IB Topic 1: Quantitative Chemistry 1.5 Solutions

Distinguish between the terms solute, solvent, solution and concentration (g dm-3 and mol dm-3)

Solve problems involving concentration, amount of solute and volume of solution.

Distinguish between the terms solute, solvent, solution and concentration (g dm-3 and mol dm-3)

Solution: Homogeneous mixtures of two or more substances. Most common is solid, liquid or gas dissolved in a liquid (usually water). These are called aqueous solutions. Can have other solutions such as solid-solid (alloy) or gas-gas (air).

Solute: The dissolved particles. Usually the substance in the least amount.

Solvent: The dissolving medium. Usually the substance in the greater amount.

Distinguish between the terms solute, solvent, solution and concentration (g dm-3 and mol dm-3)

Properties of SolutionsSolubility

Solubility is the amount of solute that dissolves in a given amount of solvent at a given temperature to produce a saturated solution. Units: grams solute/100 g solvent

NaCl: solubility of 36.2 g/ 100 g water at 25 oC

Any amount less than that is an unsaturated solution.

A solution that contains more solute than it should theoretically is supersaturated.

Distinguish between the terms solute, solvent, solution and concentration (g dm-3 and mol dm-3)

Concentration: Measure of the amount of solute dissolved in a given amount of solvent.

g dm-3 (g/dm3) Is the number of grams of solute dissolved per dm3 of

solution

mol dm-3 (mol/dm3) Molarity (M) is the number of moles of solute dissolved per dm3 of

solution. M = mol dm-3. Use [ ] to signify concentration in molarity.

Solve problems involving concentration, amount of solute and volume of solution.

Find the concentration in g dm-3 and mol dm-3 of a solution containing 2.00 g sodium hydroxide in 125 cm3 of solution.

g dm-3

125 cm3 = 0.125 dm3

2.00 g/0.125 dm3 = 16.0 g dm-3

mol dm-3

2.00 g NaOH = 2.00 g/ 40.01 g mol-1 = .0500 mol .0500 mol/0.125 dm3 = 0.400 mol dm-3

Solve problems involving concentration, amount of solute and volume of solution.

Calculate the amount of hydrochloric acid (in mol & g) present in 23.65 cm3 of 0.100 mol dm-3 HCl(aq)?

Solve problems involving concentration, amount of solute and volume of solution.

Calculate the amount of hydrochloric acid (in mol & g) present in 23.65 cm3 of 0.100 mol dm-3 HCl(aq)?

Molarity = mol dm-3 so Molarity x dm3 = mol

mol = 0.100 mol dm-3 x .02365 dm-3 = .002365 mol HCl

grams = .002365 mol x 36.46 g mol-1 = .0862 g HCl

Solve problems involving concentration, amount of solute and volume of solution.

What volume of a 1.25 mol dm-3 potassium permanganate solution, KMnO4(aq), contains 28.6 grams KMnO4?

Solve problems involving concentration, amount of solute and volume of solution.

What volume of a 1.25 mol dm-3 potassium permanganate solution, KMnO4(aq), contains 28.6 grams KMnO4?

Molarity = mol dm-3 so dm-3 = mol/Molarity

mol = 28.6 g / 158.04 g mol-1 = .181 mol

dm-3 = .181 mol/1.25 mol dm-3 = .145 dm3

Solve problems involving concentration, amount of solute and volume of solution.

What will be the concentration of the solution formed by mixing 200 cm3 of 3.00 mol dm-3 HCl(aq) with 300 cm3 of 1.50 mol dm-3 HCl(aq)?

Solve problems involving concentration, amount of solute and volume of solution.

What will be the concentration of the solution formed by mixing 200 cm3 of 3.00 mol dm-3 HCl(aq) with 300 cm3 of 1.50 mol dm-3 HCl(aq)?

Find total moles.200 dm-3 x 3.00 mol dm-3 = .600 mol HCl.300 dm-3 x 1.50 mol dm-3 = .450 mol HClTotal moles = 1.050 mol

Find total volume: .200 dm3 + .300 dm3 = .500 dm3

Concentration: 1.050 mol/ .500 dm3 = 2.10 mol dm-3

Terms to Know Element Compound Empirical formula Molecular formula Structural formula Percent composition Percentage yield Molar mass Mole Avogadro’s constant Chemical equation

Molar ratio Ionic equation Solid Liquid Gas Ideal gas equation Molar volume of a gas Solute Solvent Solution Concentration