Two questions about the pH of buffers after solutions are added questions

[TeenieBopper]

I have two questions about pH of buffers. It's a general chemistry class, but the questions aren't similar to anything we've had yet.

Homework Statement

A buffer is prepared by adding 0.96 L of 0.96 M HCl to 762 mL of 1.4 M NaHCOO. What is the pH of this buffer? [Ka(HCOOH) = 1.7 × 10-4]

Homework Equations

pH=pKa+log([A-]/[HA]

The Attempt at a Solution

It's the fact that it's HCl that's really throwing me off. I thought a buffer could only be made with a weak acid and it's conjugate base (and vice versa). I can figure out the pH of the HCl and the NaHCOO solutions without a problem, but it's the combining that's throwing me.

Homework Statement

A certain monoprotic weak acid with Ka = 0.37 can be used in various industrial processes. (a) What is the [H+] for a 0.234 M aqueous solution of this acid and (b) what is its pH?

Homework Equations

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

The Attempt at a Solution

I did an ICE table and set up the equation .37=x^2/(.234-x) to get [H+]. Once I had that, -log[H+] gave me the pH. I got [H+]=.1625 and pH=.79, which doesn't make sense at all considering it's supposed to be a weak acid, but I'm not sure what I'm doing wrong.

 

[Borek]

How does HCOO^- react with a strong acid?

Acid with Ka=0.37 is quite strong.

 

[TeenieBopper]

How does HCOO^- react with a strong acid?

Does it create Formic Acid (HCOOH)? Do I need to determine the amount of moles of HCl and NaHCOO and then that will give me how much NaHCOO is left along with how much Formic acid is made? Which I can then plug into the pH=pKA + log([A-]/[HA]) with A- being the cation of HCOOH?

Acid with Ka=0.37 is quite strong.

So my math is probably correct then?

 

[Borek]

Does it create Formic Acid (HCOOH)? Do I need to determine the amount of moles of HCl and NaHCOO and then that will give me how much NaHCOO is left
along with how much Formic acid is made?

Correct so far.

Which I can then plug into the pH=pKA + log([A-]/[HA])

Yes, that's what you should do.

with A- being the cation of HCOOH?

but this part doesn't make any sense.

So my math is probably correct then?

It IS correct.

 

[TeenieBopper]

but this part doesn't make any sense.

Okay, so i did the math, and after the formation of the Formic Acid, , I have .5352 M formic acid and .0843 sodium formate. After doing an ICE table, I find I have .003702+.009726=.013428 M COOH-. So the pH should be:

pH= -log(1.7E-4)+log(.013428/.0843)

Correct?

 

[Borek]

I have .5352 M formic acid and .0843 sodium formate

I don't see where you got these numbers from (unless your first post contains wrong data).

After doing an ICE table

You don't need ICE table, simply assume reaction went to completion.