Asked by Anonymous

A buffer is prepared by adding 0.97 L of 1.1 M HCl to 805 mL of 1.7 M NaHCOO. What is the pH of this buffer? [Ka(HCOOH) = 1.7 × 10-4]
Answered by DrBob222
Use the Henderson-Hasselbalch equation. Post your work if you get stuck.
Answered by Anonymous
What I did the first time:
pH = -log(1.7*10^-4) + log (1.7/1.1)

I tried it this way before, but it said I got the wrong answer. Does it have something to do with the Ka value?
Answered by DrBob222
No. You substituted molarity of the individual solutions BEFORE they were mixed. After mixing the molarity has changed.
initial CH3COO^- = 1.7M x 0.805L =1.3685 moles.
initial HCl = 1.1M x 0.97 L = 1.067 moles.
............CH3COO^- + H^+ ==> CH3COOH
initial mols.1.3685...1.067.....0
change.....-1.067.....-1.067.....1.067
final.......0.3015.....0........1.067

final (CH3COO^-) = 0.3015/total volume.
final (CH3COOH) = 1.067/total volume.
Those are the numbers you substitute into the HH equation.
Answered by Anonymous
total volume = 1.775 L

final concentration (CH3COO)= .169859
final concentration (CH3COOH) = .60112676

pH= -log(1.7*10^-4) + log(.169859/.60112676)

= 3.22

is this correct?
Answered by DrBob222
1.775 is right.
Concn CH3COO is right.
Concn CH3COOH is right.
Your answer of 3.22 is correct for the value of Ka you show in the problem of 1.7 x 10^-4; however, you may want to check your post and the problem from which you copied it because the Ka for CH3COOH (acetic acid) is 1.7 x 10^-5 (not 10^-4) so the correct answer for your post is 3.22 but the correct answer for an acetic acid/acetate buffer in that proportion is 4.22.
Answered by Anonymous
ok thanks!