Chapter 12, Problem 12.132QP

Interpretation Introduction

Interpretation:

A solution of – $0.360molN{a}_{3}P{O}_4,\text{ 0}{\text{.100 mol Ca(NO}}_3{\text{)}}_2$ and $\text{0}{\text{.100 mol AgNO}}_3$ are prepared in $4.000\text{ L}$ 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 12.132QP

Molarity of $N{a}^{+},P{O}_4{}^{2-}andN{O}_3{}^{-}$ ions are calculated as $0.270M,0.650M,0.0750M$ 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 – $N{a}^{+},{\text{ PO}}_4{}^{3-},{\text{ Ca}}^{2+},{\text{ Ag}}^{+}{\text{ and NO}}_3{}^{-}$.

$\begin{array}{l}No.{\text{ of moles of 3 Na}}^{+}\text{ ions}\text{= 3}\times \text{0}\text{.360 mol}\\ \text{= 1}\text{.080 mol}\\ \\ \text{No}{\text{. of moles of PO}}_4{}^{3-}\text{ ions}\text{= 0}\text{.360 mol}\\ \\ \text{no}{\text{. of moles of Ca}}^{2+}\text{ ions}\text{= 0}\text{.100 mol}\\ \\ \text{no}{\text{.of moles of Ag}}^{+}\text{ ions}\text{= 0}\text{.100 mol}\\ \\ \text{no}{\text{.of moles of NO}}^{3-}\text{ ions}\text{= 0}\text{.100 mol + 2(0}\text{.100 mol)}\\ \text{= 0}\text{.300 mol}\end{array}$

Calcium ions and Silver ions get precipitated by phosphate ions that two compounds are precipitated.  The precipitation is represented in equation as follows –

$\begin{array}{l}C{a}^{2+}(aq){\text{ + 2 PO}}_4{}^{3-}(aq)\stackrel{}{\to }{\text{ Ca}}_3{(}_P(s)\\ 3A{g}^{+}(aq){\text{ + PO}}_4{}^{3-}(aq)\stackrel{}{\to }{\text{ Ag}}_3{\text{PO}}_4(s)\end{array}$

$N{a}^{+}$ 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 –

$0.360\text{ mol}-\frac{2}{3}\times 0.100\text{ - }\frac{1}{3}\times 0.100\text{ mol = 0}\text{.2600 mol}$

Volume of the solution is $4.000\text{ L}$.  As ${\text{Ca}}^{2+}{\text{and Ag}}^{+}$ ions are precipitated, only $N{a}^{+},P{\text{O}}_4{}^{3-}{\text{and NO}}_3{}^{-}$ ions are left over in the solution.  Molarity of each ion in the solution is calculated as,

$\begin{array}{l}{\text{Molarity of Na}}^{+}\text{ ions}\text{= }\frac{no.\text{ of }molesof{\text{ Na}}^{+}\text{ ions}}{volume\text{ of solution in L}}\\ \text{= }\frac{0.1080{\text{ mol Na}}^{+}}{\text{4}\text{.000 L}}\text{ = 0}\text{.270 M}\\ {\text{Molarity of PO}}_4{}^{3-}\text{ ions}\text{= }\frac{no.\text{ of }molesof\text{ P}{O}_4{}^{3-}\text{ ions}}{4.000\text{ L}}\\ \text{= }\frac{0.2600{\text{ mol Na}}^{+}}{\text{4}\text{.000 L}}\text{ = 0}\text{.650 M}\\ {\text{Molarity of NO}}_3{}^{-}\text{ ions}\text{= }\frac{no.\text{ of }molesof{\text{ NO}}_3{}^{-}\text{ ions}}{volume\text{ of solution in L}}\\ \text{= }\frac{0.300{\text{ mol NO}}_3{}^{-}}{\text{2}\text{.000 L}}\text{ = 0}\text{.0750 M}\end{array}$

Conclusion

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