# Henderson Hasselbalch

#### circulus vitios

##### Full Member
10+ Year Member
7+ Year Member
A buffer made by mixing 100 mL of 0.5 M HOAc (Ka = 1.8E-5) with 25 mL of 1.0 M KOH has a pH approximately equal to which of the following values?

A. 0.2
B. 4.7
C. 7.0
D. 9.3

It's a weak base and a strong acid in a 2:1 mole ratio. "This means that pH = pKa." pH = pKA if amounts or concentrations of base and acid are equal. Last time I checked, 2 and 1 aren't equal and so pH shouldn't be equal to pKa?

#### slz1900

##### Full Member
7+ Year Member
Since you've got 100 mL of 0.5 M weak acid and 25 mL of 1 M strong base, half of your weak acid gets neutralized. That means half is protonated and half is deprotonated, in other words, [A-]/[HA] = 1, so ph = pKa + log [A-]/[HA] simplifies to ph=pKa + log 1 which is the same as pH = pKa. It's probably better to not memorize ratios like 2 to 1 and instead actually calculate it to avoid confusion.

#### BryanNextStep

##### Full Member
It's a weak base and a strong acid in a 2:1 mole ratio. "This means that pH = pKa." pH = pKA if amounts or concentrations of base and acid are equal. Last time I checked, 2 and 1 aren't equal and so pH shouldn't be equal to pKa?

You get pH = pKa at the half-equivalence point, which is the spot in the middle of the "buffer region" or the flat region of a titration curve. The half-equivalence point is the point at which the concentration of a weak acid is equal to the concentration of its own conjugate base.

The 2:1 ratio you're talking about there is the relationship between a weak acid and a base you've added to the solution.

It sounds like you're confusing those two scenarios which is making the problem work out funny in your head.

You're starting with 0.05 moles of acetic acid. Then you're adding 0.025 moles of hydroxide to it.

So the hydroxide deprotonates 0.025 moles of your acetic acid to make 0.025 moles of water.

That leaves you with 0.025 moles of acetic acid and 0.025 moles of potassium acetate (the deprotonated from of acetic acid). Since you now have equal amounts of a weak acid and its own conjugate base, you're now at the half equivalence point.

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