}\text{ ions}}{2.000\text{ L}}\\ \text{= }\frac{0.150{\text{ mol Na}}^{+}}{\text{2}\text{.000 L}}\text{ = 0}\text{.0750 M}\\ {\text{Molarity of NO}}_3{}^{-}\text{ ions}\text{= }\frac{number\text{ of }molesof{\text{ NO}}_3{}^{-}\text{ ions}}{volume\text{ of solution in L}}\\ \text{= }\frac{0.450{\text{ mol Na}}^{+}}{\text{2}\text{.000 L}}\text{ = 0}\text{.225 M}\end{array}$

Conclusion

Molarity of various ions present in a solution is calculated using number of moles of the ions.