Question

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. $0.0200\mathrm{mol}$ of sodium phosphate in $10.0\mathrm{mL}$ of solution b. $0.300\mathrm{mol}$ of barium nitrate in $600.0\mathrm{mL}$ of solution c. $1.00\mathrm{g}$ of potassium chloride in $0.500\mathrm{L}$ of solution d. $132\mathrm{g}$ of ammonium sulfate in $1.50\mathrm{L}$ of solution

Short answer

The concentrations of the ions present in each solution are: a. Sodium phosphate, Na₃PO₄: $$6.0\mathrm{M}$$ Na⁺ and $$2.0\mathrm{M}$$ PO₄³⁻. b. Barium nitrate, Ba(NO₃)₂: $$0.500\mathrm{M}$$ Ba²⁺ and $$1.00\mathrm{M}$$ NO₃⁻. c. Potassium chloride, KCl: $$0.0268\mathrm{M}$$ K⁺ and $$0.0268\mathrm{M}$$ Cl⁻. d. Ammonium sulfate, (NH₄)₂SO₄: $$1.33\mathrm{M}$$ NH₄⁺ and $$0.667\mathrm{M}$$ SO₄²⁻.

Step-by-step solution

Solution a: Calculate the concentrations of ions in sodium phosphate solution

First, convert 10.0 mL volume to L: 10.0 mL * (1 L / 1000 mL) = 0.0100 L Now, calculate the concentrations of ions: Na₃PO₄ → 3Na⁺ + PO₄³⁻ Concentration of Na⁺ = [(0.0200 mol * 3) / 0.0100 L] = 6.0 M Concentration of PO₄³⁻ = [(0.0200 mol * 1) / 0.0100 L] = 2.0 M

Solution b: Calculate the concentrations of ions in barium nitrate solution

First, convert 600.0 mL volume to L: 600.0 mL * (1 L / 1000 mL) = 0.600 L Now, calculate the concentrations of ions: Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻ Concentration of Ba²⁺ = [(0.300 mol * 1) / 0.600 L] = 0.500 M Concentration of NO₃⁻ = [(0.300 mol * 2) / 0.600 L] = 1.00 M

Solution c: Calculate the concentrations of ions in potassium chloride solution

First, calculate the moles of KCl in the solution: Molar mass of KCl = 39.10 (K) + 35.45 (Cl) = 74.55 g/mol moles of KCl = 1.00 g / 74.55 g/mol = 0.0134 mol Now, calculate the concentrations of ions: KCl → K⁺ + Cl⁻ Concentration of K⁺ = [(0.0134 mol * 1) / 0.500 L] = 0.0268 M Concentration of Cl⁻ = [(0.0134 mol * 1) / 0.500 L] = 0.0268 M

Solution d: Calculate the concentrations of ions in ammonium sulfate solution

First, calculate the moles of (NH₄)₂SO₄ in the solution: Molar mass of (NH₄)₂SO₄ = (2 * (14.01 + (4 * 1.01))) + (32.07 + (4 * 16.00)) = 132.14 g/mol moles of (NH₄)₂SO₄ = 132 g / 132.14 g/mol = 1.00 mol Now, calculate the concentrations of ions: (NH₄)₂SO₄ → 2NH₄⁺ + SO₄²⁻ Concentration of NH₄⁺ = [(1.00 mol * 2) / 1.50 L] = 1.33 M Concentration of SO₄²⁻ = [(1.00 mol * 1) / 1.50 L] = 0.667 M

Key concepts

Ion Concentration Calculation

When dealing with solutions of strong electrolytes, understanding how to calculate the concentration of ions is essential. Strong electrolytes dissociate completely in water, splitting into their constituent ions. To find the concentration of ions, you need to follow these steps:

• Identify the formula for the electrolyte and determine how it dissociates into ions. For instance, sodium phosphate (${\text{Na}}_3{\text{PO}}_4$) dissociates into three sodium ions (${\text{Na}}^{+}$) and one phosphate ion (${\text{PO}}_4^{3-}$).
• Convert the given volume of the solution to liters if necessary, as this will be needed to calculate molarity.
• Calculate the concentration of each ion by using the formula:$$\text{Concentration of ion}=\frac{\text{moles of ion}}{\text{volume of solution in L}}$$
• Multiply the moles of the compound by the number of each type of ion it produces upon dissociation.

Understanding this process enables you to accurately quantify the amount of each ion present in a chemical solution.

Molarity

Molarity is a key concept when dealing with solutions and is defined as the number of moles of solute per liter of solution. This is an important metric because it tells us how concentrated a solution is. When calculating molarity, the following steps are involved:

• First, you must convert the volume of the solution to liters if it isn't already.
• Next, determine the number of moles of the solute. If given in grams, the moles can be calculated using the molar mass of the solute. For example, if you have 132 g of ammonium sulfate ($(N{H}_4{)}_{2}S{O}_4$), you use its molar mass (132.14 g/mol) to find 1.00 mole of the compound.
• Finally, use the formula:$$\text{Molarity (M)}=\frac{\text{moles of solute}}{\text{liters of solution}}$$

This calculation tells us the strength of a solution, which is crucial in chemistry when reactions depend on the concentration of solutions.

Chemical Solutions

Chemical solutions consist of a solute, which is the substance being dissolved, and a solvent, which is the medium that dissolves the solute. In most cases, the solvent is water, especially when dealing with aqueous solutions. Here are some basics:

• The complexity of chemical solutions arises from the varied nature of solutes and their interactions with the solvents.
• Understanding the properties of chemical solutions, such as conductivity and pH, can vary widely depending on the concentration and type of solute used.
• In the context of strong electrolytes, as the solute dissolves, it fully dissociates into its constituent ions.

The dissolution and dissociation process impacts how chemical solutions are used in practical settings, such as in titrations or other chemical reactions, showcasing the versatile and complex nature of these fluids.

Chemical Equations

Chemical equations are representations of chemical reactions, where the reactants transform into products. For strong electrolytes, these equations represent the dissociation into ions. Here's how they work:

• The chemical formula of a compound shows the ratio of each element that composes the compound. For instance, potassium chloride ($\text{KCl}$) shows a 1:1 ratio of potassium to chloride ions.
• A balanced chemical equation respects the conservation of mass, meaning the number of each type of atom on the reactants side is equal to the number on the products side.
• In a dissociation equation, emphasis is put on displaying the ions separately, such as ${\text{Ba(NO}}_3{)}_2\to {\text{Ba}}^{2+}+2{\text{NO}}_3^{-}$, indicating that one barium nitrate turns into one barium ion and two nitrate ions.

Understanding chemical equations allows one to predict and balance chemical reactions effectively, especially when calculating the outcome of reactions and the resulting concentrations of products.