Chapter 4, Problem 4.89PAE

Interpretation Introduction

Interpretation:

The concentration of $C{a}^{2+}$ in the original solution should be found out if total dried product weight ( $Ca{F}_2(s),{\text{ PbF}}_2(s)$ ) is 3.141 g when 50 mL of the original sample containing $C{a}^{2+}and{\text{ Pb}}^{2+}$ is reacted with 53.8 mL of $NaF$ solution.

Concept introduction:

Calcium and lead ions form precipitates when reacted with fluoride ions.

Answer to Problem 4.89PAE

Solution:

Concentration of $C{a}^{2+}$ in the original solution = 0.404 M.

Explanation of Solution

Reactions given below take place to result the given products.

$\begin{array}{l}C{a}^{2+}(aq)+2NaF(aq)\to Ca{F}_2(s)+2N{a}^{+}(aq)\\ P{b}^{2+}(aq)+2NaF(aq)\to Pb{F}_2(s)+2N{a}^{+}(aq)\end{array}$

If “a” number of moles of $C{a}^{2+}$ and “b” number of moles of $P{b}^{2+}$ were present in the original solution,

$\begin{array}{l}Totalnumberofmolesof usedupinreaction=2a+2b \\ Buttotalnumberofmolesof usedupinthereaction=0.528mold{m}^{-3}\times \frac{53.8}{1000}d{m}^3\\ \text{ = 0}\text{.0284 mol}\end{array}$

Therefore,

$\begin{array}{l}2a+2b=0.0284\to equation\text{ 1}\\ b=\text{0}\text{.0142-a}\to \text{equation 2}\\ \end{array}$

Considering the mass of the product,

$\begin{array}{l}Massofprecipitate{\text{ = (number of moles of CaF}}_2\times molar{\text{ mass of CaF}}_2)+(NumberofmolesofPb{F}_2\times molar{\text{ mass of PbF}}_2)\\ \text{ = (a}\times 78)\text{ + (b}\times \text{245)}\end{array}$

Therefore,

$78a+245b=3.141\to \text{equation 3}$

Substituting equation 2 in equation 3,

$\begin{array}{l}78a+245(0.0142-a)=\text{ 3}\text{.141}\\ \text{ 3}\text{.479- 3}\text{.141= 167a}\\ \text{ a = 2}\text{.02}\times {\text{10}}^{-3}\end{array}$

Therefore number of moles of $C{a}^{2+}$ in original sample= $2.02\times {10}^{-3}mol$

Volume of the original sample = 50 mL.

Concentration of $C{a}^{2+}$ in original sample,

$Concentration=\frac{Total\text{ nu}mberofmoles}{volume\text{ of solution}}$

$\begin{array}{l}=\frac{2.02\times {10}^{-3}mol}{50/1000\text{ L}}\\ =0.0404\text{ M}\end{array}$

Conclusion

Concentration of a given ion in a solution with another similar ion can be found out by allowing the ions to precipitate and by finding the mass of the formed precipitate.