1. Titration Calculation
Learning Objectives:
All (Grade C)
Most (Grade B)
Some (Grade A/A*)
Know how to calculate moles of a solution
TBAT manipulate the formula to calculate concentration or volume
TBAT to use titration data to calculate unknown concentrations
Calculate the RFM of :
a)HCl b) KOH c) NaOH

2. Recap….calculation Moles
Remember the formula:
Moles = Mass / Mr
Calculate moles in 50g of KOH(s)
Calculate the mass in 0.75 moles of KOH(s)

3. To convert to dm3 you take the volume in cm3 and divide by 1000
Moles of solutions
When calculation the moles of solution the formula is:
Moles = volume(dm3)×concentration (mol.dm-3)
Q1 : Calculate the number of moles in 2 dm3 of 1.5 mol.dm-3 of HCl
Q2: Calculate the number of moles in 20cm3 of 1.5 mol.dm-3 of HCl
To convert to dm3 you take the volume in cm3 and divide by 1000
20 /1000 = 0.02 dm3

4. Concentration Example
10.0 cm3 of a solution of potassium hydroxide was titrated with a 0.10 M solution of hydrochloric acid cm3 of the acid was required for neutralization. Calculate the concentration of the potassium hydroxide solution.
STEP 1 – Write what you know
KOH(aq) HCl (aq)
Conc = ? Conc = 0.10 M
Vol = 10.0cm3 Vol = 13.5cm3

5. STEP 2 – Write a symbol equation to identify ratios
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
STEP 3 – Calculate moles for known substance
Moles of HCl(aq) = vol x conc
remember vol has to be in dm3
( 13 / 1000 ) x 0.1 = moles
From the equation, it is a 1:1 ratio so the moles
Moles of NaOH is
Now calculate conc of NaOH
If the equation was HCl + 2NaOH,what would the moles of NaOH be?
Conc of NaOH = 0.14 (M) mol.dm-3

6. Extra.. You may be asked to then convert it from mol.dm-3 to g.dm -3
If the KOH has a conc of 0.218M
Step 1 – work out 1 mole in grams
( K =39.1, O=16, H=1)
So in 1M you would have 56.1g
But we have only 0.218M so simply…
Step x gives you
12.3 g.dm -3

7. Graded questions Q – Determine the Conc of acid
B-30cm3 of 0.10M NaOH neutralised 25cm3 of hydrochloric acid. A- 50cm3 of 0.2M NaOH neutralised 20cm3 of sulphuric acid. A* cm3 of 0.05M Ba(OH)2 neutralised 40.0cm3 of nitric acid.
EXT: try to convert to g.dm-3

8. B-NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
= 0.12M
A - 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l)
= 0.25M
A* Ba(OH)2(aq) + 2HNO3(aq)  Ba(NO3)2(aq) + 2H2O(l)
=0.0625M

9. REVIEW Learning Objectives: All (Grade C) Most (Grade B)
Learning Objectives:
All (Grade C)
Most (Grade B)
Some (Grade A/A*)
Know how to calculate moles of a solution
TBAT manipulate the formula to calculate concentration or volume
TBAT to use titration data to calculate unknown concentrations

10. Results 25 cm3 of the 1 M NaOH reacted with cm3 of the HCl.
In the box, write your average for the accurate titrations (at least two, but not more than three).

11. Over to you…
In a particular experiment 10.0 cm3 of a M solution of sodium ethanedioate was found to react with 8.7 cm3 of sulphuric acid. Find the concentration of the concentration of the sulphuric acid.
Na2C2O4 + H2SO4 → Na2SO4 + (CO2H)2 
Moles of Sodium Ethanedioate = 0.005
Moles of Sulphuric Acid = 0.005
Conc of Acid = 0.005/ (8.7/1000) = M