Please answer part 1 and part 2 A buffer solution is made from a weak add (or ba
ID: 483416 • Letter: P
Question
Please answer part 1 and part 2 A buffer solution is made from a weak add (or base) and one of its salts, which provides more anion (or cation). If a small amount of add or base is added, the pH does not change significantly; it is protected or buffered. For example, in a acetate/acetic add buffer, the following equilibria must be considered: CH_3COO (aq) + H_2O(l) doubleheadarrow CH_3COOH(aq) + OH^-(aq) CH_3COOH(aq) + H_2O(l) doubleheadarrow H_3O^+(aq) + CH_3COO^-(aq) if acid (i.e H_3O^+) is added reaction (2) is driven to the left, acetate ion removes the proton forming unionized acetic acid and the pH does not fall significantly. Similarly the addition of OH^- causes reaction (1) to move to the left, the pH does not rise significantly. But, there are limits to the amount of acid or base that may be added without upsetting the stable Your lab teaching assistant will inform you whether to make "Buffer A" or "Buffer B" Do not make both. Preparation of "Buffer A": Prepare solutions 1 and 2 as directed below. Pipette 20 00 mL of 0.100 M acetic acid and 20.00 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.) Pipette 20.00 mL of 0.100 M acetic acid into another 100 mL beaker and add 20.00 mL of distilled water with a pipette. This produces 40.00 mL of a solution in which the concentration of the acetic acid is half that of the stock solution. Prepare the solution called "Buffer A" by mixing the solutions prepared in steps 1 and 2 above. The total volume of "Buffer A" should be 80.00 mL. What are the molarities of CH_3COOH and CH_3COO in "Buffer A" described above? Pipette 40.00 mL of 0.100 M acetic acid into a 100mL beaker Add 10 00 mL of distilled water. Add 30.00 mL of 0.100 M NaOH solution from a burette. This produces 80.00 mL of solution that shall be denoted "Buffer B" What are the molarities of CH_3COOH and CH_3COO^- in "Buffer B"?Explanation / Answer
Part 1:
After mixing solutions 1 and 2, final volume is 80 mL
Total volume of acetic acid initially = 20+20 = 40 mL
Moles of acid initially = 0.1*0.04 = 0.004
Total volume of base initially = 20+0 = 20 mL
Moles of base initially = 0.1*0.02 = 0.002
Moles of acid left after reaction with base = 0.004-0.002 = 0.002
Moles of CH3COO- produced = moles of base added = 0.002
So, Molarities are:
[CH3COOH] = 0.002/0.08 = 0.025 M
[CH3COO-] = 0.002/0.08 = 0.025 M
Part 2:
Moles of acid taken initially = Molarity*Volume = 0.1*0.04 = 0.004 moles
Moles of base taken = 0.1*0.03 = 0.003 moles
moles of acid left after reaction = 0.004-0.003 = 0.001 moles
moles of CH3COO- produced after reaction = moles of base added = 0.003
Final volume = 80 mL = 0.08 L
So, Molarities are:
[CH3COOH] = 0.001/0.08 = 0.0125 M
[CH3COO-] = 0.003/0.08 = 0.0375 M