Chapter 3, Problem 3.64PAE

3.62 What is the molarity of each ion present in aqueous solutions prepared by dissolving 20.00 g of the following compounds in water to make 4.50 L of solution?

(a) cobalt(III) chloride, (b) nickel(III) sulfate, (c) sodium permanganate, (d) iron(II) bromide

Interpretation Introduction

To determine: the molarity of cobalt (III) chloride solution.

Explanation of Solution

Molar mass of cobalt (III) chloride i.e. $CoC{l}_3$ is:

$\begin{array}{c}M{W}_{CoC{l}_3}=M{W}_{Co}+3M{W}_{Cl}\\ =58.9+3\times 35.5\\ =165.4g/mol\end{array}$

Now, number of moles of $CoC{l}_3$ present in 20.00 g of compound is:

$n=\frac{Massofthecompound}{Molarmass}=\frac{20.0g}{165.4g/mol}=0.12moles$

Thus, molarity of $CoC{l}_3$ is:

$M=0.12\times \frac{1L}{4.5L}=0.027\text{}M$

$CoC{l}_3$ dissolves in water as:

$CoC{l}_3(s)\stackrel{aq.}{\to }C{o}^{+3}(aq)+3C{l}^{-}(aq)$

Thus, 1 mole of $CoC{l}_3$ produces 1 mole of $C{o}^{+3}$ and 3 moles of $C{l}^{-}$ ions.

Hence,

Molarity of $C{o}^{+3}$ is same as the molarity of $CoC{l}_3$.

Molarity of $C{l}^{-}$ is thrice that of $CoC{l}_3$.

$\begin{array}{l}{M}_{C{o}^{+3}}=M_{CoC{l}_3}=0.027M\\ M_{C{l}^{-}}=3\times M_{CoC{l}_3}=3\times 0.027=0.081M\end{array}$

Interpretation Introduction

To determine:

The molarity of nickel (III) sulfate solution.

Explanation:

Molar mass of nickel (III) sulfate i.e. $N{i}_2{(S{O}_4)}_3$ is:

$\begin{array}{c}M{W}_{N{i}_2{(S{O}_4)}_3}=2M{W}_{Ni}+3(M{W}_s+4M{W}_O)\\ =2\times 58.7+3(32.1+4\times 16.0)\\ =117.4+3(96.1)\\ =405.7g/mol\end{array}$

Now, number of moles of $N{i}_2{(S{O}_4)}_3$ present in 20.00 g of compound is:

$n=\frac{Massofthecompound}{Molarmass}=\frac{20.0g}{405.7g/mol}=0.049moles$

Thus, molarity of $N{i}_2{(S{O}_4)}_3$ is:

$M=0.049\times \frac{1L}{4.5L}=0.011\text{}M$

$N{i}_2{(S{O}_4)}_3$ dissolves in water as:

$N{i}_2{(S{O}_4)}_3(s)\stackrel{aq.}{\to }2N{i}^{+3}(aq)+3S{O}_4^{2-}(aq)$

Thus, 1 mole of $N{i}_2{(S{O}_4)}_3$ produces 2 moles of $N{i}^{+3}$ and 3 moles of $S{O}_4^{2-}$ ions.

Hence,

Molarity of $N{i}^{+3}$ is twice the molarity of $N{i}_2{(S{O}_4)}_3$.

Molarity of $S{O}_4^{2-}$ is thrice that of $N{i}_2{(S{O}_4)}_3$.

$\begin{array}{l}{M}_{N{i}^{+3}}=2\times M_{N{i}_2{(S{O}_4)}_3}=2\times 0.011=0.022M\\ M_{S{O}_4^{2-}}=3\times M_{N{i}_2{(S{O}_4)}_3}=3\times 0.011=0.033M\end{array}$

Explanation of Solution

Molar mass of nickel (III) sulfate i.e. $N{i}_2{(S{O}_4)}_3$ is:

$\begin{array}{c}M{W}_{N{i}_2{(S{O}_4)}_3}=2M{W}_{Ni}+3(M{W}_s+4M{W}_O)\\ =2\times 58.7+3(32.1+4\times 16.0)\\ =117.4+3(96.1)\\ =405.7g/mol\end{array}$

Now, number of moles of $N{i}_2{(S{O}_4)}_3$ present in 20.00 g of compound is:

$n=\frac{Massofthecompound}{Molarmass}=\frac{20.0g}{405.7g/mol}=0.049moles$

Thus, molarity of $N{i}_2{(S{O}_4)}_3$ is:

$M=0.049\times \frac{1L}{4.5L}=0.011\text{}M$

$N{i}_2{(S{O}_4)}_3$ dissolves in water as:

$N{i}_2{(S{O}_4)}_3(s)\stackrel{aq.}{\to }2N{i}^{+3}(aq)+3S{O}_4^{2-}(aq)$

Thus, 1 mole of $N{i}_2{(S{O}_4)}_3$ produces 2 moles of $N{i}^{+3}$ and 3 moles of $S{O}_4^{2-}$ ions.

Hence,

Molarity of $N{i}^{+3}$ is twice the molarity of $N{i}_2{(S{O}_4)}_3$.

