Question

Using an acetic acid/sodium acetate buffer solution, what ratio of acid to conjugate base will you need to maintain the pH at 5.00?

To solve, first I found the pH by setting 5= -log[H+] giving me [H+}=10^-5
Then I set this equal to $K_a*\frac{[Acid]}{[Base]}$ according to the property $[H+]=K_a*\frac{[Acid]}{[Base]}$

But then when solving for $\frac{[Acid]}{[Base]}$ I got 5/9, or the inverse of the actual answer which is 9/5 which can be gotten from the Henderson-Hasselbalch equation. What is wrong with my way of doing it? Thank you.

Answer 1

Using an acetic acid/sodium acetate buffer solution, what ratio of acid to conjugate base will you need to maintain the pH at 5.00?

To solve, first I found the pH by setting 5= -log[H+] giving me [H+}=10^-5

Then I set this equal to K_a*\frac{[Acid]}{[Base]}K

a

∗

[Base]

[Acid]

according to the property [H+]=K_a*\frac{[Acid]}{[Base]}[H+]=K

a

∗

[Base]

[Acid]

$But then when solving for \frac{[Acid]}{[Base]}$

I got 5/9, or the inverse of the actual answer which is 9/5 which can be gotten from the Henderson-Hasselbalch equation. What is wrong with my way of doing it? Thank you.

$\: \:$

Sorry, I didn't understand ur question..

Answer 2

Answer:

CH3COONa ⇌Na+ + CH3COO-

CH3COOH ⇌ H+ + CH3COO-

Calculation of pH acetate buffer,

Let, [CH3COO-] and [CH3COOH] denotes [A-] and [HA]

pH = p Ka + log [A-]/[HA]

5.3 = 4.74 + log [A-]/[HA]

log [A-]/[HA] = 5.3 - 4.74

log [A-]/[HA] = 0.56

[A-]/[HA] = antilog 0.56

[A-]/[HA] = 3.63

Hope it helps u !!