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

1. Mar 25, 2012

### 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.

1. The problem statement, all variables and given/known data
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]

2. Relevant equations
pH=pKa+log([A-]/[HA]

3. 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.

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Ka=[H+][A-]/[HA]
pH=-log[H+]

3. 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.

2. Mar 25, 2012

### Staff: Mentor

How does HCOO- react with a strong acid?

Acid with Ka=0.37 is quite strong.

3. Mar 25, 2012

### TeenieBopper

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?

So my math is probably correct then?

4. Mar 25, 2012

### Staff: Mentor

Correct so far.

Yes, that's what you should do.

but this part doesn't make any sense.

It IS correct.

5. Mar 25, 2012

### TeenieBopper

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?

6. Mar 25, 2012

### Staff: Mentor

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

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