Question

What is the molarity of each ion in a solution prepared by dissolving \(0.550 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}, 1.188 \mathrm{~g}\) of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(0.223 \mathrm{~g}\) of \(\mathrm{Li}_{2} \mathrm{SO}_{4}\) in water and diluting to a volume of \(100.00 \mathrm{~mL}\) ?

Short answer

Molarities: \(\text{Na}^+ = 0.2949 \text{ M}\), \(\text{SO}_4^{2-} = 0.0590 \text{ M}\), \(\text{PO}_4^{3-} = 0.0725 \text{ M}\), \(\text{Li}^+ = 0.0406 \text{ M}\).

Step-by-step solution

Determine Molar Masses

First, we need to determine the molar mass of each compound: - Sodium sulfate (\(\text{Na}_2\text{SO}_4\)) has a molar mass of \(2(23.0)+32.1+4(16.0) = 142.1 \text{ g/mol}.\)- Sodium phosphate (\(\text{Na}_3\text{PO}_4\)) has a molar mass of \(3(23.0)+31.0+4(16.0) = 163.9 \text{ g/mol}.\)- Lithium sulfate (\(\text{Li}_2\text{SO}_4\)) has a molar mass of \(2(6.9)+32.1+4(16.0) = 109.9 \text{ g/mol}.\)

Convert Mass to Moles

Convert the mass of each compound to moles using their molar masses:- For \(\text{Na}_2\text{SO}_4\): \(\dfrac{0.550\text{ g}}{142.1 \text{ g/mol}} = 0.00387 \text{ mol}.\)- For \(\text{Na}_3\text{PO}_4\): \(\dfrac{1.188\text{ g}}{163.9 \text{ g/mol}} = 0.00725 \text{ mol}.\)- For \(\text{Li}_2\text{SO}_4\): \(\dfrac{0.223\text{ g}}{109.9 \text{ g/mol}} = 0.00203 \text{ mol}.\)

Calculate Ion Concentration for Na

Each compound dissociates in water releasing ions. Calculate the molarity of each ion separately using the dissolved moles and the total solution volume: - From \(\text{Na}_2\text{SO}_4\), 2 moles of \(\text{Na}^+\) are released for each mole of compound, so the contribution is \(2\times0.00387\text{ mol} = 0.00774\text{ mol}.\)- From \(\text{Na}_3\text{PO}_4\), 3 moles of \(\text{Na}^+\) for each mole, contributing \(3\times0.00725\text{ mol} = 0.02175\text{ mol}.\)- Total moles of \(\text{Na}^+\) are \(0.00774 + 0.02175 = 0.02949 \text{ mol}.\)Convert to molarity: \(\dfrac{0.02949\text{ mol}}{0.1 \text{ L}} = 0.2949 \text{ M}.\)

Calculate Ion Concentration for SO4 and PO4

Calculate the molarity of \(\text{SO}_4^{2-}\):- From \(\text{Na}_2\text{SO}_4\), 1 mole of \(\text{SO}_4^{2-}\) is released per mole, contributing \(0.00387 \text{ mol}.\)- From \(\text{Li}_2\text{SO}_4\), 1 mole of \(\text{SO}_4^{2-}\) per mole, contributing \(0.00203 \text{ mol}.\)- Total moles of \(\text{SO}_4^{2-}\) are \(0.00387 + 0.00203 = 0.00590 \text{ mol}.\)Convert to molarity: \(\dfrac{0.00590\text{ mol}}{0.1 \text{ L}} = 0.0590 \text{ M}.\)Calculate the molarity of \(\text{PO}_4^{3-}\):- From \(\text{Na}_3\text{PO}_4\), 1 mole of \(\text{PO}_4^{3-}\) is released per mole, contributing \(0.00725 \text{ mol}.\)Convert to molarity: \(\dfrac{0.00725\text{ mol}}{0.1 \text{ L}} = 0.0725 \text{ M}.\)

Calculate Ion Concentration for Li

Calculate the molarity of \(\text{Li}^+\):- From \(\text{Li}_2\text{SO}_4\), 2 moles of \(\text{Li}^+\) are released per mole, contributing \(2\times0.00203 \text{ mol} = 0.00406 \text{ mol}.\)Convert to molarity: \(\dfrac{0.00406 \text{ mol}}{0.1 \text{ L}} = 0.0406 \text{ M}.\)

Key concepts

Ion Concentration

Understanding ion concentration is essential in chemistry, especially when dealing with mixtures in solutions. It refers to the concentration of individual ions in a solution, usually expressed in molarity (M), which is moles per liter.
When compounds dissolve in water, they dissociate into ions. For example, sodium sulfate (\(\text{Na}_2\text{SO}_4\)) dissociates into two sodium ions (\(\text{Na}^+\)) and one sulfate ion (\(\text{SO}_4^{2-}\)). Similarly, sodium phosphate (\(\text{Na}_3\text{PO}_4\)) releases three sodium ions (\(\text{Na}^+\)) and one phosphate ion (\(\text{PO}_4^{3-}\)).

To determine the ion concentration, follow these steps:

• Calculate the moles of each compound added to the solution.
• Use stoichiometry to determine how many moles of each ion result from the dissociation.
• Convert the moles of ions to molarity by dividing by the volume of the solution in liters.

Ion concentration helps predict how a solution will behave in chemical reactions, as different ions can react to form new compounds.

Molar Mass

Molar mass is a fundamental concept in chemistry, necessary for converting between grams and moles of a substance. It's the total mass of all atoms in a molecule, expressed in grams per mole (g/mol).
To find the molar mass of a compound like sodium sulfate (\(\text{Na}_2\text{SO}_4\)):

• Identify the number of each type of atom in the molecule.
• Multiply the atomic mass of each element by the number of atoms of that element in the molecule.
• Sum the masses for all elements to get the compound's molar mass.For example, sodium (\(23.0\ \text{g/mol}\)) appears twice, sulfur (\(32.1\ \text{g/mol}\)) once, and oxygen (\(16.0\ \text{g/mol}\)) four times, giving sodium sulfate a molar mass of \(142.1\ \text{g/mol}\).

Molar mass allows chemists to convert between the mass of a substance and the amount of substance in moles, which is crucial for equations and reactions involving the scalability of chemical reactions.

Dissociation in Water

When compounds are added to water, they often dissociate, splitting into individual ions. This process is key to understanding reactions in aqueous solutions.

For example, sodium sulfate (\(\text{Na}_2\text{SO}_4\)) dissociates into its constituent ions:- Two sodium ions (\(\text{Na}^+\))- One sulfate ion (\(\text{SO}_4^{2-}\))Lithium sulfate (\(\text{Li}_2\text{SO}_4\)) does similarly by releasing two lithium ions (\(\text{Li}^+\)) and one sulfate ion (\(\text{SO}_4^{2-}\)).

Here's how to assess dissociation:

• Identify the ionic species present in the solid compound.
• Determine the ratio in which each ion appears per mole of compound (stoichiometry).
• Understand the stoichiometry to predict the number of ions produced when the compound dissolves.

Dissociation is crucial in predicting the properties and behaviors of solutions, such as conductivity and reactivity in processes like titrations or reaction mechanisms.