Chapter 12, Problem 12.95QP

A sample of potassium aluminum sulfate 12-hydrate. KAl(SO₄)₂·12H₂O, containing 101.5 mg is dissolved in 1.000 L of solution Calculate the following for the solution:

1. a The molarity of KAl(SO₄)₂.
2. b The molarity of SO₄²⁻.
3. c The molality of KAl(SO₄)_­2, assuming that the density of the solution is 1.00 g/mL

Interpretation Introduction

Interpretation:

From a sample of $KAl{(S{O}_4)}_2{\text{.12H}}_2\text{O}$, Molarity of $KAl{(S{O}_4)}_2,{\text{ SO}}_4{}^{2-}$ and molality of $KAl{(S{O}_4)}_2$ has to be calculated.

Concept Introduction:

Molality is one of the parameters in expressing the concentration of a solution.  It is expressed as,

$\text{Molality = }\frac{number\text{ of }molesof\text{ solute}}{mass\text{ of solvent in kg}}$

Molarity is one of the in expressing the 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 12.95QP

Molarity of $KAl{(S{O}_4)}_2$ is calculated as $0.0002140M.$

Molarity of $S{O}_4{}^{2-}$ is calculated as $0.000428M.$

Molality of $KAl{(S{O}_4)}_2$ is calculated as $0.215m.$

Explanation of Solution

Given that volume of solution is $1.000\text{ L}$ in which $101.5\text{ mg}$ of $KAl{(S{O}_4)}_2{\text{.12H}}_2\text{O}$ is dissolved.  To determine the molar concentration of the given compounds/ion, we need to know their number of moles.

Mass and molar mass of $KAl{(S{O}_4)}_2{\text{.12H}}_2\text{O}$ are $101.5\text{ mg(= 0}\text{.1015 g)}$ and $474.4\text{ g/mol}$ respectively.

Calculate the number of moles of $KAl{(S{O}_4)}_2{\text{.12H}}_2\text{O}$

$\begin{array}{l}no.{\text{ of moles of KAl(SO}}_4{\text{)}}_2{\text{.12H}}_2\text{O}\text{= }\frac{mass}{molar\text{ mass }}\\ =\frac{0.1015\text{ g}}{474.4\text{ g/mol}}\\ =\text{ 0}\text{.0002140 mol}\end{array}$

One mole of $KAl{(S{O}_4)}_2{\text{.12H}}_2\text{O}$ contains one mole of $KAl{(S{O}_4)}_2$. Hence its molarity of $KAl{(S{O}_4)}_2$ is calculated as –

$\begin{array}{l}\text{Molarity = }\frac{number\text{ of }moles}{volume\text{ of solution in L}}\\ \text{ = }\frac{0.0002140\text{ mol}}{1.000\text{ L}}\text{ = 0}\text{.0002140 M}\end{array}$

One mole of $KAl{(S{O}_4)}_2{\text{.12H}}_2\text{O}$ contains two moles of $S{O}_4{}^{2-}$. Hence its molarity of $S{O}_4{}^{2-}$ is calculated as –

$\begin{array}{l}\text{Molarity}\text{= }\frac{2\times number\text{ of }moles{\text{ of K(AlSO}}_4)}{volume\text{ of solution in L}}\\ \text{= }\frac{2\times 0.0002140\text{ mol}}{1.000\text{ L}}\text{ = 0}\text{.000428 M}\end{array}$

Given that density of solution is $1.00\text{ g/mL}$.  Mass of $1.000\text{ L}$ solution is equivalent to $1.000\text{ g}\text{.}$  mass of the solute $KAl{(S{O}_4)}_2{\text{.12H}}_2\text{O}$ is $0.1015\text{ g}$.  The solute is in hydrated form that molecules has 12 water molecules that mass of those twelve water molecules has to be reduced to get actual mass of the solute.  It is known that $0.04625\text{ g}$ is mass of water in the solute.  Hence the actual mass of solute is $0.1015\text{ g}-0.04625\text{ g = 0}\text{.05525 g}$.

$\begin{array}{l}Massofthesolution\text{= mass of solute + mass of solvent}\\ 1.000\text{ g}\text{= 0}\text{.05525 g + mass of solvent}\\ \text{mass of solvent}\text{= 1}\text{.000 g - 0}\text{.05525 g}\\ \text{= 0}\text{.94475 g = 0}\text{.00099475 kg}\end{array}$

Hence molality of $KAl{(S{O}_4)}_2$ is,

$\begin{array}{l}\text{Molality}\text{= }\frac{number\text{ of }moles{\text{ of K(AlSO}}_4)}{mass\text{ of solvent in kg}}\\ \text{= }\frac{0.0002140\text{ mol}}{0.00099475\text{ kg}}\text{ = 0}\text{.215 m}\end{array}$

Conclusion

Number of moles of the solute $KAl{(S{O}_4)}_2{\text{.12H}}_2\text{O}$ is the key parameter to determine the molarity of $KAl{(S{O}_4)}_2,{\text{ SO}}_4{}^{2-}$ and molality of $KAl{(S{O}_4)}_2$.