Quantitative Chemistry

A.S. 90763 (2.3)

Year 12 Chemistry

Knowing the concentration

• Why?

Concentration of Solution

• n = amount of solute (moles)

• c = concentration of solution (moles/litre)

• V = volume of solution (litres)

nc V

• c=n/V• =0.025/0.050• =0.5 molL-1

• n = cV• = 0.3 x 0.021• = 0.0063 moles

nc V

Problem 1. Calculate concentration if Problem 1. Calculate concentration if 0.025 moles of HCl are present in 0.025 moles of HCl are present in 50ml of solution.50ml of solution.

Problem 2. Calculate moles of NaProblem 2. Calculate moles of Na22COCO33

present in 21mL of 0.3molLpresent in 21mL of 0.3molL-1-1 solution solution

• 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(Na2CO3) = 106gmol-1

nc V

mn M

Practice

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

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.

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?

Prepare a standard solution

• Your turn…

• Complete Experiment 5 in your lab manual.

• We are using potassium hydroxide instead of sodium carbonate.

Volumetric Analysis

• Acid Base Titrations– Used to determine an

unknown concentration

ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

HH22CC22OO44(aq) + 2 NaOH(aq) ---> Na(aq) + 2 NaOH(aq) ---> Na22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)

acidacid basebase

Carry out this Carry out this neutralisationneutralisation reaction using a reaction using a TITRATION.TITRATION.

Oxalic acid,Oxalic acid,

HH22CC22OO44

Setup for titrating an acid with a baseSetup for titrating an acid with a base

TitrationTitrationTitrationTitration1.1. Use Use pipettepipette to measure your (unknown to measure your (unknown

concentration) solution into the flask.concentration) solution into the flask.2.2. Add known solution from the Add known solution from the buretburet to to

find the average titre.find the average titre.3.3. Indicator shows when exact Indicator shows when exact

stoichiometric reaction has occurred. stoichiometric reaction has occurred. (Acid = Base)(Acid = Base)

4.4. Calculate moles (n) of known solution.Calculate moles (n) of known solution.5.5. Use balanced equation to find the Use balanced equation to find the

unknown amount of moles.unknown amount of moles.6.6. Calculate the concentration of Calculate the concentration of

unknown.unknown.

PROBLEM #1: Standardise a solution of PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its NaOH — i.e., accurately determine its concentration.concentration.

PROBLEM #1: Standardise a solution of PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its NaOH — i.e., accurately determine its concentration.concentration.

Titre 1 2 3 4 Av.

Vol. ml

25.0 25.4 29.4 25.2 25.2

35 mL of NaOH is 35 mL of NaOH is pipettedpipetted

into a flask and neutralised into a flask and neutralised

(by titration) with (by titration) with ?? mL of mL of

0.0998 M HCl. What is the 0.0998 M HCl. What is the

concentration of the NaOH? concentration of the NaOH?

25.2

CONCORDANCE

3 consistent results

2. Find average titre of HCl (L)

NaOH + HCl NaCl + H20

1:1 ratio, therefore 0.025 moles NaOH reacts with 0.025 moles HCl

PROBLEM #1: Standardise a solution of PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its NaOH — i.e., accurately determine its concentration.concentration.

PROBLEM #1: Standardise a solution of PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its NaOH — i.e., accurately determine its concentration.concentration.

4. Calculate moles (n) of HCl

nc V

n = cV

= 0.0998 x 0.0252

= 0.025 moles HCl used

5. Use balanced equation to find the unknown amount of moles NaOH

• c=n/V• = 0.025/0.035

• = 0.71molL-1

PROBLEM #1: Standardise a solution of PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its NaOH — i.e., accurately determine its concentration.concentration.

PROBLEM #1: Standardise a solution of PROBLEM #1: Standardise a solution of NaOH — i.e., accurately determine its NaOH — i.e., accurately determine its concentration.concentration.

6. Calculate the concentration of NaOH nc V

Your turnYour turn

20 mL of Al(OH)20 mL of Al(OH)33 is neutralised (by titration) with 18.6mL is neutralised (by titration) with 18.6mL

of 0.2 M HCl. What is the concentration of the Al(OH)of 0.2 M HCl. What is the concentration of the Al(OH)33??

Concentration of Al(OH)3 = 0.062molL-1

PROBLEM: You have 50.0 mL of 3.0 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

PROBLEM: You have 50.0 mL of 3.0 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

Add water to the 3.0 M solution to lower its Add water to the 3.0 M solution to lower its

concentration to 0.50 Mconcentration to 0.50 M i.e.i.e. Dilute it!Dilute it!

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how But how much much water water do we do we add?add?

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?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

c V

PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?

Conclusion:Conclusion:add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of

0.50 M NaOH.

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

Diluting Solutions

Finding the new concentrationYour 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 4x10-3molL-1

X concentration = diluted

concentration

0.05molL-1

No need to convert ml to L if both units are the same

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 Cu2+ if 20ml of 0.64molL-1

of CuSO4 is diluted to 100ml?

0.128molL-1

• 25ml CuSO4 solution was diluted to 200ml. The new concentration is 7.36x10-4 molL-1. What was the original concentration?

5.89x10-3 molL-1

Density

• Convert L g and g L• E.g. Calculate the volume of 6

moles of ethanol given the density of ethanol (CH3CH2OH) is 0.789gml-1

• M = 46gmol-1

• m = Mn = 46 x 6 = 276g• 276g/0.789gml-1 = 349.8ml or

0.350L

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