6.50 mL of 0.100 M H2C2O4 is required to neutralize Titration Calculation 6.50 mL of 0.100 M H2C2O4 is required to neutralize 10.0 mL of KOH solution in a titration. Calculate the base concentration. H2C2O4 + 2KOH  K2C2O4 + 2H2O 0.00650 L 0.0100 L 0.100 M ? M

Titration Calculation 1. 6.50 mL of 0.100 M H2C2O4 is required to neutralize 10.0 mL of KOH solution in a titration. Calculate the base concentration. H2C2O4 + 2KOH  K2C2O4 + 2H2O 0.00650 L 0.0100 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.

Titration Calculation 1. 6.50 mL of 0.100 M H2C2O4 is required to neutralize 10.0 mL of KOH solution in a titration. Calculate the base concentration. H2C2O4 + 2KOH  K2C2O4 + 2H2O 0.00650 L 0.0100 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] = 0.00650 L H2C2O4

Titration Calculation 1. 6.50 mL of 0.100 M H2C2O4 is required to neutralize 10.0 mL of KOH solution in a titration. Calculate the base concentration. H2C2O4 + 2KOH  K2C2O4 + 2H2O 0.00650 L 0.0100 L 0.100 M ? M [KOH] = 0.00650 L H2C2O4 x 0.100 mole 1 L

Titration Calculation 1. 6.50 mL of 0.100 M H2C2O4 is required to neutralize 10.0 mL of KOH solution in a titration. Calculate the base concentration. H2C2O4 + 2KOH  K2C2O4 + 2H2O 0.00650 L 0.0100 L 0.100 M ? M [KOH] = 0.00650 L H2C2O4 x 0.100 mole x 2 mole KOH 1 L 1 mole H2C2O4

Titration Calculation 1. 6.50 mL of 0.100 M H2C2O4 is required to neutralize 10.0 mL of KOH solution in a titration. Calculate the base concentration. H2C2O4 + 2KOH  K2C2O4 + 2H2O 0.00650 L 0.0100 L 0.100 M ? M [KOH] = 0.00650 L H2C2O4 x 0.100 mole x 2 mole KOH 1 L 1 mole H2C2O4 0.0100 L

Titration Calculation 1. 6.50 mL of 0.100 M H2C2O4 is required to neutralize 10.0 mL of KOH solution in a titration. Calculate the base concentration. H2C2O4 + 2KOH  K2C2O4 + 2H2O 0.00650 L 0.0100 L 0.100 M ? M [KOH] = 0.00650 L H2C2O4 x 0.100 mole x 2 mole KOH 1 L 1 mole H2C2O4 0.0100 L = 0.130 M

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

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

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

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

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.

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

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

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

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

Setup for titrating an acid with a base

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

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 H2SO4 + 2 KOH ==> K2SO4 + 2 H2O Answer: 35.0 mL

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 0.00 8.95 17.41 Final 8.95 17.41 25.85 Volume Added 8.95 mL 8.46 mL 8.44 mL reject average 8.46 + 8.44 = 8.45 mL 2

Titration Calculation 8.45 mL of 0.200 M H2SO4 is required to neutralize 25.0 mL of KOH solution in a titration. Calculate the base concentration. H2SO4 + 2KOH  K2SO4 + 2H2O 0.00845 L 0.0250 L 0.200 M ? M

Titration Calculation 2. 8.45 mL of 0.200 M H2SO4 is required to neutralize 25.0 mL of KOH solution in a titration. Calculate the base concentration. H2SO4 + 2KOH  K2SO4 + 2H2O 0.00845 L 0.0250 L 0.200 M ? M [KOH] = 0.00845 L H2SO4 x 0.200 mole x 2 mole KOH 1 L 1 mole H2SO4 0.0250 L = 0.135 M