Question

Determine the concentration of the ion indicated in each solution. (a) \(\left[\mathrm{K}^{+}\right]\) in \(0.238 \mathrm{M} \mathrm{KNO}_{3} ;\) (b) \(\left[\mathrm{NO}_{3}\right]\) in \(0.167 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ;(\mathrm{c})\left[\mathrm{Al}^{3+}\right]\) in \(0.083 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3};\) (d) \(\left[\mathrm{Na}^{+}\right]\) in \(0.209 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\).

Short answer

The concentrations are as follows: (a) \([ \mathrm{K}^{+} ]= 0.238 \mathrm{M} \), (b) \([ \mathrm{NO}_{3}^{-} ]= 0.334 \mathrm{M} \), (c) \([ \mathrm{Al}^{3+} ]= 0.166 \mathrm{M} \), (d) \([ \mathrm{Na}^{+} ]= 0.627 \mathrm{M} \)

Step-by-step solution

Understanding Dissociation

For an ionic compound, when it dissolves in water, it dissociates into its ions. The stoichiometry of the reaction will tell how many of each ion are produced.

Determine the concentration of \( \mathrm{K}^{+} \) ions in \(0.238 \mathrm{M} \mathrm{KNO}_{3}\)

Potassium nitrate (\( \mathrm{KNO}_{3} \)) dissociates in water to yield potassium (\( \mathrm{K}^{+} \)) and nitrate (\( \mathrm{NO}_{3}^{-} \)) ions in a 1:1 ratio. Therefore, the concentration of \( \mathrm{K}^{+} \) ions will also be \(0.238 \mathrm{M}.\)

Determine the concentration of \( \mathrm{NO}_{3}^{-} \) ions in \(0.167 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)

The calcium nitrate (\( \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \)) dissociates in water yielding calcium (\( \mathrm{Ca}^{2+} \)) and nitrate ions (\( \mathrm{NO}_{3}^{-} \)), but due to the subscript '2' in the formula, two nitrate ions are produced per molecule. Therefore, the concentration of \( \mathrm{NO}_{3}^{-} \) ions will be \(2 \times 0.167 = 0.334 \mathrm{M}. \)

Determine the concentration of \( \mathrm{Al}^{3+} \) ions in \(0.083 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)

The aluminum sulfate (\( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \)) dissociates in water to give aluminum (\( \mathrm{Al}^{3+} \)) and sulfate (\( \mathrm{SO}_{4}^{2-} \)) ions. There are two aluminum ions for each formula unit in solution. Therefore, the concentration of \( \mathrm{Al}^{3+} \) ions will be \(2 \times 0.083 = 0.166 \mathrm{M}. \)

Determine the concentration of \( \mathrm{Na}^{+} \) ions in \(0.209 \mathrm{M} \mathrm{Na}_{3}\mathrm{PO}_{4}\)

The sodium phosphate (\( \mathrm{Na}_{3}\mathrm{PO}_{4} \)) dissociates in water to yield sodium (\( \mathrm{Na}^{+} \)) and phosphate (\( \mathrm{PO}_{4}^{3-} \)) ions. There are three sodium ions for each formula unit dissolved. Therefore, the concentration of \( \mathrm{Na}^{+} \) ions will be \(3 \times 0.209 = 0.627 \mathrm{M}.\)

Key concepts

Molarity Calculations

Molarity is a measure of how much solute is dissolved in a given volume of solution. It's expressed as moles of solute per liter of solution (mol/L), and it's denoted by the capital letter M. Calculating molarity involves knowing the amount of solute and the total volume of the solution it's in.
For example, if you're given a chemical like potassium nitrate (KNO3), which dissociates in water, you'll be told its molarity, say 0.238 M. This number tells you that there are 0.238 moles of KNO3 in every liter of solution.

• If you were to transfer 1 L of this solution into another container, you would have 0.238 moles of KNO3 in that container.

Molarity is a handy way to express concentration because it directly links the amount of substance (moles) with the volume ( V ).

Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the quantities of reactants and products in chemical reactions. When dealing with dissociation of ionic compounds, stoichiometry can help you figure out the ratio of ions produced.
For example, consider calcium nitrate, Ca(NO3)2. When it dissolves in water, it dissociates into 1 calcium ion ( Ca2+ ) and 2 nitrate ions ( NO3- ). The subscript '2' next to NO3 indicates that for every 1 formula unit of calcium nitrate, 2 nitrate ions are released. Thus, if you have a 0.167 M solution of Ca(NO3)2, the concentration of NO3- ions will be 2 times 0.167 M, which is 0.334 M.

• The stoichiometric ratio shows us that double the concentration of nitrate ions is generated from calcium nitrate.

Knowing these ratios is essential for determining the concentrations of individual ions in a solution.

Ion Concentration Determination

Determining the concentration of ions in a solution is an essential skill in chemistry. Once a compound dissolves in water, it separates into its constituent ions. Understanding the dissociation and the result of the stoichiometric relationships allows you to calculate ion concentrations.
Let's look at aluminum sulfate: Al2(SO4)3. When it dissociates, it forms 2 aluminum ions ( Al3+ ) and 3 sulfate ions ( SO4^2- ) for each formula unit. This means if you start with a 0.083 M solution of Al2(SO4)3, the concentration of Al3+ ions will be 2 times 0.083 M, resulting in 0.166 M.

• The subscript '2' in the formula indicates that there are 2 aluminum ions per formula unit, and this directly informs the calculation of concentration

Understanding how a compound dissociates and using stoichiometry to determine these values form the basis of being able to calculate the concentrations of ions in a variety of chemical solutions.