DILUTION CALCULATIONS

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

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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14.3 Dilution of solutions calculations

calculating dilutions - volumes involved etc.

• In conjunction with this page it be important to study

  • Introducing Molarity, volumes and the concentration of solutions

  • Concentration of solution calculations

  • before tackling this section and note the triangle of relationships you need to know!

• 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

  • -

• 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

  • -

• 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): M_r(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): M_r(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 M_r = 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

      •   section 7. introducing molarity and section 12. on dilution

      • -

• 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.

    • -

• 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)

    • M_r = (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) / M_r,   so mass = mol x M_r,

    • 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.)

OTHER CALCULATION PAGES

1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

2. Calculating relative formula/molecular mass of a compound or element molecule

3. Law of Conservation of Mass and simple reacting mass calculations

4. Composition by percentage mass of elements in a compound

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

   • Reacting masses, concentration of solution and volumetric titration calculations (NOT using moles)

7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law

10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

13. Electrolysis products calculations (negative cathode and positive anode products)

14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

    • 14.1 % purity of a product 14.2a % reaction yield 14.2b atom economy 14.3 dilution of solutions

    • 14.4 water of crystallisation calculation  14.5 how much of a reactant is needed?

15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

17. Radioactivity & half-life calculations including dating materials

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