posted by DrFunk on .
A beaker with 195 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 8.20 mL of a 0.260 it M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
Call base = B (acetate ion).
Call acid = A (acetic acid).
A + B = 0.1 M
pH = pKa + log (B/A).
Two equations and two unknowns. Solve for A and B.
You know pH and pKa, so solve for B/A.
Then A + B = 0.1
That is the first part of the problem.
For the second part, recognize that
HCl + B^- ==>A + Cl^- (HB is the acid, A).
You know B and you know how much HCl is added, so redo the Henderson-Hasselbalch equation to solve for the new pH.
Post your work if you get stuck.
who knows,but are you good at chemistry, because I need help baddly