(a) What is the pH of a 2.0 molar solution of acetic acid. Ka of acetic acid = 1.8 x 10¯5
(b) A buffer solution is prepared by adding 0.10 liter of 2.0 molar acetic acid solution to 0.1 liter of a 1.0 molar sodium hydroxide solution. Compute the hydrogen ion concentration of the buffer solution.
(c) Suppose that 0.10 liter of 0.50 molar hydrochloric acid is added to 0.040 liter of the buffer prepared in (b). Compute the hydrogen ion concentration of the resulting solution.
How do you do this problem?
Please help!
Thank you SOO much in advance
a) Write the ionization of acetic acid. To simplify, CH3COOH, I will write it as HAc.
HAc ==> H^+ + Ac^-
Ka = (H^+)(Ac^-)/(HAc)
Using the ICE method,
(H^+)= y
(Ac-) = y
(HAc) = 2-y
Solve for y = (H^+) and convert to pH where pH = - log(H^+)
b)HAc + NaOH ==> NaAc + H2O
Calculate mols HAc. Calculate mols NaOH. Determine how much NaAc is formed and how much HAc is unreacted. That provides a buffer of NaAc and HAc. Then use the Henderson-Hasslebalch equation to calculate pH, and convert to (H^+).
c) Recalculate (acid) and (base) done in b) using the added 0.1 L of 0.50 M HCl and use the H-H equation to solve for pH. Convert to (H^+).
Post your work if you get stuck.
For b) Is the molarity of acetic acid 1 and the molarity of NaOH .5 ?
Do you mean before or after the addition of NaoH??
The problem states that the molarity of HAc is 2 M and the molarity of NaOH is 1 M.
HAc + NaOH ==> NaAc + H2O
mols HAc initially = 2 M x 0.1 L = 0.2 mols.
mols NaOH initially = 1 M x 0.1 L = 0.1 mols. After the reaction,
HAc + NaOH ==> NaAc + H2O
0.2 mol + 0.1 => 0        0
-0.1 mol -0.1 mol=>+0.1 mol + 0.1 mol
0.1 mol + 0 mol ==>0.1 mol + 0.2 liter
Line 1 under the equation gives mols INITIALLY before HAc and NaOH react.
Line 2 gives the CHANGE in mols.
Line 3 give the FINAL mols after the reaction has occurred.
(HAc) after rxn = 0.1 mol/0.2 L = ??
(Ac^-) after rxn = 0.1 mol/0.2 L = ??
Plug these values into the HH equation and calculate pH.
is the HH equation products over reactants and also I do not get part c at all.
so is the H+ concentration is 1 ? for part b
Don't you have a text?
The HH equation is
pH = pKa + log (base)/(acid)
No, the (H^+) isn't 1.
my text does no use the HH equation.
It does products over reactants and plugs in the numbers to solve to the [H+]. I never learned the HH equation.
You may use the following if you don't use the HH equation.
Ka=(H^+)(Ac^-)/(HAc).
I posted the (HAc)=0.1 mol/0.2 L = ??
Also, (Ac^-) = 0.1 mol/0.2 L = ??
Plug those into Ac and HAc in the Ka equation and solve for H+.
So the H+ is 1.8 x 10 ^-5 for part b
and how would you start part C
Do you find the moles of HCL which would be .05 moles and what would you do with the .04 liter of the buffer prepared.
Thank you so much for all the help so far
I wonder if a typo has been made in part c? I wonder if that is 0.1 L of 0.05 M HCl added to the buffer mixture of part b? If not, then the addition of 0.1L of 0.5 M produces a solution which is not buffered any more. All of the buffering capacity has been used up. But here is how you do it.
In part b, we had 0.1 mol HAc/0.2 L = 0.5 mol/L.
We had (Ac^-) also at 0.1 mol/0.2 L = 0.5 mol/L.
The new solution in part c takes 0.04 L of that solution. So how many mols HAc and Ac are in part c?
HAc = 0.5 mol/L x 0.04 L = 0.02 mols.
Ac^- = 0.5 mol/L x 0.04 L = 0.02 mols.
We now add 0.1 L x 0.5M HCl to that so we are adding 0.05 mol H^+ from the HCl.
The H^+ reacts with the Ac^- to produce HAc (and will use all of the Ac^-) as follows:
Ac^- + H^+ ==> HAc
0.02 mol Ac^- will react with 0.02 mol H^+ to produce another 0.02 mol HAc making the total mols HAc now 0.02 + 0.02 = 0.04 mol HAc.
All of the Ac^- will be used up.
We started with 0.05 mol HCl and we have used 0.02 to make HAc so we have 0.03 mol HCl remaining.
SO, in the final solution (now 0.04 L + 0.1 L = 0.14L) we have
0.04 mol HAc.
no mols Ac^- from the NaAc.
0.03 mol H^+ from HCl.
We will get H^+ from two sources. We will get a few from HAc but it is a weak acid and won't contribute much. We will have 0.03 mol from HCl. I would just ignore those from HAc and calculate H^+ from HCl alone which will be 0.03 mol/0.14 L = ??
JISKHA
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