[LHM]

What is the pH of the solution that results when 10.0 mL of 0.10 M HF (K_a=6.7 * 10^-4) and 10.0 mL of 0.040 M NaOH are mixed?
A) 1.0
B) 1.2
C) 1.5
D) 3.0

Again, I can't seem to make my answer match any of the answer choices. I first assumed that the NaOH and HF would react until the NaOH was completely consumed, so I had 0.010 L * 0.10 M HF - 0.010 L * 0.040 M NaOH = 0.0006 moles of HF leftover. There's now 20.0 mL, so I then did 0.0006 mol HF/0.020 L = 0.03 M HF. And then after that it was just 6.7 * 10^-4 = x²/(0.03 - x). I didn't really feel like using the quadratic formula, so I used a calculator to get that x = 0.0041608. -log(0.0041608)= 2.4

I've tried figuring out where I went wrong, but I don't really see where, so if someone wouldn't mind pointing it out to me, I would really appreciate it.

[Borek]

One magic word: buffer.

[rabolisk]

What is the reaction between HF and NaOH?

[LHM]

I did forget about the concentration of F^- in the solution, so I tried the problem again, and this time I got one of the answers, but not the right answer (which is C), so I think I'm still doing something wrong.

I got that the first reaction between HF and NaOH is a neutralization reaction, which would be HF + NaOH H₂O + NaF. And then I had that 0.010 L*0.10 M HF - 0.010 L * 0.040 M NaOH = 0.0006 mol HF leftover again, but that this produced 0.010 L * 0.040 M = 0.0004 mol NaF. I divided both 0.0006 mol HF and 0.004 mol F^- by 0.020 L to get 0.03 M HF and 0.02 M F^-.

And then I set up the other equation for HF + H₂O  H₃O⁺ + F^-. It's given that K_a= 6.7*10^-4. And after the ICE chart, K_a= (0.02-x)(x)/(0.03+x). I assumed that x was small enough to let 0.02-x=0.02 and 0.03+x=0.03. Using a calculator, I then got that the pH = 3.0, which is D, but the right answer is C?

[PhilipCheney]

What is the buffer

[Borek]

What is the buffer

http://lmgtfy.com/?q=chemical+buffer

[LHM]

I did forget about the concentration of F^- in the solution, so I tried the problem again, and this time I got one of the answers, but not the right answer (which is C), so I think I'm still doing something wrong.

I got that the first reaction between HF and NaOH is a neutralization reaction, which would be HF + NaOH H₂O + NaF. And then I had that 0.010 L*0.10 M HF - 0.010 L * 0.040 M NaOH = 0.0006 mol HF leftover again, but that this produced 0.010 L * 0.040 M = 0.0004 mol NaF. I divided both 0.0006 mol HF and 0.004 mol F^- by 0.020 L to get 0.03 M HF and 0.02 M F^-.

And then I set up the other equation for HF + H₂O  H₃O⁺ + F^-. It's given that K_a= 6.7*10^-4. And after the ICE chart, K_a= (0.02-x)(x)/(0.03+x). I assumed that x was small enough to let 0.02-x=0.02 and 0.03+x=0.03. Using a calculator, I then got that the pH = 3.0, which is D, but the right answer is C?

I would really appreciate it if someone could tell me what I'm still doing wrong here.

[Borek]

You were already told this is a buffer question. Buffer questions are solved using Henderson-Hasselbalch equation.

[rabolisk]

Are you sure the right answer is C?

[Borek]

Good point. Funny thing:

I assumed that x was small enough to let 0.02-x=0.02 and 0.03+x=0.03

That yields exactly the same result as using Henderson-Hasselbalch equation. Not without a reason.

[LHM]

The answer key's been wrong before, so I'm not sure that the answer's C. But I didn't want to assume that I was right if I was actually wrong, so that's why I ask. Oh, and for some reason I prefer to do buffer problems without the Henderson-Hasselbalch equation even though it's easier.