Molarity of $S{O}_4^{2-}$ is thrice that of $N{i}_2{(S{O}_4)}_3$.

$\begin{array}{l}{M}_{N{i}^{+3}}=2\times M_{N{i}_2{(S{O}_4)}_3}=2\times 0.011=0.022M\\ M_{S{O}_4^{2-}}=3\times M_{N{i}_2{(S{O}_4)}_3}=3\times 0.011=0.033M\end{array}$

Conclusion

Therefore, knowing the number of moles of ionic solute for any volume of a given solution with its dissolution equation, the molarity could be calculated for the individual ion species formed in the solution.

Interpretation Introduction

To determine:

The molarity of sodium permanganate solution.

Explanation:

Molar mass of sodium permanganate i.e. $NaMn{O}_4$ is:

$\begin{array}{c}M{W}_{NaMn{O}_4}=M{W}_{Na}+M{W}_{Mn}+4M{W}_O\\ =23.0+54.9+4\times 16.0\\ =141.9g/mol\end{array}$

Now, number of moles of $NaMn{O}_4$ present in 20.00 g of compound is:

$n=\frac{Massofthecompound}{Molarmass}=\frac{20.0g}{141.9g/mol}=0.14moles$

Thus, molarity of $NaMn{O}_4$ is:

$M=0.14\times \frac{1L}{4.5L}=0.03\text{}M$

$NaMn{O}_4$ dissolves in water as:

$NaMn{O}_4(s)\stackrel{aq.}{\to }N{a}^{+}(aq)+Mn{O}_4^{-}(aq)$

Thus, 1 mole of $NaMn{O}_4$ produces 1 mole of $N{a}^{+}$ and 1 mole of $Mn{O}_4^{-}$ ions.

Hence,

Molarity of $N{a}^{+}$ and $Mn{O}_4^{-}$ ions is same as the molarity of $NaMn{O}_4$.

$M_{N{a}^{+}}=M_{Mn{O}_4^{-}}=M_{NaMn{O}_4}=0.03M$

Explanation of Solution

To determine:

The molarity of sodium permanganate solution.

Molar mass of sodium permanganate i.e. $NaMn{O}_4$ is:

$\begin{array}{c}M{W}_{NaMn{O}_4}=M{W}_{Na}+M{W}_{Mn}+4M{W}_O\\ =23.0+54.9+4\times 16.0\\ =141.9g/mol\end{array}$

Now, number of moles of $NaMn{O}_4$ present in 20.00 g of compound is:

$n=\frac{Massofthecompound}{Molarmass}=\frac{20.0g}{141.9g/mol}=0.14moles$

Thus, molarity of $NaMn{O}_4$ is:

$M=0.14\times \frac{1L}{4.5L}=0.03\text{}M$

$NaMn{O}_4$ dissolves in water as:

$NaMn{O}_4(s)\stackrel{aq.}{\to }N{a}^{+}(aq)+Mn{O}_4^{-}(aq)$

Thus, 1 mole of $NaMn{O}_4$ produces 1 mole of $N{a}^{+}$ and 1 mole of $Mn{O}_4^{-}$ ions.

Hence,

Molarity of $N{a}^{+}$ and $Mn{O}_4^{-}$ ions is same as the molarity of $NaMn{O}_4$.

$M_{N{a}^{+}}=M_{Mn{O}_4^{-}}=M_{NaMn{O}_4}=0.03M$

Conclusion

Therefore, knowing the number of moles of ionic solute for any volume of a given solution with its dissolution equation, the molarity could be calculated for the individual ion species formed in the solution.

Interpretation Introduction

To determine:

The molarity of iron(II) bromide solution.

Explanation of Solution

Molar mass of iron(II) bromide i.e. $FeB{r}_2$ is:

$\begin{array}{c}M{W}_{FeB{r}_2}=M{W}_{Fe}+2M{W}_{Br}\\ =55.8+2\times 79.9\\ =215.6g/mol\end{array}$

Now, number of moles of $FeB{r}_2$ present in 20.00 g of compound is:

$n=\frac{Massofthecompound}{Molarmass}=\frac{20.0g}{215.6g/mol}=0.093moles$

Thus, molarity of $FeB{r}_2$ is:

$M=0.093\times \frac{1L}{4.5L}=0.02\text{}M$

$FeB{r}_2$ dissolves in water as:

$FeB{r}_2(s)\stackrel{aq.}{\to }F{e}^{2+}(aq)+2B{r}^{-}(aq)$

Thus, 1 mole of $FeB{r}_2$ produces 1 mole of $F{e}^{2+}$ and 2 moles of $B{r}^{-}$ ions.

Hence,

Molarity of $F{e}^{2+}$ is same as the molarity of $FeB{r}_2$.

Molarity of $B{r}^{-}$ is twice the molarity of $FeB{r}_2$

$M_{F{e}^{2+}}=M_{FeB{r}_2}=0.02M$

$M_{B{r}^{-}}=2\times M_{FeB{r}_2}=2\times 0.02=0.04M$

Conclusion

Therefore, knowing the number of moles of ionic solute for any volume of a given solution with its dissolution equation, the molarity could be calculated for the individual ion species formed in the solution.