GCSE & A level Chemistry Calculations: Calculating concentrations on dilution

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Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations 14. Other GCSE chemical calculations - e.g. working out dilutions needed or the resulting concentration from diluting a stock solution This is a BIG website, you need to take time to explore it [SEARCH BOX] Keywords: Quantitative chemistry calculations Diluting solutions. Help for problem solving dilution calculations. How do you do solution dilution calculations? Using dilution factors to solve concentration problems - fully worked out example calculations for diluting solutions. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses.  These revision notes and practice questions on how to do solution dilution chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.

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14.2b atom economy calculations  *   14.4 water of crystallisation calculations

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14.3 Dilution of solutions calculations

calculating dilutions - volumes involved etc.  • It is important to know how to accurately dilute a more concentrated solution to a specified solution of lower concentration. It involves a bit of logic using ratios of volumes. The diagram above illustrates some of the apparatus that might be used when dealing with solutions.

• Dilution calculation Example 14.3.1

• A purchased standard solution of sodium hydroxide had a concentration of 1.0 mol/dm3.

• (a) How would you prepare 100 cm3 of a 0.1 mol/dm3 solution to do a titration of an acid?

• The required concentration is 1/10th of the original solution.

• To make 1dm3 (1000 cm3) of the diluted solution you would take 100 cm3 of the original solution and mix with 900 cm3 of water.

• The total volume is 1dm3 but only 1/10th as much sodium hydroxide in this diluted solution,

• so the concentration is 1/10th, 0.1 mol/dm3.

• To make only 100 cm3 of the diluted solution you would dilute 10cm3 by mixing it with 90 cm3 of water.

• How to do this in practice is described at the end of Example 14.3.2 below and a variety of accurate/'less inaccurate' apparatus is illustrated above.

• (b) Given a standard concentrated solution of hydrochloric acid of concentration 2.0 mol/dm3, how much of this solution in cm3, is needed to make up 250 cm3 of a more dilute standard solution of concentration 0.20 mol/dm3? Also, briefly describe the procedure to make this solution.

• molarity = moles / volume, mol = molarity x volume, volume = moles / molarity

• 1st consider the solution to be made up:

• 250 cm3 = 250/1000 = 0.25 dm3

• mol HCl needed = 0.20 x 0.25 = 0.05 mol HCl

• 2nd consider the original standard solution:

• volume = moles / molarity = 0.05 / 2.0 = 0.025 dm3

• Therefore, volume of original standard solution needed is 0.025 x 1000 = 25.0 cm3.

• This would be measured out with a 50 ml burette or better, a calibrated 25.0 cm3 pipette, and carefully transferred into a 250 cm3 calibrated flask and topped up with deionised water to the calibration mark.

• For more details see steps and in section 11(d) How to make up a standard solution

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• Dilution calculation Example 14.3.2

• Given a stock solution of sodium chloride of 2.0 mol/dm3, how would you prepare 250cm3 of a 0.5 mol/dm3 solution?

• The required 0.5 mol/dm3 concentration is 1/4 of the original concentration of 2.0 mol/dm3.

• To make 1dm3 (1000 cm3) of a 0.5 mol/dm3 solution you would take 250 cm3 of the stock solution and add 750 cm3 of water.

• Therefore to make only 250 cm3 of solution you would mix 1/4 of the above quantities

• i.e. mix 62.5 cm3 of the stock solution plus 187.5 cm3 of pure water.

• This can be done, but rather inaccurately, using measuring cylinders and stirring to mix the two liquids in a beaker.

• It can be done much more accurately by using a burette or a pipette to measure out the stock solution directly into a 250 cm3 graduated-volumetric flask.

• Topping up the flask to the calibration mark (meniscus should rest on it). Then putting on the stopper and thoroughly mixing it by carefully shaking the flask holding the stopper on at the same time!

• For picture details see  11(d) How to make up a standard solution

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• Dilution calculation Example 14.3.3

• In the analytical laboratory of a pharmaceutical company a laboratory assistant was asked to make 250 cm3 of a 2.0 x 10-2 mol dm-3 (0.02M) solution of paracetamol (C8H9NO2).

• (a) How much paracetamol should the laboratory assistant weigh out to make up the solution?

• Atomic masses: C = 12, H = 1, N = 14, O = 16

• method (i): Mr(paracetamol) = (8 x 12) + (9 x 1) + (1 x 14) + (2 x 16) = 151

• 1000 cm3 of 1.0 molar solution needs 151g

• 1000 cm3 of 2.0 x 10-2 molar solution needs 151 x 2.0 x 10-2/1 = 3.02g

• (this is just scaling down the ratio from 151g : 1.0 molar)

• Therefore to make 250 cm3 of the solution you need 3.02 x 250/1000 = 0.755 g

• -

• method (ii): Mr(paracetamol) = 151

• moles = molarity x volume in dm3

• mol paracetamol required = 2.0 x 10-2 x 250/1000 = 5.0 x 10-3 (0.005)

• mass = mol x Mr = 5.0 x 10-3 x 151 = 0.755 g

• -

• (b) Using the 2.0 x 10-2 molar stock solution, what volume of it should be added to a 100cm3 volumetric flask to make 100 cm3 of a 5.0 x 10-3 mol dm-3 (0.005M) solution?

