Question

What is the molarity of each ion present in aqueous solutions of the following compounds prepared by dissolving \(20.00 \mathrm{~g}\) of each compound in water to make \(4.50 \mathrm{~L}\) of solution? (a) cobalt(III) chloride (b) nickel(III) sulfate (c) sodium permanganate (d) iron(II) bromide

Short answer

(a) 20.00 g of Cobalt(III) Chloride (\(\text{CoCl}_3\)) dissolved in 4.50 L of water. The molarity of \(\text{Co}^{3+}\) ions is 0.0269 M and the molarity of \(\text{Cl}^-\) ions is 0.0807 M. (b) 20.00 g of Nickel(III) Sulfate (\(\text{Ni}_2(\text{SO}_4)_3\)) dissolved in 4.50 L of water. The molarity of \(\text{Ni}^{3+}\) ions is 0.02196 M and the molarity of \(\text{SO}_4^{2-}\) ions is 0.03294 M. (c) 20.00 g of Sodium Permanganate (\(\text{NaMnO}_4\)) dissolved in 4.50 L of water. The molarity of \(\text{Na}^+\) ions is 0.0313 M and the molarity of \(\text{MnO}_4^-\) ions is 0.0313 M. (d) 20.00 g of Iron(II) Bromide (\(\text{FeBr}_2\)) dissolved in 4.50 L of water. The molarity of \(\text{Fe}^{2+}\) ions is 0.0206 M and the molarity of \(\text{Br}^-\) ions is 0.0413 M.

Step-by-step solution

(a) Cobalt(III) Chloride

1. Cobalt(III) Chloride has the chemical formula \(\text{CoCl}_3\). 2. Molar mass of \(\text{CoCl}_3 = \text{Co} + 3(\text{Cl}) = 58.93 + 3(35.45) = 165.28 \mathrm{g/mol}\). 3. Moles of \(\text{CoCl}_3 = \frac{20.00 \mathrm{g}}{165.28 \mathrm{g/mol}} = 0.121 \mathrm{mol}\). 4. Molarity = \(\frac{0.121 \mathrm{mol}}{4.50 \mathrm{L}} = 0.0269 \mathrm{M}\). 5. For every 1 mole of \(\text{CoCl}_3\), we get 1 mole of \(\text{Co}^{3+}\) and 3 moles of \(\text{Cl}^-\), so their molar concentrations are \(0.0269 \mathrm{M}\) and \(3(0.0269) = 0.0807 \mathrm{M}\), respectively.

(b) Nickel(III) Sulfate

1. Nickel(III) Sulfate has the chemical formula \(\text{Ni}_2(\text{SO}_4)_3\). 2. Molar mass of \(\text{Ni}_2(\text{SO}_4)_3 = 2(\text{Ni}) + 3( \text{S} + 4(\text{O})) = 2(58.69) +3(32.07 + 4(16.00)) = 405.62 \mathrm{g/mol}\). 3. Moles of \(\text{Ni}_2(\text{SO}_4)_3 =\frac{20.00 \mathrm{g}}{405.62 \mathrm{g/mol}} = 0.0493 \mathrm{mol}\). 4. Molarity = \(\frac{0.0493 \mathrm{mol}}{4.50 \mathrm{L}} = 0.01098 \mathrm{M}\). 5. For every 1 mole of \(\text{Ni}_2(\text{SO}_4)_3\), we get 2 moles of \(\text{Ni}^{3+}\) and 3 moles of \(\text{SO}_4^{2-}\), so their molar concentrations are \(2(0.01098) = 0.02196 \mathrm{M}\) and \(3(0.01098) = 0.03294 \mathrm{M}\), respectively.

