[NYM]

0,0500 M HCOOH (aq), pH = 2,54 at 25 *C

1) Find the "dissociation thingy" (don't know what it is called in English), the propotion between HCOOH and CHOO^-, called alpha:
alpha = 10^-2,54/0,0500M = 5,8%

2) Find K_a:
Ostwald's dilution law: K_a = c_A * alpha²/(1-alpha) = 1,77*10^-4
 or pK_a = 3,75

3) 4,08 g of HCOONa is added to 1 L of the solution, V is approximately constant.
Find pH.
n(HCOONa) = 7,15*10^-2 mole + 10^-2,54 mole = 7,44*10^-2 mole
n(HCOOH) = 0,0500 mole - 10^-2,54 mole =4,71*10^-2 mole
since V = 1 L.
Then I use the Henderson-Hasselbalch equation..

4) 40 mL of 1 M HCl is added to the solution, find pH.
n(HCl) = 4*10^-2 mole and therefore
n(HCOOH) = 4,71*10^-2 mole  - 4*10^-2 mole = 0,71*10^-2 mole
and
n(HCOONa) = 4,71*10^-2 mole + 7,44*10^-2 mole = "something"
And then I just use the Henderson-Hasselbalch equation.

Is this correct?

Then I thought... what if n(HCl) > n(HCOOH)? How do I calculate pH then?

[Borek]

"dissociation thingy"

dissociation percentage or dissociation fraction, see introduction to acid/base equilibrium

5,8% (...) pK_a = 3,75

Both OK.

n(HCOONa) = 7,15*10^-2 mole + 10^-2,54 mole = 7,44*10^-2 mole
n(HCOOH) = 0,0500 mole - 10^-2,54 mole =4,71*10^-2 mole
since V = 1 L.

Nope. Adding HCOO^- you move equilibrium of the dissociation reaction of the acid to the left so you can't assume it becomes summed with the HCOO^- from the acid dissociation. Plus, HCOO^- added gets hydrolized - it is not-so weak conjugated base. The easiest approach is to assume HCOOH concentration from the original acid and HCOO^- from salt added. See pH of buffers.

n(HCOOH) = 4,71*10^-2 mole  - 4*10^-2 mole = 0,71*10^-2 mole

Do I read correctly that you are neutralizing HCOOH with HCl? Acid with acid?

Then I thought... what if n(HCl) > n(HCOOH)? How do I calculate pH then?

Simple - pH is determined by excess hydrochloric acid.

[NYM]

dissociation percentage or dissociation fraction, see introduction to acid/base equilibrium

Ah, okay. I couldn't find it on Wiki.

Nope. Adding HCOO^- you move equilibrium of the dissociation reaction of the acid to the left so you can't assume it becomes summed with the HCOO^- from the acid dissociation. Plus, HCOO^- added gets hydrolized - it is not-so weak conjugated base. The easiest approach is to assume HCOOH concentration from the original acid and HCOO^- from salt added. See pH of buffers.

Okay, but is there any way to calculate the exact pH?

Do I read correctly that you are neutralizing HCOOH with HCl? Acid with acid?

Oh yeah, I messed that one up. Anyhow, it's the same, just the other way around:
n(HCl) = 4*10^-2 mole and therefore
n(HCOOH) = 4,71*10^-2 mole  + 4*10^-2 mole = something
and
n(HCOONa) = 7,16*10^-2 mole - 4*10^-2 mole  = "something"

Simple - pH is determined by excess hydrochloric acid.

Of course

Thanks.

[Borek]

Okay, but is there any way to calculate the exact pH?

Browse pH calculation lectures at ChemBuddy. There is no simple, bulletproof method.

Oh yeah, I messed that one up. Anyhow, it's the same, just the other way around:
n(HCl) = 4*10^-2 mole and therefore
n(HCOOH) = 4,71*10^-2 mole  + 4*10^-2 mole = something
and
n(HCOONa) = 7,16*10^-2 mole - 4*10^-2 mole  = "something"

Almost OK - the idea is correct, but you are using numbers that you have found in the previous step - and they are wrong.

[NYM]

Yeah, I corrected the whole assignment now

Browse pH calculation lectures at ChemBuddy. There is no simple, bulletproof method.

Nice website
Hmm, I thought about something with

n(CHOOH) = 10^-2,54 + x
n(HCOONa) = 0,0715 - x
n(H₃O⁺) = 0,0715 - x

And then I use K_a and write the expression for the acid dissociation constant:

K_a = (0,0715 - x)²/10^-2,54 + x
which yields a quadratic equation.

No/Yes?