[skibum143]

What is the pH of a buffer prepared by mixing 154 mL of 0.442 M HCl and 0.475 L of 0.400 M sodium acetate. (Acetic Acid: pKa = 4.74)

First, I used M1V2 = M2V2 to find the Molarity of HCL and sodium acetate once mixed:
HCL = 0.154(0.442) = 0.692M2  M2 = 0.108
sodium acetate = 0.400(0.475) = 0.629M2  M2 = 0.302

Then I used the ICE equation for:
   HCL + C2H3NaO2 -> NaCl + CH3COOH
I  0.108   0.302            -          0
C   -x         -x                        +x
E 0.108+x 0.302+x                   x

Then I found the Ka using the pKa = 10^-4.74 = 1.82E-5

Then I used Ka to solve for x:  Ka = (x) / (.108)(.302)  (disregard x in the denominator bc is very small)
I got x to be 5.94E-7
Which meant the pH was -log(5.94E-7) = 6.23

But this was wrong and I'm not sure where I'm going astray - maybe in solving for x? Or the equation setup?

Thanks for any *delete me*!!

[Borek]

Don't solve buffer questions using ICE table, use Henderson-Hasselbalch equation.

I got x to be 5.94E-7
Which meant the pH was -log(5.94E-7) = 6.23

What is x? You have assigned it to something when constructing ICE table, was it concentration of H⁺, or something else?

[skibum143]

I see... so I was using the wrong equation. So when I use the Henderson-Hasselback, I get:

pH = Pka + log (base/acid)
pH = 4.74 + log (0.302/0.108) which gives me 5.19, but my HW is telling me this is wrong. Am I doing something else incorrectly?

Thanks so much for your *delete me*

[Borek]

Why 0.302?

[skibum143]

I used the M1V1 = M2V2 formula for sodium acetate C2H3NaO2 to find the molarity after the combination
.400*.475 = .629M2
M2 = .302

It's the same way I calculated the Molarity of the HCL. Am I doing something wrong here?
First, I used M1V2 = M2V2 to find the Molarity of HCL and sodium acetate once mixed:
HCL = 0.154(0.442) = 0.692M2  M2 = 0.108
sodium acetate = 0.400(0.475) = 0.629M2  M2 = 0.302

[Borek]

Hint: acetate is a weak base, HCl is a strong base.

[skibum143]

So weak bases don't dissociate completely, would that mean there would be more acetate remaining? I don't know how to calculate that numerically?

[skibum143]

Would the molarity of the acetate stay closer to 0.4 rather than going down to 0.302?

[Borek]

Acetate will get protonated to acetic acid. You may safely assume reaction proceeds quantitatively to the end.

[skibum143]

so if it proceeds to the end, why is 0.302 incorrect? i guess i'm not following. i thought using the m1v1 = m2v2 would give me the correct base/acid ratio to use in the hh equation?

[Borek]

0.302 is an initial concentration of acetate, not final, after the reaction with strong acid.

[skibum143]

Sorry, I'm still lost. If 0.302 is the initial concentration of acetate, how do you calculate the final concentration? Do you subtract x (5.94E-7) <- found in my ICE table calculation?

That wouldn't really change the value, it would still be 0.302.

I don't understand where to go from here...

[Borek]

CH₃COO^- + H⁺ -> CH₃COOH

You may safely assume this reaction goes to completion. One of the substances is a limiting reagent.

[skibum143]

I see, so it reacts with the acid and you have to account for that.

I have one other question: for
What is the pH of a buffer prepared by mixing 54.2 mL of 0.050 M sodium bicarbonate and 19.8 mL of 0.10 M NaOH. (Carbonic Acid: pKa1 = 6.35, pKa2 = 10.33)

Do you take the average of pKa1 and pKa2 to get the pKa? And after the sodium bicarbonate and NaOH have reacted to form carbonic acid, will the carbonic acid continue to react (like in the previous question) so you would have to subtract the amount of initial base (NaOH) from the amount of initial acid?

[Borek]

Do you take the average of pKa1 and pKa2 to get the pKa?

No, you have to check reaction stoichiometry, calculate which acid/conjugate pair is present in the solution, and select appropriate pKa value.

And after the sodium bicarbonate and NaOH have reacted to form carbonic acid

They don't form carbonic acid as a product.