Chapter 3, Problem 18QAP

What is the molarity of each ion present in aqueous solutions of the following compounds prepared by dissolving 28.0 g of each compound in water to make 785 mL of solution?

(a) potassium oxide

(b) sodium hydrogen carbonate

(c) scandium(III) iodite

(d) magnesium phosphate

Interpretation Introduction

(a)

Interpretation:

The molarity of all the ions in the solution should be calculated.

Concept introduction:

The number of moles of a substance is related to mass and molar mass as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass of the substance.

Also, according to Avogadro’s law in 1 mol of a substance there are $6.023\times {10}^{23}$ units of that substance. Here, $6.023\times {10}^{23}$ is known as Avogadro’s number and denoted by symbol $N_A$.

The molarity of a solution is defined as number of moles of solute in 1 L of the solution. It is mathematically represented as follows:

$M=\frac{n}{V}$

Here, n is number of moles of solute and V is volume of solution in L. Thus, the unit of molarity is mol/L.

Answer to Problem 18QAP

The molarity of ${\text{K}}^{+}$ and ${\text{O}}^{2-}$ ions is 0.756 M and 0.378 M respectively.

Explanation of Solution

The given compound is potassium oxide. The mass of compound is 28.0 g and volume of solution is 785 mL. The number of moles of compound can be calculated as follows:

$n=\frac{m}{M}$

Molar mass of potassium oxide is 94.2 g/mol thus,

$\begin{array}{c}n=\frac{28.0\text{ g}}{94.2\text{ g/mol}}\\ =0.297\text{ mol}\end{array}$

Thus, number of moles of potassium oxide is 0.297 mol. Molarity of solution can be calculated as follows;

$M=\frac{n}{V}$

Convert the volume from mL to L.

$V=785\text{ mL}\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=0.785\text{ L}$

Putting the values,

$\begin{array}{c}M=\frac{0.297\text{ mol}}{0.785\text{ L}}\\ =0.378\text{ mol/L}\\ \text{=}0.378\text{ M}\end{array}$

Therefore, molarity of solution is $0.378\text{ M}$.

The formula of potasisum oxide is ${\text{K}}_{\text{2}}\text{O}$ thus, it has 2 ${\text{K}}^{+}$ and 1 ${\text{O}}^{2-}$ ions. Since, there are 2 mol of K in 1 mol of ${\text{K}}_{\text{2}}\text{O}$ the molarity of ${\text{K}}^{+}$ will be twice the molarity of ${\text{K}}_{\text{2}}\text{O}$.

$\begin{array}{c}{M}_{{\text{K}}^{+}}=2\times 0.378\text{ M}\\ \text{=0}\text{.756 M}\end{array}$

And, molarity of ${\text{O}}^{2-}$ will be equal to the molarity of ${\text{K}}_{\text{2}}\text{O}$.

Thus,

$M_{{\text{O}}^{2-}}=0.378\text{ M}$

Interpretation Introduction

(b)

Interpretation:

The molarity of all the ions in the solution should be calculated.

Concept introduction:

The number of moles of a substance is related to mass and molar mass as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass of the substance.

Also, according to Avogadro’s law in 1 mol of a substance there are $6.023\times {10}^{23}$ units of that substance. Here, $6.023\times {10}^{23}$ is known as Avogadro’s number and denoted by symbol $N_A$.

The molarity of a solution is defined as number of moles of solute in 1 L of the solution. It is mathematically represented as follows:

$M=\frac{n}{V}$

Here, n is number of moles of solute and V is volume of solution in L. Thus, the unit of molarity is mol/L.

Answer to Problem 18QAP

Molarity of both ${\text{Na}}^{+}$ and ${\text{HCO}}_3^{-}$ is 0.4204 M.

Explanation of Solution

The given compound is sodium hydrogen carbonate. The mass of compound is 28.0 g and volume of solution is 785 mL. The number of moles of compound can be calculated as follows:

$n=\frac{m}{M}$

Molar mass of sodium hydrogen carbonate is 84.0 g/mol thus,

$\begin{array}{c}n=\frac{28.0\text{ g}}{84.0\text{ g/mol}}\\ =0.33\text{ mol}\end{array}$

Thus, number of moles of sodium hydrogen carbonate is 0.33 mol. Molarity of solution can be calculated as follows;

$M=\frac{n}{V}$

Convert the volume from mL to L.

$V=785\text{ mL}\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=0.785\text{ L}$

Putting the values,

$\begin{array}{c}M=\frac{0.33\text{ mol}}{0.785\text{ L}}\\ =0.4204\text{ mol/L}\\ \text{=}0.4204\text{ M}\end{array}$

Therefore, molarity of solution is $0.4204\text{ M}$.

The molecular formula of sodium hydrogen carbonate is ${\text{NaHCO}}_{\text{3}}$ thus, the ions present in the solution are ${\text{Na}}^{+}$ and ${\text{HCO}}_3^{-}$.

The molarity of sodium hydrogen carbonate is equal to the molarity of ${\text{Na}}^{+}$ and ${\text{HCO}}_3^{-}$.

