Question

What is the molarity of each ion present in aqueous solutions of the following compounds prepared by dissolving \(28.0 \mathrm{~g}\) of each compound in water to make \(785 \mathrm{~mL}\) of solution? (a) potassium oxide (b) sodium hydrogen carbonate (c) scandium(III) iodite (d) magnesium phosphate

Short answer

Answer: The molarities of the ions present are: (a) [K⁺] = 0.756 M, [O²⁻] = 0.378 M (b) [Na⁺] = 0.424 M, [HCO₃⁻] = 0.424 M (c) [Sc³⁺] = 0.0802 M, [IO₂⁻] = 0.240 M (d) [Mg²⁺] = 0.345 M, [PO₄³⁻] = 0.230 M

Step-by-step solution

Find the moles of each compound

Find the moles of each compound by dividing their mass by their molar mass. Use the periodic table to obtain the atomic masses. (a) Potassium oxide (K₂O): Molar mass of K₂O = 2*39.10 (K) + 16.00 (O) = 94.20 g/mol Moles(K₂O) = 28.0g / 94.20 g/mol ≈ 0.297 mol (b) Sodium hydrogen carbonate (NaHCO₃): Molar mass of NaHCO₃ = 22.99 (Na) + 1.008 (H) + 12.01 (C) + 3 * 16.00 (O) = 84.008 g/mol Moles(NaHCO₃) = 28.0 g / 84.008g/mol ≈ 0.333 mol (c) Scandium(III) iodite (Sc(IO₂)₃): Molar mass of Sc(IO₂)₃ = 44.956 (Sc) + 3 * (127.60 (I) + 2*16.00 (O)) = 444.156 g/mol Moles(Sc(IO₂)₃) = 28.0g / 444.156 g/mol ≈ 0.0630 mol (d) Magnesium phosphate (Mg₃(PO₄)₂): Molar mass of Mg₃(PO₄)₂ = 3 * 24.305 (Mg) + 2 * (30.974 (P) + 4 * 16.00 (O)) = 310.177 g/mol Moles(Mg₃(PO₄)₂) = 28.0g / 310.177 g/mol ≈ 0.0903 mol

Convert the volume to liters

Convert the volume of the solution from milliliters to liters: Volume(L) = 785 mL * (1 L / 1000 mL) = 0.785 L

Find the molarities of each compound

Divide the moles of each compound by the volume of the solution to find their molarities: (a) [K₂O] = 0.297 mol / 0.785 L ≈ 0.378 M (b) [NaHCO₃] = 0.333 mol / 0.785 L ≈ 0.424 M (c) [Sc(IO₂)₃] = 0.0630 mol / 0.785 L ≈ 0.0802 M (d) [Mg₃(PO₄)₂] = 0.0903 mol / 0.785 L ≈ 0.115 M

Calculate molarity of ions from stoichiometry

Determine the molarity of each ion in the solution according to the stoichiometry from dissociation: (a) K₂O dissociates as: K₂O -> 2 K⁺ + O²⁻ [K⁺] = 2 * 0.378 M = 0.756 M [O²⁻] = 0.378 M (b) NaHCO₃ dissociates as: NaHCO₃ -> Na⁺ + HCO₃⁻ [Na⁺] = 0.424 M [HCO₃⁻] = 0.424 M (c) Sc(IO₂)₃ dissociates as: Sc(IO₂)₃ -> Sc³⁺ + 3 IO₂⁻ [Sc³⁺] = 0.0802 M [IO₂⁻] = 3 * 0.0802 M = 0.240 M (d) Mg₃(PO₄)₂ dissociates as: Mg₃(PO₄)₂ -> 3 Mg²⁺ + 2 PO₄³⁻ [Mg²⁺] = 3 * 0.115 M = 0.345 M [PO₄³⁻] = 2 * 0.115 M = 0.230 M The molarity of each ion present is as follows: (a) [K⁺] = 0.756 M, [O²⁻] = 0.378 M (b) [Na⁺] = 0.424 M, [HCO₃⁻] = 0.424 M (c) [Sc³⁺] = 0.0802 M, [IO₂⁻] = 0.240 M (d) [Mg²⁺] = 0.345 M, [PO₄³⁻] = 0.230 M

Key concepts

Stoichiometry

Stoichiometry is a term that may sound complex, but it's like the recipe for a cake in chemistry. Just like you need a certain amount of each ingredient to make the perfect cake, stoichiometry tells you how much reactant you need to produce a certain amount of product.

When we talk about stoichiometry in relation to molarity calculations, we're looking at how compounds break apart in solution. The moles of each component in a dissolved compound are directly related to the compound's coefficients in the chemical equation. For example, in case (a) with potassium oxide, the stoichiometry is critical to determine that for every 1 mole of K₂O, there are 2 moles of K⁺ and 1 mole of O²⁻ in the solution. Generally speaking, understanding stoichiometry is crucial for identifying the proportions of reactants and products in chemical reactions.

Dissolving Compounds in Solution

Dissolving compounds in solution may seem like magic, but it's really a bit like making instant coffee—you add the solid to the water and it's ready to go. In chemistry, when a solid, such as sodium hydrogen carbonate from example (b), is dissolved in water, we break it down into ions. These ions are the actual substances we often need to analyze.

The concentration of these ions in a solution is given by molarity, which is moles of solute per liter of solution. Think of it just like how strong your coffee is based on how much instant coffee you put in your mug. The more compound you dissolve, the 'stronger' (or more concentrated) the solution becomes. Managing the dissolving process and understanding the resulting changes in concentration is a key skill in chemistry.

Molar Mass Calculation

Knowing molar mass is like knowing the weight of a bag of chocolates—you need it to share the bag equally with your friends. In chemistry, molar mass is the weight of one mole of a substance, and is usually given in grams per mole (g/mol).

To calculate it, just like summing up the total weight of different types of chocolates in the bag, we sum the atomic masses of all the atoms in a molecule. This step is fundamental in stoichiometry as it allows us to convert grams of a substance to moles, which we then use in molarity calculations, as seen in the earlier examples. And vice versa, if we know how many moles we need, we can find out how many grams to weigh out.

Chemical Dissociation

When chemical compounds dissolve, they go through a 'break-up' phase called dissociation. Just like a dance team splits into individual dancers, ionic compounds split into their ions when dissolved in water. Scandium(III) iodite (Sc(IO₂)₃) from example (c) breaks up into Sc³⁺ ions and IO₂⁻ ions.

This knowledge isn't just academic; it affects how the substances behave in the solution. For instance, these individual ions can now conduct electricity or react with other substances in the solution. Understanding this concept is crucial when studying chemistry because it helps us predict how a solution will react under different conditions, determine what kind of reactions might occur, and calculate the concentrations of individual ions.