Question

Calculate the concentrations of each of the ions in (a) \(0.25 M \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{2}\), (b) \(0.10 \mathrm{M} \mathrm{CuSO}_{4}\), (c) \(0.16 \mathrm{M}\) \(\mathrm{Na}_{3} \mathrm{PO}_{4},\) (d) \(0.075 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)

Short answer

Ion concentrations: (a) 0.25 M \( \mathrm{Cr}^{2+} \), 0.50 M \( \mathrm{NO}_{3}^{-} \); (b) 0.10 M \( \mathrm{Cu}^{2+} \), 0.10 M \( \mathrm{SO}_{4}^{2-} \); (c) 0.48 M \( \mathrm{Na}^{+} \), 0.16 M \( \mathrm{PO}_{4}^{3-} \); (d) 0.15 M \( \mathrm{Al}^{3+} \), 0.225 M \( \mathrm{SO}_{4}^{2-} \).

Step-by-step solution

Understand the formula for Chromium(II) nitrate

Recognize that the chemical formula of Chromium(II) nitrate is \( \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{2} \). This means each formula unit is composed of one Chromium ion (\( \mathrm{Cr}^{2+} \)) and two nitrate ions (\( \mathrm{NO}_{3}^{-} \)).

Calculate ion concentrations for Chromium(II) nitrate

Since the initial concentration of Chromium(II) nitrate is 0.25 M, and the compound dissociates into one Chromium ion and two nitrate ions, the concentration of \( \mathrm{Cr}^{2+} \) is also 0.25 M and the concentration of \( \mathrm{NO}_{3}^{-} \) ions is 0.25 M * 2 = 0.50 M.

Understand the formula for Copper(II) sulfate

Identify that the chemical formula of Copper(II) sulfate is \( \mathrm{CuSO}_{4} \). It dissociates into one Copper ion (\( \mathrm{Cu}^{2+} \)) and one sulfate ion (\( \mathrm{SO}_{4}^{2-} \)).

Calculate ion concentrations for Copper(II) sulfate

Given the 0.10 M concentration of Copper(II) sulfate, upon dissociation, it results in the same concentration of 0.10 M for both Copper and sulfate ions, since there is a 1:1 ratio between them.

Understand the formula for Sodium phosphate

Examine that the chemical formula for Sodium phosphate is \( \mathrm{Na}_{3}PO_{4} \). It breaks down into three sodium ions (\( \mathrm{Na}^{+} \)) and one phosphate ion (\( \mathrm{PO}_{4}^{3-} \)).

Calculate ion concentrations for Sodium phosphate

With a starting concentration of 0.16 M for Sodium phosphate, the dissociation yields a concentration of 0.16 M * 3 = 0.48 M for Sodium ions and 0.16 M for phosphate ions.

Understand the formula for Aluminum sulfate

Notice that the chemical formula for Aluminum sulfate is \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \). Upon dissociation, it produces two aluminum ions (\( \mathrm{Al}^{3+} \)) and three sulfate ions (\( \mathrm{SO}_{4}^{2-} \)).

Calculate ion concentrations for Aluminum sulfate

Given the concentration of Aluminum sulfate is 0.075 M, the concentration of Aluminum ions will be 0.075 M * 2 = 0.15 M, and for sulfate ions, it will be 0.075 M * 3 = 0.225 M.

Key concepts

Chemical Formula Dissociation

Understanding chemical formula dissociation is crucial when examining how salts and certain molecular compounds behave in solution. Dissociation is the process by which ionic compounds separate into their individual ions when they dissolve in water. This happens because water molecules are polar, meaning they have a positive end and a negative end, which allows them to surround and pull apart the ions that make up the compound.

For instance, when Chromium(II) nitrate, represented by the formula \( \mathrm{Cr}(\mathrm{NO}_{3})_{2} \), dissolves in water, it separates into one Chromium ion \( \mathrm{Cr}^{2+} \) and two nitrate ions \( \mathrm{NO}_{3}^{-} \). Each ion becomes surrounded by water molecules, which stabilize the ions in solution. It's important to note that the subscript number in a chemical formula indicates the number of ions present when the compound dissociates. As such, a balanced equation showing the dissociation of Chromium(II) nitrate into its ions is written as follows:\[ \mathrm{Cr}(\mathrm{NO}_{3})_{2} \rightarrow \mathrm{Cr}^{2+} + 2\mathrm{NO}_{3}^{-} \]

Molarity

Molarity, denoted by the symbol \( M \), measures a solution’s concentration in moles of solute per liter of solution. It determines how many moles, the unit for amount of substance, are present in one liter of solution. Calculating the molarity of an aqueous solution involves determining the number of moles of the solute and dividing by the volume of the solution in liters.

The formula to calculate molarity is:\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]. When applied to the dissociation problem, if you have a 0.25 M solution of Chromium(II) nitrate, this means there are 0.25 moles of Chromium(II) nitrate dissolved in every liter of the solution. For compounds that dissociate into multiple ions, such as \( \mathrm{Cr}(\mathrm{NO}_{3})_{2} \), we must also consider the stoichiometry to calculate the molarity of each ion after dissociation.

Stoichiometry

Stoichiometry revolves around the quantitative relationships between the substances as they are used and produced in chemical reactions. By using stoichiometry, we can predict the amounts of products formed or reactants needed in a given chemical reaction.

For a dissociation reaction, the stoichiometry tells us the ratio of ions produced from the compound. For instance, the compound \( \mathrm{Na}_{3}\mathrm{PO}_{4} \) dissociates to produce three sodium ions \( \mathrm{Na}^{+} \) for every one phosphate ion \( \mathrm{PO}_{4}^{3-} \). This 3:1 ratio is vital for calculating the concentration of each ion type in solution.Using the molarity of the original compound (e.g., 0.16 M \( \mathrm{Na}_{3}\mathrm{PO}_{4} \)) and the stoichiometric ratios, we can calculate that the sodium ion concentration will be triple that of the original compound (0.16 M multiplied by 3), and the phosphate ion concentration will remain the same as that of the original compound due to the 1:1 ratio.