Concentration Concentration of a solution is a measure of the amount of a particular solute in a given amount of the solution. Chemists use different measures of concentration depending on the particular situation –

Units of concentration % m/m (%w/w)- grams in every 100grams % m/v (%w/v)-grams in every 100mL %v/v - mL in every 100mL ppm-Milligrams per litre express the concentration of ppb- micrograms per litre very dilute solutions Molarity- moles per litre

Amount in mol per litre of solution Molarity (M) tells us the amount in mole in one litre of solution. M = n / V

Practise (m/m) Example: 0.2%w/w indicates 0.2 g of solute in 100g of solvent. What does 7% w/w indicate? Example: 40g of sugar solution contains 8g of sugar. What is the concentration (w/w) of this solution? Concentration = 8 / 40 x 100 = 20%w/w 20g of sugar solution contains 2g of sugar. What is the concentration (w/w) of this solution?

Practise (v/v) Percentage by volume is usually used when the solute is a liquid Example: A 200ml of champagne contains 28ml of alcohol. What is the concentration (v/v) of alcohol in this solution? Concentration = 28 / 200 x100 = 14%(v/v) If 10ml of cordial is present in a 200ml cordial solution what is the concentration (v/v) of the cordial solutions?

Percentage by mass/ volume (w/v) Example: 2%w/v indicates 2 g of solute in 100ml of solvent. What does 30% w/v indicate?

Percentage by mass/ volume (w/v) What is the concentration of 5 g of NaCl in a 200ml NaCl solution?(w/v) Concentration = mass of solute (g) x 100% volume of solvent (ml) = 5 / 200 x 100% = 2.5%w/v

Mass of solute per litre of solution(mgL -1 ) Example: A 250ml glass of orange flavoured mineral water contains 4 mg of sulfate. What is the concentration in (mg L -1 ) of sulfate in the mineral water? Step one: Convert 250ml to a volume in litres One litre is equivalent to 1000ml. Therefore 250 ml is equivalent to (250/1000) 0.250L Step two: Use equation for concentration in mg/L Concentration (mg/L) = mass of solute (mg) / volume of solvent (L) = 4 / =16 mg/L or 16 mg L -1

Practise Express the concentration of each of the following solutions in the units stated in brackets. a 49.2 g of an iron ore sample contains 32.1 g of iron (% m/m). b A M solution of KNO3 (% m/v).

Calculate the molarity (concentration in M) of each of the following solutions. a 2.0 mol of NaCl is dissolved in 8.0 L of water. b 0.25 mol of MgCl2 is dissolved in 500 mL of water. c 30.0 g of sucrose (C12H22O11) is dissolved in 200 mL of water.

Dilutions When a solution of accurately known concentration is diluted, we can use the relationship: C 1 V 1 = C 2 V 2

Worked example A student adds 200 mL of deionised water to 50 mL of 0.20 M HCl solution. Determine the concentration of the diluted hydrochloric acid solution. Solution c1 = 0.20 M c2 = ? V1 = 50 mL V2 = ( ) = 250 mL