Therefore, molarity of both ${\text{Na}}^{+}$ and ${\text{HCO}}_3^{-}$ is 0.4204 M.

Interpretation Introduction

(c)

Interpretation:

The molarity of all the ions in the solution should be calculated.

Concept introduction:

The number of moles of a substance is related to mass and molar mass as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass of the substance.

Also, according to Avogadro’s law in 1 mol of a substance there are $6.023\times {10}^{23}$ units of that substance. Here, $6.023\times {10}^{23}$ is known as Avogadro’s number and denoted by symbol $N_A$.

The molarity of a solution is defined as number of moles of solute in 1 L of the solution. It is mathematically represented as follows:

$M=\frac{n}{V}$

Here, n is number of moles of solute and V is volume of solution in L. Thus, the unit of molarity is mol/L.

Answer to Problem 18QAP

Molarity of Sc³+ and I- is 0.068 M and 0.204 M respectively.

Explanation of Solution

The given compound is scandium (III) iodite. The mass of compound is 28.0 g and volume of solution is 785 mL. The number of moles of compound can be calculated as follows:

$n=\frac{m}{M}$

Molar mass of scandium (III) iodite is 521.66 g/mol thus,

$\begin{array}{c}n=\frac{28.0\text{ g}}{521.66\text{ g/mol}}\\ =0.054\text{ mol}\end{array}$

Thus, number of moles of scandium (III) iodide is 0.054 mol. Molarity of solution can be calculated as follows;

$M=\frac{n}{V}$

Convert the volume from mL to L.

$V=785\text{ mL}\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=0.785\text{ L}$

Putting the values,

$\begin{array}{c}M=\frac{0.054\text{ mol}}{0.785\text{ L}}\\ =0.068\text{ mol/L}\\ \text{=}0.068\text{ M}\end{array}$

Therefore, molarity of solution is $0.068\text{ M}$.

The molecular formula of scandium (III) iodite is $\text{Sc}{\left({\text{IO}}_2\right)}_3$. Thus, it will have 1 Sc³+ and 3 IO₂ - ions. The molarity of Sc³+ is equal to $\text{Sc}{\left({\text{IO}}_2\right)}_3$ and that of IO₂ - is 3 times the molarity of $\text{Sc}{\left({\text{IO}}_2\right)}_3$

Thus, molarity of IO₂ - is $3\times 0.068\text{ M}=0.204\text{ M}$.

Therefore, molarity of Sc³+ and I- is 0.068 M and 0.204 M respectively.

Interpretation Introduction

(d)

Interpretation:

The molarity of all the ions in the solution should be calculated.

Concept introduction:

The number of moles of a substance is related to mass and molar mass as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass of the substance.

Also, according to Avogadro’s law in 1 mol of a substance there are $6.023\times {10}^{23}$ units of that substance. Here, $6.023\times {10}^{23}$ is known as Avogadro’s number and denoted by symbol $N_A$.

The molarity of a solution is defined as number of moles of solute in 1 L of the solution. It is mathematically represented as follows:

$M=\frac{n}{V}$

Here, n is number of moles of solute and V is volume of solution in L. Thus, the unit of molarity is mol/L.

Answer to Problem 18QAP

The molarity of ${\text{Mg}}^{2+}$ and ${\text{PO}}_{\text{4}}{}^{3-}$ is $0.4068\text{ M}$ and $0.2712\text{ M}$ respectively.

Explanation of Solution

The given compound is magnesium phosphate. The mass of compound is 28.0 g and volume of solution is 785 mL. The number of moles of compound can be calculated as follows:

$n=\frac{m}{M}$

Molar mass of magnesium phosphate is 262.85 g/mol thus,

$\begin{array}{c}n=\frac{28.0\text{ g}}{262.85\text{ g/mol}}\\ =0.1065\text{ mol}\end{array}$

Thus, number of moles of potassium oxide is 0.297 mol. Molarity of solution can be calculated as follows;

$M=\frac{n}{V}$

Convert the volume from mL to L.

$V=785\text{ mL}\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=0.785\text{ L}$

Putting the values,

$\begin{array}{c}M=\frac{0.1065\text{ mol}}{0.785\text{ L}}\\ =0.1356\text{ mol/L}\\ \text{=}0.1356\text{ M}\end{array}$

Therefore, molarity of solution is $0.1356\text{ M}$.

The formula of magnesium phosphate is ${\text{Mg}}_3{\left({\text{PO}}_{\text{4}}\right)}_2$. Thus, it has 3 magnesium ions and 2 phosphate ions.

The molarity of ${\text{Mg}}^{2+}$ is $3\times 0.1356=0.4068\text{ M}$ and that of ${\text{PO}}_{\text{4}}{}^{3-}$ ion is $2\times 0.1356=0.2712\text{ M}$.

Therefore, the molarity of ${\text{Mg}}^{2+}$ and ${\text{PO}}_{\text{4}}{}^{3-}$ is $0.4068\text{ M}$ and $0.2712\text{ M}$ respectively.