Henderson Hasselbalch Buffers and Volumes

  • #1
Meadow_Lark
8
0

Homework Statement



You have 725 mL of an 0.55 acetic acid solution. What volume (V) of 1.30 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.99? ( The pKa of acetic acid is 4.76.)
[/B]

Homework Equations



pH = pKa + log A/HA

Ratio of A- to HA = 10^(pH-pKA)[/B]


The Attempt at a Solution



- pH = pKa + log A/HA
- Ratio A/HA = 10^(pH-pKA) = 10^(4.99-4.76) = 1.70
- Mols of acid = 0.725 mL * 0.55 M = 0.3988 mM
- HA = (0.3988 - 1.30 * V) / (0.725 + V)
- A- = (1.30 * V) / (0.725 + V)
- A-/HA = (1.30 * V) / (0.725 + V) = 0.3988 = (1.30 * V) / (0.725 + V)

It has been a while since i've had general chemistry, and now i'm taking biochemistry and need to review acids, buffers, pH, pKa, molarity and volumes. I get to the last step and from here I completely forget what to do. Any help/hints would be appreciated.
 

Answers and Replies

  • #2
Borek
Mentor
29,101
3,711
pH = pKa + log A/HA

OK

Ratio A/HA = 10^(pH-pKA) = 10^(4.99-4.76) = 1.70

OK

Mols of acid = 0.725 mL * 0.55 M = 0.3988 mM

Yes and no - watch your units.

HA = (0.3988 - 1.30 * V) / (0.725 + V)

OK

A- = (1.30 * V) / (0.725 + V)

OK

A-/HA = (1.30 * V) / (0.725 + V) = 0.3988 = (1.30 * V) / (0.725 + V)

No. But the mistake has nothing to do with the chemistry, just check your algebra. You started right with the A-/HA ratio, but instead of plugging correct formulas/numbers (which you correctly calculated earlier) you did something strange. Plug them as it should be done and solve for V, that's all.
 

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