# Chemistry

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]

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1. Use the Henderson-Hasselbalch equation. Post your work if you get stuck.

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2. 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?

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3. 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.

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4. 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?

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5. 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.

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6. ok thanks!

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