Preparation of Buffer A: 1. Pipette 20.00mL of 0.100 M acetic acid and 20.00mL o
ID: 489495 • Letter: P
Question
Preparation of Buffer A:
1. Pipette 20.00mL of 0.100 M acetic acid and 20.00mL of 0.100 MNaOH from the burette into a 100mL beaker. (Remember that the finalvolume of this solution is the sum of the volumes of the aceticacid and NaOH solutions that have been mixed.)
2. Pipette 20.00 mL of 0.100 M acetic acid into another 100mLbeaker and add 20.00 mL of distilled water with a pipette. Thisproduces 40.00mL of a solution in which the concentration of theacetic acid is half of the stock solution.
What are the molarities of CH3COOH and CH3COO in "Buffer A"?
Preparation of Buffer B: Pipette 40.00mL of 0.100M acetic acid into a 100mL beaker. Add10.00mL of distilled water. Add 30.00mL of 0.100 M NaOH solutionfrom a burette. This produces 80.00mL of solution that shall bedenoted "Buffer B".
What are the molarities of CH3COOH and CH3COO in "Buffer B"?
Explanation / Answer
Ans for Buffer A
Number of moles n = (Concentration)(Volume)
So, n = (0.1M)(0.02L) = 0.002 moles of acetic acid
Similarly n = (0.1M)(0.02L) = 0.002 moles of NaOH
Now, in case 1. Total volume is 0.04L, so concentration of acetic acid = 0.002 / 0.04 L=0.05 M
CH3COOH + NaOH ---> CH3COONa +H2O, this shows that 1 mole of acetic acid gives 1 mole of sodium acetate and thus concentration of sodium acetate = 0.002 / 0.04 =0.05 M.
Now, in case 2. Total volume is 0.08L, so concentration of acetic acid = 0.002 / 0.08 =0.025 M = concentration of sodium acetate.
Ans for Buffer B
n = (0.04)(0.1) = 0.004 moles of acetic acid, so concentration of acetic acid = 0.004 / 0.08 =0.05 M = concentration of sodium acetate.