pH = pKa + log(b/a)
5.00 = 4.74 + log b/a
solve for b/a which has two unknowns. The second equation you need is
b + a = 0.1M
Solve the two equations simultaneously for CH3COOH and CH3COO^- and that's the first part of the problem. Post back with this work if you need help finishing with the second part.
I don't get this at all.
A beaker with 105mL of an acetic acid buffer with a pH of 5.00 is sitting on a bench top. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.40 mL of a 0.470 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
3 answers
need more help
need more help
2 answers