(c) Sodium Permanganate

1. Sodium Permanganate has the chemical formula \(\text{NaMnO}_4\). 2. Molar mass of \(\text{NaMnO}_4 = \text{Na} + \text{Mn} + 4(\text{O}) = 22.99 + 54.94 + 4(16.00) = 141.93 \mathrm{g/mol}\). 3. Moles of \(\text{NaMnO}_4 = \frac{20.00 \mathrm{g}}{141.93 \mathrm{g/mol}} = 0.141 \mathrm{mol}\). 4. Molarity = \(\frac{0.141 \mathrm{mol}}{4.50 \mathrm{L}} = 0.0313 \mathrm{M}\). 5. For every 1 mole of \(\text{NaMnO}_4\), we get 1 mole of \(\text{Na}^+\) and 1 mole of \(\text{MnO}_4^-\), so their molar concentrations are \(0.0313 \mathrm{M}\) and \(0.0313 \mathrm{M}\), respectively.

(d) Iron(II) Bromide

1. Iron(II) Bromide has the chemical formula \(\text{FeBr}_2\). 2. Molar mass of \(\text{FeBr}_2 = \text{Fe} + 2(\text{Br}) = 55.85 + 2(79.90) = 215.65 \mathrm{g/mol}\). 3. Moles of \(\text{FeBr}_2 = \frac{20.00 \mathrm{g}}{215.65 \mathrm{g/mol}} = 0.0927 \mathrm{mol}\). 4. Molarity = \(\frac{0.0927 \mathrm{mol}}{4.50 \mathrm{L}} = 0.0206 \mathrm{M}\). 5. For every 1 mole of \(\text{FeBr}_2\), we get 1 mole of \(\text{Fe}^{2+}\) and 2 moles of \(\text{Br}^-\), so their molar concentrations are \(0.0206 \mathrm{M}\) and \(2(0.0206) = 0.0413 \mathrm{M}\), respectively.

Key concepts

Molar Concentration

Molar concentration, commonly known as molarity, is a measure of the concentration of a solute in a solution. It is expressed as the number of moles of solute per liter of solution. To calculate molarity, you can use the formula:
\[ Molarity (M) = \frac{moles \ of \ solute}{volume \ of \ solution \ in \ liters} \]
In our textbook example, to find the molarity of ions in a solution, one would start by calculating the moles of the compound dissolved. This is done by dividing the mass of the compound (in grams) by its molar mass (in grams per mole). The resulting value is then divided by the volume of the solution to get the molarity.

• Determine the molar mass of the compound.
• Calculate the moles of the compound by dividing the given mass by the molar mass.
• Divide the moles of compound by the volume of the solution to find the molarity.

The concept is critical in stoichiometry as it allows for the calculation of reactants and products in a chemical reaction. Understanding the molarity concept helps in preparing solutions with precise concentrations, which is essential in various chemical and biological experiments.

Dissolving Ionic Compounds

When ionic compounds dissolve in water, they dissociate into their constituent ions. The process of dissolving encompasses the physical separation and dispersion of the ions throughout the solvent. To predict the concentration of each ion in solution, it is necessary to know the compound’s chemical formula and its solubility behavior.

• Ion Dissociation: An ionic compound like cobalt(III) chloride \((\text{CoCl}_3)\) dissociates into one cobalt ion \((\text{Co}^{3+})\) and three chloride ions \((\text{Cl}^-)\).
• Cation and Anion Molarity: The molarity of each ion is related to the molarity of the original compound. For each mole of the compound dissolved, there will be a stoichiometric amount of cations and anions released.

Water’s polar nature makes it an excellent solvent for ionic compounds. The ions are attracted to the partial charges on the water molecules, encouraging them to separate and interact with the solvent. This interaction is essential for processes such as biological functions and industrial applications.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the conservation of mass and the principles of a balanced equation.

• Stoichiometric Coefficients: These numbers in front of compounds in a balanced equation tell us the ratio at which molecules react or form. When calculating molar concentrations of ions, the stoichiometric coefficients indicate how many moles of each ion are produced per mole of ionic compound dissolved.
• Reaction Ratios: The balanced chemical equation provides the ratio of reactants to products, which allows chemists to predict the amounts of substances consumed and produced in a reaction.

Applying stoichiometry in our exercise involves understanding that when an ionic compound dissolves, the ratio of ions produced comes directly from the coefficients in the compound's formula. Hence, stoichiometry allows us to account for these ratios when calculating the molarity of ions, ensuring accurate and predictable results in chemical reactions and solutions.