WJEC Chemistry for AS Level Student Book: 2nd Edition (Draft)

WJEC Chemistry for AS Level

Concentrations of solutions The concentration of a solution measures how much of a dissolved substance is present per unit volume of a solution. A solution with a large quantity of solute in a small quantity of solvent is described as concentrated. A solution with a small quantity of solute in a large quantity of solvent is described as dilute. The concentration of a solution can be stated in many ways, e.g. the concentration of ions in a bottle of mineral water is stated in mg/litre, but the most convenient way is to state the amount in moles of a solid present in 1dm 3 of solution. i.e. concentration = amount of moles of solute volume of solution  or  c = n V and the unit is moles per cubic decimetre or mol dm −3 . This expression can be rearranged in two ways to find the amount in moles or the volume, e.g. 2.1g of sodium hydrogencarbonate, NaHCO 3 , is dissolved in 250cm 3 of water. What is the concentration of the solution in mol dm −3 ? Molar mass of NaHCO 3  = 84 Amount of moles NaHCO 3  = 2.1 84  = 0.025 Volume of water is 250cm 3  = 250 1000  = 0.25dm 3 Concentration = 0.025 0.25  = 0.100mol dm −3 Another way of expressing concentration is by using solubility. Solubility is usually measured in g/100g water and since the density of water is 1gcm −3 , this is the same as g/100cm 3 . Therefore to change solubility into concentration in moldm −3 the following steps are used. Step 1 Change mass into moles (divide by molar mass). Step 2 Change 100cm 3 into dm 3 (divide by 1000). Step 3 Divide the moles by the volume. For example, the solubility of potassium nitrate, KNO 3 , is 31.6 g/100g water. What is the concentration in mol dm −3 ? Step 1 Acid–base titration calculation Volumetric analysis is a means of finding the concentration of a solution. An acid–base titration is a type of volumetric analysis. In an acid–base titration, the reacting volumes of an acidic solution and a basic solution are accurately determined. If the concentration of one of these solutions is known, by using the stoichiometric ratios of the solutions, the concentration of the unknown solution can be determined. Again, there are three main steps to follow: Step 1 Find the amount of moles of the solution for which you know the concentration (e.g. acid). Step 2 Use a balanced chemical equation to give the stoichiometric (mole) ratio between the acid and base. Step 3 Calculate the concentration of the second solution (e.g. base) from the known volume and amount in moles. Molar mass KNO 3  = 101.1 Amount in moles = 31.6 101.1  = 0.313 Step 2 100cm 3 water = 0.100dm 3 Step 3 Concentration = 0.313 0.100  = 3.13mol dm −3

Study point Solute is the dissolved substance. The liquid in which the solute dissolves is the solvent (usually water).

Maths tip Remember 1 litre = 1dm 3 = 1000cm 3 . 1cm 3 = 1ml. To change cm 3 into dm 3 you must divide by 1000.

Stretch & challenge The concentration of magnesium ions in a sample of Welsh mineral water is 15mg/litre. Calculate the concentration in mol dm –3 .

Study point

Learn the equations that connect amount of substance and concentration for a solution: n  =  cV  or  c  =  n V  or  V  =  n c Remember if v is given in cm 3 divide by a 1000 to change it into dm 3 .

13 Knowledge check Calculate the concentration, in mol dm –3 , of 0.037g of calcium hydroxide in 200cm 3 of solution.

Link DRAFT Acid-base titrations pages 83–87 42

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