Chapter 12, Problem 12.96QP

A sample of aluminum sulfate 18-hydrate, Al₂(SO₄)₃. 18H₂O, containing 125.0 mg is dissolved in 1.000 L of solution. Calculate the following for the solution:

1. a The molarity of Al₂(SO₄)₃.
2. b The molarity of SO₄²⁻.
3. c The molality of Al₂(SO₄)₃, assuming that the density of the solution is 1.00 g/mL.

Interpretation Introduction

Interpretation:

From a sample of $A{l}_2{(S{O}_4)}_3{\text{.18H}}_2\text{O}$, Molarity of $A{l}_2{(S{O}_4)}_3,{\text{ SO}}_4{}^{2-}$ and molality of $A{l}_2{(S{O}_4)}_3$ 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 12.96QP

Molarity of $A{l}_2{(S{O}_4)}_3$ is calculated as $1.8756\times 10^{-4}M.$

Molarity of $S{O}_4{}^{2-}$ is calculated as $5.627\times 10^{-4}M.$ S

Molality of $A{l}_2{(S{O}_4)}_3$ is calculated as $0.200m.$

Explanation of Solution

Given that volume of solution is $1.000\text{ L}$ in which $125\text{ mg}$ of $A{l}_2{(S{O}_4)}_3{\text{.18H}}_2\text{O}$ is dissolved.  To determine the molar concentration of the given compounds/ion, we need to know their respective number of moles.

Mass and molar mass of $A{l}_2{(S{O}_4)}_3{\text{.12H}}_2\text{O}$ is $125\text{ mg(= 0}\text{.125 g)}$ and $666.46\text{ g/mol}$ respectively.

Calculate the number of moles of $A{l}_2{(S{O}_4)}_3{\text{.18H}}_2\text{O}$

$\begin{array}{l}no.{\text{ of moles of Al}}_2{{\text{(SO}}_4}_{\text{)}}{\text{.18H}}_2\text{O}\text{= }\frac{mass}{molar\text{ mass }}\\ =\frac{0.125\text{ g}}{666.46\text{ g/mol}}\\ =\text{ 1}\text{.8756}\times {\text{10}}^{-4}\text{ mol}\end{array}$

One mole of $A{l}_2{(S{O}_4)}_3{\text{.18H}}_2\text{O}$ contains one mole of $A{l}_2{(S{O}_4)}_3$. Hence its molarity of $A{l}_2{(S{O}_4)}_3$ is calculated as –

$\begin{array}{l}\text{Molarity }\text{= }\frac{number\text{ of }moles}{volume\text{ of solution in L}}\\ \text{= }\frac{\text{1}\text{.8756}\times {\text{10}}^{-4}\text{ mol}}{1.000\text{ L}}\text{ = 1}\text{.8756}\times {\text{10}}^{-4}\text{ M}\end{array}$

One mole of $A{l}_2{(S{O}_4)}_3{\text{.18H}}_2\text{O}$ contains three moles of $S{O}_4{}^{2-}$. Hence its molarity of $S{O}_4{}^{2-}$ is calculated as –

$\begin{array}{l}\text{Molarity}\text{= }\frac{3\times number\text{ of }moles{\text{ of Al}}_2{{\text{(SO}}_4)}_3}{volume\text{ of solution in L}}\\ \text{= }\frac{3\times 1.8756\times {10}^{-4}\text{ mol}}{1.000\text{ L}}\text{ = 5}\text{.627}\times {\text{10}}^{-4}\text{ M}\end{array}$

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 $A{l}_2{(S{O}_4)}_3{\text{.18H}}_2\text{O}$ is $0.125\text{ g}$.  The solute is in hydrated form that molecules has $18$ water molecules that mass of those $18$ water molecules has to be reduced to get actual mass of the solute.  It is known that $0.06082\text{ g}$ is mass of water in the solute.  Hence the actual mass of solute is $0.125\text{ g}-0.06082\text{ g = 0}\text{.06418 g}$.

$\begin{array}{l}Massofthesolution\text{= mass of solute + mass of solvent}\\ 1.000\text{ g}\text{= 0}\text{.06418 g + mass of solvent}\\ \text{mass of solvent}\text{= 1}\text{.000 g - 0}\text{.06418 g}\\ \text{= 0}\text{.93582 g = 0}\text{.00093582 kg}\end{array}$

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

$\begin{array}{l}\text{Molality}\text{= }\frac{number\text{ of }moles{\text{ of K(AlSO}}_4)}{mass\text{ of solvent in kg}}\\ \text{= }\frac{1.8756\times {10}^{-4}\text{ mol}}{0.00093582\text{ kg}}\text{ = 0}\text{.200 m}\end{array}$

Conclusion

Number of moles of the solute $A{l}_2{(S{O}_4)}_3{\text{.18H}}_2\text{O}$ is the key parameter to determine the molarity of $A{l}_2{(S{O}_4)}_3,{\text{ SO}}_4{}^{2-}$ and molality of $A{l}_2{(S{O}_4)}_3$.