• The ratio of the two molarities is stock/diluted = 2.0 x 10-2/5.0 x 10-3 = 4.0 or a dilution factor of 1/4 (0.02/0.005).

• Therefore 25 cm3 (1/4 of 100) of the 2.0 x 10-2 molar solution is added to the 100 cm3 volumetric flask prior to making it up to 100 cm3 with pure water to give the 5.0 x 10-3 mol dm-3 (0.005M) solution.

• There are more questions involving molarity in

• Dilution calculation Example 14.3.4

• You are given a stock solution of concentrated ammonia with a concentration of 17.9 mol dm-3 (conc. ammonia! ~18M!)

• (a) What volume of the conc. ammonia is needed to make up 1dm3 of 1.0 molar ammonia solution?

• Method (i) using simple ratio argument.

• The conc. ammonia must be diluted by a factor of 1.0/17.9 to give a 1.0 molar solution.

• Therefore you need (1.0/17.9) x 1000 cm3 = 55.9 cm3 of the conc. ammonia.

• If the 55.9 cm3 of conc. ammonia is diluted to 1000 cm3 (1 dm3) you will have a 1.0 mol dm-3 (1M) solution.

• -

• Method (ii) using molar concentration equation - a much better method that suits any kind of dilution calculation involving molarity.

• molarity = mol / volume (dm3), therefore mol = molarity x volume in dm3

• Therefore you need 1.0 x 1.0 = 1.0 moles of ammonia to make 1 dm3 of 1.0M dilute ammonia.

• Volume = mol / molarity

• Volume of conc. ammonia needed = 1.0 / 17.9 = 0.0559 dm3 (55.9 cm3) of the conc. ammonia is required,

• and, if this is diluted to 1 dm3, it will give you a 1.0 mol dm-3 dilute ammonia solution.

• -

• (b) What volume of conc. ammonia is needed to make 5 dm3 of a 1.5 molar solution?

• molarity = mol / volume (dm3), therefore mol = molarity x volume in dm3

• Therefore you need 1.5 x 5 = 7.5 moles of ammonia to make 5 dm3 of 1.5M dilute ammonia.

• Volume (of conc. ammonia needed) = mol / molarity

• Volume of conc. ammonia needed = 7.5 / 17.9 = 0.419 dm3 (419 cm3) of the conc. ammonia is required,

• and, if this is diluted to 5 dm3, it will give you a 1.5 mol dm-3 dilute ammonia solution.

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• Dilution calculation 14.3.5

• Oleic has the structural formula CH3(CH2)7CH=CH(CH2)7COOH, molecular formula C18H34O2

• It is not very soluble in water, and solutions of it are often prepared using an organic solvent like petroleum ether.

• (a) Calculate the molecular mass of oleic acid (Relative atomic masses C = 12, H = 1, O =16)

• Mr = (18 x 12) + (34 x 1) + (2 x 16) = 282

• (b) A stock solution of oleic acid in a solvent has a molarity of 4.0 x 10-4 mol dm-3.

• 2 cm3 of this solution is added to 8 cm3 of the solvent to make a more dilute solution.

• 1 cm3 of this diluted solution is further diluted and mixed with 9.0 cm3 of the same solvent.

• What is the final molarity of the doubly diluted solution?

• The first dilution is 2 ==> 10 (or 1 ==> 5 by ratio)

• so concentration = 4.0 x 10-4 / 5 = 0.8 x 10-4 = 8.0 x 10-5 mol dm-3

• The 2nd dilution is 1 ==> 10 by ratio

• so concentration = 8.0 x 10-5 / 10  = 8.0 x 10-6 mol dm-3 (final molarity)

• -

• (c) What mass of oleic acid would be in 5.0 cm3 of this final diluted solution?

• Molarity is mol per dm3 (litre) or mol per 1000 cm3 (ml)

• Therefore in each cm3 of the final dilution there are 8.0 x 10-6 / 1000 = 8.0 x 10-9 mol

• Therefore in 5.0 cm3 there are 5 x 8.0 x 10-9 =  40.0 x 10-9 = 4.0 x 10-8 mol

• mol = mass (g) / Mr,   so mass = mol x Mr,

• Therefore mass of oleic acid in the 5.0 cm3 of solution

• = 4.0 x 10-8 x 282 = = 1128 x 10-8 = 1.128 x 10-5 = 1.1 x 10-5 g (= 11 µg, 2 s.f.)

14.2b atom economy calculations  *   14.4 water of crystallisation calculations

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling

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