Concentration of Solutions 1 of 27 Boardworks Ltd 2016
Concentration of Solutions 2 of 27 Boardworks Ltd 2016
Measuring concentrations 3 of 27 Boardworks Ltd 2016 It is not enough to say that one concentration is higher or lower than another. Concentrations usually need to be measured accurately. There are two ways of measuring concentration: mass per unit volume, e.g. grams per decimetre cubed (g/dm 3 ) moles per unit volume, e.g. moles per decimetre cubed (mol/dm 3 ). A cubic decimetre (dm 3 ) is a unit of volume. 1dm 3 is equivalent to 1000cm 3. divide by 1000 1cm 3 1dm 3 multiply by 1000
Volume unit conversions 4 of 27 Boardworks Ltd 2016
5 of 27 Boardworks Ltd 2016 Concentrations in g/dm 3 The following equation gives concentration in g/dm 3 : concentration = mass dissolved (g) volume of solution (dm 3 ) If 1.0g of solid sodium hydroxide is dissolved in 250 cm 3 of water, what is the concentration in g/dm 3? 1g concentration = 250cm 3 1 g = = 0.25dm 3 1 0.25 g dm 3 convert the units = 4g/dm 3
Calculating concentrations in g/dm 3 6 of 27 Boardworks Ltd 2016
7 of 27 Boardworks Ltd 2016 Concentrations in mol/dm 3 To calculate concentration in mol/dm 3 : concentration = number of moles (mol) volume of solution (dm 3 ) This equation can be added to a formula triangle to rearrange the formula: c = n v n = c v = n c v
Concentrations in mol/dm 3 example 8 of 27 Boardworks Ltd 2016 If 1.0g of solid sodium hydroxide (NaOH) is dissolved in 250 cm 3 of water, what is the concentration in mol/dm 3? c = n v? = 0.25dm 3 The information in the question provides the volume but not the number of moles. The following formula is required: mass 1 number of moles = = = 0.025mol molar mass 40 Remember, molar mass = relative formula mass (RFM) Now substitute 0.025 mol into the original formula: 0.025 mol c = = 0.1mol/dm 3 0.25 dm 3
Concentrations in mol/dm 3 : practise 9 of 27 Boardworks Ltd 2016
Rearranging formulae example 1 10 of 27 Boardworks Ltd 2016 What volume of 0.80mol/dm 3 potassium bromide solution contains 1.6moles of potassium bromide (KBr)? v = n / c This calculation is simply a matter of substituting the values into the rearranged formula: v = n / c = 1.6 / 0.8 = 2dm 3
Rearranging formulae example 2 11 of 27 Boardworks Ltd 2016 How many moles of copper sulfate are there in 250cm 3 of 0.2mol/dm 3 copper sulfate solution (CuSO 4 )? n = c v Step 1: convert the units 250cm 3 = 0.25dm 3 Step 2: substitute into the formula n = c x v = 0.2 0.25 = 0.05mol
Calculating mass of solute 12 of 27 Boardworks Ltd 2016 What mass of copper sulfate (CuSO 4 ) was used to make 250cm 3 of 0.2mol/dm 3 copper sulfate solution? Step 1: Calculate the number of moles of copper sulfate Step 2: Rearrange the moles formula to give the mass value This is simply an extension of the previous calculation, from which the moles of copper sulfate was found to be 0.05mol. mass of copper sulfate = number of moles molar mass = 0.05 159.6 = 8.0g
Concentration of Solutions 13 of 27 Boardworks Ltd 2016
Titrations 14 of 27 Boardworks Ltd 2016 A titration is a technique used to accurately determine the concentration of a substance in solution. During a titration, a solution of known concentration, called a standard solution, is added to a solution of unknown concentration. The purpose of a titration is to determine the volume of solution required to reach an endpoint. An endpoint is an observable physical change, such as a colour change. Why is measuring the volume useful?
Finding an unknown concentration 15 of 27 Boardworks Ltd 2016 Volume can be used to determine the concentration by using the following relationship: concentration = number of moles (mol) volume of solution (dm 3 ) An unknown concentration value can be found if the following values are known: number of moles in solution volume of solution. These values can be found through a titration.
Titration equipment 16 of 27 Boardworks Ltd 2016 The following laboratory equipment is required to perform a titration: safety filler burette stand pipette conical flask beaker Use to measure a set volume of alkali that is then added to the conical flask. Filled with the standard solution likely to be acid with a known concentration Indicator then added either Phenolphthalein or Methyl Orange
Titration process 17 of 27 Boardworks Ltd 2016 Stages: 1. Small amount of acid added (a drop at a time) from the burette. 2. Swirl conical flask 3. Stop adding acid when a colour change is seen: 1. Phenolphthalein Pink (alkali) Colourless (acid) 2. Methyl Orange Yellow (alkali) Red (acid) 4. The colour change signifies the end point 5. Record the added volume of acid (standard solution) to neutralise. 6. Repeat a minimum of three times so that a mean can be calculated.
Indicators 18 of 27 Boardworks Ltd 2016 The endpoint of a titration is often marked by a colour change. This is provided by an indicator solution. Indicators are substances which change colour according to the ph of a solution. A small amount of indicator solution is added to one of the solutions during a titration. When the indicator changes colour, the endpoint of the titration has been reached.
Performing a titration 19 of 27 Boardworks Ltd 2016
Titration apparatus 20 of 27 Boardworks Ltd 2016
Concentration of Solutions 21 of 27 Boardworks Ltd 2016
Using titration results 22 of 27 Boardworks Ltd 2016 How are the results of a titration used to calculate the unknown molar concentration of a solution? Step 1: Write a balanced equation for the reaction. Step 2: Calculate the number of moles of standard solution by rearranging the concentration formula: moles = concentration volume Step 3: Use the balanced equation to determine the number of moles for the solution under investigation. Step 4: Use the concentration formula to determine the unknown molar concentration: molar concentration = moles volume
Titration calculations: worked example (1) 23 of 27 Boardworks Ltd 2016 The endpoint of a titration was reached when 20cm 3 of 0.1mol/dm 3 sodium hydroxide was added to 25 cm 3 of hydrochloric acid. What is the concentration of the acid? 1. Write a balanced equation for the reaction: 2. Calculate the number of moles of sodium hydroxide: moles = concentration (mol/dm 3 ) volume (dm 3 ) = 0.1 (20.0 1000) = 0.002 mol NaOH
Titration calculations: worked example (2) 24 of 27 Boardworks Ltd 2016 3. The balanced equation from step 1 shows that one mole of HCl reacts with one mole of NaOH: Therefore 0.002 moles of NaOH will react with: 0.002 moles of HCl. 4. Use the number of moles to calculate the molar concentration of the hydrochloric acid: molar concentration = moles 0.002 = volume (dm 3 ) (25 1000) = 0.08mol/dm 3
Titration calculations: questions 25 of 27 Boardworks Ltd 2016
Titration calculations 26 of 27 Boardworks Ltd 2016 Calculating volume 25 cm 3 of 0.2 mol/dm 3 sodium hydroxide solution was neutralised by 0.15 mol/dm 3 sulphuric acid. Calculate the volume of sulphuric acid needed for the neutralisation. 2NaoH + H 2 SO 4 Na 2 SO 4 + 2H 2 O
Concentration of Solutions 27 of 27 Boardworks Ltd 2016
Glossary 28 of 27 Boardworks Ltd 2016
Multiple-choice quiz 29 of 27 Boardworks Ltd 2016