A beaker with 200 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.40 mL of a 0.400 \it M \rm HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
I've tried using the Henderson Hasselbalch equation but the answer isn't coming out right.
4 answers
Perhaps I can help if you will decipher how much and what molarity HCl was added. \it M \rm doesn't make any sense to me.
It's supposed to be 0.400 M HCl.
Can you show me what you've done? You say you don't get the right answer. Do you nave the answer?
pH = pKa + log(B/A) where B = base and A = acid.
5.00 = 4.76 + log(B/A)
Solve for B/A. There are two unknowns here.
Then the problem states that
A + B = 0.1M. A second equation with the same two unknowns. Solve the two equations simultaneously for (A) and (B).
Then millimoles A = 200 mL x (A) = ??
mmoles B = 200 x B = ??
.........base + HCl ==> acid
You add 6.40 mL x 0.400 M HCl = xx mmoles.
Then the acid is increased by xx mmoles and the base is decreased by xx mmoles. Then plug these numbers back into the HH equation to find final pH. My answer is 4.77.
5.00 = 4.76 + log(B/A)
Solve for B/A. There are two unknowns here.
Then the problem states that
A + B = 0.1M. A second equation with the same two unknowns. Solve the two equations simultaneously for (A) and (B).
Then millimoles A = 200 mL x (A) = ??
mmoles B = 200 x B = ??
.........base + HCl ==> acid
You add 6.40 mL x 0.400 M HCl = xx mmoles.
Then the acid is increased by xx mmoles and the base is decreased by xx mmoles. Then plug these numbers back into the HH equation to find final pH. My answer is 4.77.