Preparation of Buffers: Regardless of which buffer you an directed below 1. Pipe
ID: 1084391 • Letter: P
Question
Preparation of Buffers: Regardless of which buffer you an directed below 1. Pipette 10.00 mL of 0.100 M acetic acid and 10.00 mL of 0.100 M NaOH from the burette into a 100 mL beaker. (Remember that the final volume of this solution is the sum of the volumes of the acetic acid and NaOH solutions that have been mixed so the total volume of this solution will be 20.00 mL.) 2. Pipette 10.00 mL of 0.100 M acetic acid into another 100 mL beaker and add 10.00 mL of distilled water with a pipette. This produces 20.00 mL of a solution in which the concentration of the acetic acid is half that of the stock solution.Explanation / Answer
1)
in the mixture molarity of CH3COOH = millimoles / total volume
= 10 x 0.100 / 10 + 10
= 0.05 M
in the mixture molarity of NaOH = 10 x 0.1 / 20
= 0.05 M
CH3COOH + NaOH ---------------> CH3COONa + H2O
0.05 0.05 0 0 ----------------> initial
-0.05 -0.05 + 0.05 +0.05 ---------------> change
0.00 0.00 0.05 0.05 -----------------------> after reaction
molaritie of CH3COOH = 0.00 M
molaritie of CH3COO- = 0.05 M
2)
after adding distilwater molarity of CH3COOH = millimoles / total volume
= 10 x 0.100 / 10 + 10
= 0.05 M
stock solution molarity = 0.1 M is given .
so calculate concentration = 1/2 * stock solution molarity