Chem (continuation)
posted by Marisol .
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 = 141.54 = 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 0.5490.01=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 HH 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 .2053g 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!
Respond to this Question
Similar Questions

Health
I posted this question yesterday and I got confused. The determinant of total energy requirement that may most appropriately be manipulated in a weight loss program is: 1. basal metabolic rate 2. physical activity 3. dietary thermogensis … 
~Chem
I was looing over this question and the answer I got was not correct, I noticed that somone already posted this question up before but maybe if I provide the answer I could get an understanding of the steps. @462degrees the reaction … 
CHEM (DrBob222)
I had posted the question to that previous chem question, regarding Mo & V...they said that it should have been correct. What I am thinking that I was missing...must have been that I did not include 4s & then just not fill it. I had … 
chem
so i posted this question yesturday and got some help. which is the stuff below. can you look at the last thing i said which was the answer i got by doing what DrBob222 said to do and tell me if its correct.(the question is the first … 
grade 9 GEOGRAPHY
I posted a question earlier about the sensitive tourist I am a little confused on what the question is asking exactly. "Sensitve tourist" suggestions so that visitors can minimize their impact on the local environment and community. … 
chem
I am still having a little problem with my original question. However, this is what I have done so far. I am not sure what to do with the density of the solution of 0.78 g/mL. The answer I came up with is 5.47 mol/kg. I found the density … 
Math CONTINUATION OF QUESTION FROM bobpursley
Thank you for answering the question before bobpursley, I understand how you did the deriv of inside, outside, but how do I now do the second derivative since you have the extra 18x^2 in the numerator? 
Math
This is a continuation of a math problem I posted earlier. I'm a little confused by this question. From your equation, when during the day would the temperature be 30C? 
Calc pt2
Original question: Posted by Lost One on Tuesday, March 5, 2013 at 12:04pm. Let f(x) = (2/3)x^3  2x + 1 with a restricted domain of [1,infinity]. What is the value of (f1)'(x) when x = 13? 
English
Homework Help: English Posted by rfvv on Tuesday, March 12, 2013 at 6:29pm. Posted by rfvv on Tuesday, March 12, 2013 at 5:35pm. Posted by rfvv on Monday, March 5, 2012 at 7:39pm. Try hard to succeed. (What is the part of speech of …