1. Quantitative Chemistry A.S. 90763 (2.3) Year 12 Chemistry

2. Knowing the concentration Why?

3. Concentration of Solution n = amount of solute (moles) c = concentration of solution (moles/litre) V = volume of solution (litres) n cV

4. c=n/V =0.025/0.050 =0.5 molL -1 n = cV = 0.3 x 0.021 = 0.0063 moles n cV Problem 1. Calculate concentration if 0.025 moles of HCl are present in 50ml of solution. Problem 2. Calculate moles of Na 2 CO 3 present in 21mL of 0.3molL -1 solution

5. n = m/M = 2.4/106 = 0.023 moles c = n/V = 0.023/0.250 = 0.092molL -1 2.4g of sodium carbonate is dissolved in water to make 250ml of solution. Calculate the concentration. M(Na 2 CO 3 ) = 106gmol -1 n cV m nM

6. Practice Complete the odd numbered Q’s of activity 8A in your textbook.

7. Standard Solutions A standard solution is a solution for which the concentration is accurately known. A primary standard solution has been made from weighing a reactant and dissolving it in a known volume of solvent. A secondary standard solution is when the concentration has been determined by experimental procedure.

8. Criteria for a primary standard Be readily available in a pure form Resist absorbing water Be stable when stored Have a high molar mass – why?

9. Prepare a standard solution Your turn… Complete Experiment 5 in your lab manual. We are using potassium hydroxide instead of sodium carbonate.

10. Volumetric Analysis Acid Base Titrations –Used to determine an unknown concentration

11. ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) acid base acid base Carry out this neutralisation reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4

12. Setup for titrating an acid with a base

13. TitrationTitration 1.Use pipette to measure your (unknown concentration) solution into the flask. 2.Add known solution from the buret to find the average titre. 3.Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) 4.Calculate moles (n) of known solution. 5.Use balanced equation to find the unknown amount of moles. 6.Calculate the concentration of unknown.

14. PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its concentration. Titre1234Av. Vol. ml 25.025.429.425.2 35 mL of NaOH is pipetted into a flask and neutralised (by titration) with ? mL of 0.0998 M HCl. What is the concentration of the NaOH? 25.2 CONCORDANCE 3 consistent results 2. Find average titre of HCl (L)

15. NaOH + HClNaCl + H 2 0 1:1 ratio, therefore 0.025 moles NaOH reacts with 0.025 moles HCl PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its concentration. 4. Calculate moles (n) of HCl n cV n = cV = 0.0998 x 0.0252 = 0.025 moles HCl used 5. Use balanced equation to find the unknown amount of moles NaOH

16. c=n/V = 0.025/0.035 = 0.71molL -1 PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its concentration. 6. Calculate the concentration of NaOH n cV Your turn 20 mL of Al(OH) 3 is neutralised (by titration) with 18.6mL of 0.2 M HCl. What is the concentration of the Al(OH) 3 ? Concentration of Al(OH) 3 = 0.062molL -1

17. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M i.e. Dilute it! But how much water do we add?

18. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do ? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution moles of NaOH in FINAL solution

19. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? c V = Amount (n) of NaOH in original solution (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH n/C = Volume of final solution (0.15 mol NaOH)/(0.50 mol/L) = 0.30 L or 300 mL n cV

20. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

21. Diluting Solutions Finding the new concentration Your volume Total volume e.g. 20ml of 0.5molL -1 is diluted to 100ml. What is the final concentration? 0.02/0.1 X 0.5 =0.1molL -1 Find the original concentration of HCl when 20ml is pipeted into a flask and made up to 250ml giving a final concentration of 4x10 -3 molL -1 X concentration = diluted concentration 0.05molL -1 No need to convert ml to L if both units are the same

22. Test Yourself You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. How much water do you need to add? 250ml What is the concentration of Cu 2+ if 20ml of 0.64molL -1 of CuSO 4 is diluted to 100ml? 0.128molL -1 25ml CuSO 4 solution was diluted to 200ml. The new concentration is 7.36x10 -4 molL -1. What was the original concentration? 5.89x10-3 molL -1

23. Density Convert Lg and gL E.g. Calculate the volume of 6 moles of ethanol given the density of ethanol (CH 3 CH 2 OH) is 0.789gml -1 M = 46gmol -1 m = Mn = 46 x 6 = 276g 276g/0.789gml -1 = 349.8ml or 0.350L

24. Apple Juice 20ml apple juice was titrated against 0.1molL -1 NaOH and the average was 10.36ml. Assuming all apple juice is citric acid and that 1mole reacts with 3 moles of NaOH. Calculate the concentration of citric acid in gL -1. Apple juice must contain between 0.3 and 0.8g per 100ml. Is this considered apple juice or apple drink? M(citric acid)=192gmol -1

25. You try this dilution problem You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400 mL of 0.10 M HCl. How much of the acid and how much water will you need?