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August 20, 2014

Homework Help: Chem (continuation)

Posted by Marisol on Wednesday, March 7, 2007 at 5:57pm.

I am a little confused regarding a question I posted yesterday ("Chem" Tuesday, March 6, 2007 at 9:29pm)

I understand the answer to the first question: What would the pH of this solution be after 0.01 mol of HCl was added?
I got 6.09

I'm still kind of confused about the second question: What would the pH of this solution be after 0.02 mol of NaOH was added?

I don't understand the part where I am calculating the excess of OH^-. Can someone please explain that? I'm sorry if it is obvious, but it's just not clicking.

Thanks Dr. Bob for the previous help though!




Yes, thank you for explanation again!

If I remember the problem from yesterday:

We had 45g NaAc in 155 mL HAc; mols Ac^- = 45/82 = 0.549

mols acetic acid = mols HAc = 0.155 L x 0.1M = 0.0155 mols.

Now, if we add 0.02 mols NaOH to this solution, it will react with ALL of the acetic acid.
HAc + NaOH ==> NaAc + H2O

0.0155 mols HAc - 0.02 mol HAc = 0 mols HAc. mols NaOH in excess = 0.02 - 0.0155 = 0.0045. NaOH and Ac^- don't constitute a buffer. We simply have 0.0045 mols NaOH ( a strong base; i.e., a strong electrolyte) in 155 mL solution.
(OH^-) = 0.0045 mols/0.155L = 0.0293 mols/L = 0.0293M

pOH = 1.54
pH = 14-1.54 = 12.46
Does this help?

I was looking over the first problem again...

This is what you posted:
"When we add 0.01 mol HCl, the Ac^- part of the buffer takes over and forms an extra mol HAc; i.e.,
Ac^- + H^+ ==> HAc.
So the new mols HAc will be 0.0155 + 0.01 and the new mols Ac^- will be 0.549 = 0.01
So new (HAc) = 0.539/0.155 = ??
and new (Ac^-) = 0.0255/0.155=??
pH = 4.76 + (Ac^-)/(HAc). I have 6.08 or so but check my work."

The part where you talk about the "new" (HAc) and (Ac^-) towards the end... do you have them flipped? Shouldn't the new (HAc) = 0.0255/.155 and (Ac^-) = 0.539/0.155

The answer then being, 5.30, not 6.08? I'm sorry for the bother, I just want to be sure...

Nevermind, the answer is 6.08



"When we add 0.01 mol HCl, the Ac^- part of the buffer takes over and forms an extra mol HAc; i.e.,
Ac^- + H^+ ==> HAc.
So the new mols HAc will be 0.0155 + 0.01 and the new mols Ac^- will be 0.549 = 0.01 Should be 0.549-0.01=0.539
So new (HAc) = 0.539/0.155 = ??
and new (Ac^-) = 0.0255/0.155=??
pH = 4.76 + (Ac^-)/(HAc). I have 6.08 or so but check my work."

The part where you talk about the "new" (HAc) and (Ac^-) towards the end... do you have them flipped? Shouldn't the new (HAc) = 0.0255/.155 and (Ac^-) = 0.539/0.155

You are right. I did flip them when I typed them in but I put them into the H-H equation correctly and calculated the pH correctly.
(HAc) = 0.0255/0.155 M
(Ac^-) = 0.539/0.155 M.
pH = pKa + log(base)/(acid)
pH = 4.76+log(0.539/0.155)/(0.0255/0.155)=
4.76+log(0.539/0.0255)=4.76+1.32=6.08.



See the correction below. You are right, I did flip them.

(a)By titration, 15.0 mL of 1.008 M sodium hydroxide is needed to neutralize a .2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic? (b)An elemental analysis of the acid indicates that it is composed of 5.89%H, 70..6%C, and 23.5%O by mass. What is it's molecular formula?

Thank you for the clarification!

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