1. 6.50 mL of 0.100 M H2C2O4 is required to neutralize
Titration Calculation
6.50 mL of M H2C2O4 is required to neutralize
10.0 mL of KOH solution in a titration. Calculate the base concentration.
H2C2O4 + 2KOH  K2C2O H2O
L L
0.100 M ? M

2. Titration Calculation
mL of M H2C2O4 is required to neutralize
10.0 mL of KOH solution in a titration. Calculate the base concentration.
H2C2O4 + 2KOH  K2C2O H2O
L L
0.100 M ? M
In a titration the molarity of one chemical is determined by the knowing the molarity of the other chemical.
this is a molarity = question molarity = draw a line
[KOH] =
[KOH] square brackets means the molarity or concentration of what is inside.

3. Titration Calculation
mL of M H2C2O4 is required to neutralize
10.0 mL of KOH solution in a titration. Calculate the base concentration.
H2C2O4 + 2KOH  K2C2O H2O
L L
0.100 M ? M
You need to get moles KOH over litres KOH
You need to start with H2C2O4 and then use stoichoimetry
[KOH] = L H2C2O4

4. Titration Calculation
mL of M H2C2O4 is required to neutralize
10.0 mL of KOH solution in a titration. Calculate the base concentration.
H2C2O4 + 2KOH  K2C2O H2O
L L
0.100 M ? M
[KOH] = L H2C2O4 x mole L

5. Titration Calculation
mL of M H2C2O4 is required to neutralize
10.0 mL of KOH solution in a titration. Calculate the base concentration.
H2C2O4 + 2KOH  K2C2O H2O
L L
0.100 M ? M
[KOH] = L H2C2O4 x mole x 2 mole KOH L mole H2C2O4

6. Titration Calculation
mL of M H2C2O4 is required to neutralize
10.0 mL of KOH solution in a titration. Calculate the base concentration.
H2C2O4 + 2KOH  K2C2O H2O
L L
0.100 M ? M
[KOH] = L H2C2O4 x mole x 2 mole KOH L mole H2C2O4 L

7. Titration Calculation
mL of M H2C2O4 is required to neutralize
10.0 mL of KOH solution in a titration. Calculate the base concentration.
H2C2O4 + 2KOH  K2C2O H2O
L L
0.100 M ? M
[KOH] = L H2C2O4 x mole x 2 mole KOH L mole H2C2O4 L
= M

8. Titration Purpose: To find concentration of acid or base
Method: Acid Base neutralization

9. Titration Purpose: To find concentration of know ion
Method: acid base neutralization
Material:
Burette, Erlenmeyer flask, indicator, measuring cylinder to measure volume, known Concentration base solution, acid solution of unknown concentration

10. Titration Terms Standard solution
Solution in burette, known concentration
Sample solution
Solution in burette, known volume, unknown concentration

11. Titration Procedure: Wash burette and flask with distilled water
Remove water completely from burette and flask
Measure 15 mL of acid solution using measuring cylinder and transfer to flask
Close tap of burette and fill it to zero

12. Titration Procedure: Put 5 drops of indicator solution in flask
Hang the burette into the ring stand.
Open the tap of the burette and carefully allow Base solution to drop into the flask drop-wise.
Keep stirring the flask to mix acid and base solution.

13. Titration
Procedure:
When colour change is observed then stop and measure volume of base solution that was used to titrate the acid solution

14. Titration Calculation
Find moles of the base solution by volume and concentration of standard solution
Find moles of sample solution
Find concentration of the sample solution by moles and volume of the sample solution

15. Titration Terms Indicator
A solution that change colour when reaction is complete. This is added into the flask.

16. Titration Terms Transition point, end point, stoichiometric point
A point in time during titration process when moles of acid become equal to moles of base

17. Setup for titrating an acid with a base

18. Example 1 H2SO4 + 2 KOH ==> K2SO4 + 2 H2O
How many mL of 0.17 M KOH needed to neutralize 50 mL of 0.1 M H2SO4?
L H2SO4 => mol H2SO4 => mol KOH => L KOH
Step 1. Vol A to Moles A (H2SO4) Molarity = mol<=>vol conv factor
Step 2. Moles A to Moles B (Coeff. from chem eqn give conv factor)
Step 3. Moles B to Vol B (Molarity as conv. factor)
0.05 L H2SO4 (0.1 mol/1L) x (2 mol KOH/1 mol H2SO4) x ?
Final answer: 0.06 L = 60 mL KOH
CH. 3 -CH1100-Fall 2004

19. Example: How many mL of a 0
Example: How many mL of a 0.5 M KOH solution needed to neutralize 25 mL.0 mL of a 0.35 M H2SO4 soln?
1 H2SO KOH ==> K2SO H2O
Answer: mL

21. If you get data in a table, you need to subtract the final burette reading from the initial to get the volume of acid or base added.
A burette is a device used to deliver a volume of solution and determine the amount that was added.
If you get more than one run or trials, average the three but reject any volumes that are way off. Use some common sense.
Burette Volume in mL
Initial
Final
Volume Added 8.95 mL 8.46 mL 8.44 mL
reject
average = 8.45 mL 2

23. Titration Calculation
8.45 mL of M H2SO4 is required to neutralize 25.0 mL of KOH solution in a titration. Calculate the base concentration.
H2SO4 + 2KOH  K2SO H2O
L L
0.200 M ? M

24. Titration Calculation
mL of M H2SO4 is required to neutralize
25.0 mL of KOH solution in a titration. Calculate the base concentration.
H2SO4 + 2KOH  K2SO H2O
L L
0.200 M ? M
[KOH] = L H2SO4 x mole x mole KOH L mole H2SO4 L
= M