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!
No problem! I'm glad I could clarify things for you. Now, let's move on to your new question.
(a) To find the molar mass of the acid, we need to use the information given in the titration. We know that 15.0 mL of 1.008 M sodium hydroxide is needed to neutralize a 0.2053 g sample of the acid.
First, we need to find the number of moles of sodium hydroxide used in the titration. This can be calculated using the equation:
moles NaOH = Molarity of NaOH x Volume of NaOH (in L)
moles NaOH = 1.008 M x 0.015 L = 0.01512 mol NaOH
Since the acid is monoprotic (meaning it donates only one proton or hydrogen ion), the number of moles of acid is equal to the number of moles of NaOH used in the reaction.
moles acid = 0.01512 mol
Now, we can use the equation:
molar mass = mass of acid / moles of acid
molar mass = 0.2053 g / 0.01512 mol ≈ 13.57 g/mol
So, the molar mass of the acid is approximately 13.57 g/mol.
(b) To determine the molecular formula of the acid, we need to use the elemental composition given in the problem. The percentages of each element in the acid are:
Hydrogen (H): 5.89%
Carbon (C): 70.6%
Oxygen (O): 23.5%
We can assume that we have 100g of the acid, which allows us to convert the mass percentages to grams:
Hydrogen (H): 5.89 g
Carbon (C): 70.6 g
Oxygen (O): 23.5 g
Next, we need to calculate the number of moles of each element present in 100g of the acid. This can be done using the molar masses of each element:
moles H = mass H / molar mass H = 5.89 g / 1 g/mol = 5.89 mol
moles C = mass C / molar mass C = 70.6 g / 12 g/mol ≈ 5.88 mol
moles O = mass O / molar mass O = 23.5 g / 16 g/mol ≈ 1.47 mol
The ratio of the moles of each element gives us the empirical formula of the acid. In this case, the ratio is approximately:
H : C : O ≈ 5.89 : 5.88 : 1.47
To simplify the ratio, we can divide each of the values by the smallest value (in this case, 1.47):
H : C : O ≈ 4 : 4 : 1
Therefore, the empirical formula of the acid is C4H4O.
To determine the molecular formula, we need to know the molar mass of the empirical formula. In this case, the molar mass of C4H4O is:
(4 x molar mass C) + (4 x molar mass H) + (molar mass O)
= (4 x 12 g/mol) + (4 x 1 g/mol) + (16 g/mol)
= 64 g/mol
Since the molar mass of the acid is approximately 13.57 g/mol (as calculated in part a), we can calculate the ratio of the molecular formula to the empirical formula molar mass:
molar mass of acid / molar mass of empirical formula
= 13.57 g/mol / 64 g/mol ≈ 0.212
To get the molecular formula from the empirical formula, we need to multiply the subscripts in the empirical formula by this ratio:
C4H4O x 0.212 = C0.848H0.848O
To simplify, we can round the subscripts to the nearest whole number:
C1H1O
Therefore, the molecular formula of the acid is CH2O.
I hope this explanation helps! Let me know if you have any further questions.