[J-Mart]

I am working with equilibria with weak acids and weak bases. I am trying to determine the pH of a .1M HC2H3O2 solution, but I am not good at it what so ever. I think my major species in solution are CH3COOH, H+ and the conjugate base CH3COO-.

Setting it up like this:
                      CH3COOH(aq) <=> H+(aq)  +  CH3COO-(aq)
Initial (M):         .1                          0.0           0.0
Change (M):       -X                        +X            +X
Equilibrium(M):    .1-X                      X                 X

K(a)= [H+][CH3OO-]/[CH3COOH]
1.8E-5 = (X^2)/.1-X  (Approx. .1-X=.1) <-- This is what my book says, I don't get the logic but ok.
Then, 1.3E-3 M = X

Another thing from the book I don't quite get follows:

1.3E-3/.1 *100% = 1.3%  <-- Since this value is less than 5% the quadratic form is not needed.
[H+] = 1.3e-3
pH= -log(1.3E-3)
pH=2.89

Is that the correct pH? Is my practice right?

[Borek]

pH is OK.

See

http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation

and

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

for explanation and discussion of why and when

Approx. .1-X=.1

works.

[J-Mart]

Borek, thanks for the links and pH confirmation.

[J-Mart]

I have a follow up question. The next problem asks me to find the pH of a solution comprised of the following:

5mL 0.1M CH3COOH + 5mL H20   <---- Does the volume impact anything??

Here's my work as it stands.

The reaction is:  CH3COOH + H2O <==> CH3COO-(aq) + H3O+(aq)
Or Simply:                 CH3COOH <==> CH3COO-(aq) + H+(aq)


Setting it up again like this:
                      CH3COOH(aq) <=> H+(aq)  +  CH3COO-(aq)
Initial (M):         .1                          0.0           0.0
Change (M):       -X                        +X            +X
Equilibrium(M):    .1-X                      X                 X

K(a)= [H+][CH3OO-]/[CH3COOH]
1.8E-5 = (X^2)/.1-X  (Approx. .1-X=.1)
Then, 1.3E-3 M = X

1.3E-3/.1 *100% = 1.3%  <-- Since this value is less than 5% the quadratic form is not needed.
[H+] = 1.3e-3
pH= -log(1.3E-3)
pH=2.89

So I get the same answer... Is this correct logic or no? Thanks again.

[Borek]

5mL 0.1M CH3COOH + 5mL H20   <---- Does the volume impact anything??

Yes. It halves the concentration. It is called dilution

[J-Mart]

So the molarity would then be 0.05M instead of 0.1M?

And if it were 1mL .1M CH3COOH + 99mL H2O it would be .001M?

[Borek]

Yes.

http://www.chembuddy.com/?left=concentration&right=dilution-mixing

[J-Mart]

Ok, I'm starting to get this...

[J-Mart]

And I am tripped up again...

I now have the problem of:

5mL .1M CH3COOH + 5mL .1M HCl

Is this a buffer? I have looked over my book section on buffers and the buffer section in link posted earlier. I can't exactly interpret what it is saying for the Henderson-Hasselbalch equation.

I am also confused to the dilution or if it is even one... Here's my work, that I believe to be incorrect so far so you can see my thought process.

I determined both to be .05M. Then did the following ICE:
                      CH3COOH(aq) <=> H+(aq)  +  CH3COO-(aq)
Initial (M):         .05                          0.0           0.0
Change (M):       -X                        +X            +X
Equilibrium(M):    .05-X                      X                 X

                      HCl(aq) <=> H+(aq)      +      Cl-(aq)              (Cl becomes a spectator ion)
Initial (M):         .05                        0.0           0.0
Change (M):       -.05                     +.05            +.05
Equilibrium(M):    0                        .05                 .05

                      CH3COOH(aq) <-- H+(aq)  +  CH3COO-(aq)
Initial (M):         .05                        .05           0.0
Change (M):       .05                       -.05           -.05
Equilibrium(M):    .10                      0               -.05              <---- Having the -.05 is where I think I go wrong

                      CH3COOH(aq) <=> H+(aq)  +  CH3COO-(aq)
Initial (M):         .1                          0.0           -.05
Change (M):       -X                        +X            +X
Equilibrium(M):    .1-X                      X             .05+X

[Borek]

Is this a buffer?

Buffer must contain both weak acid and its conjugated base. Do you have them both in solution?

I determined both to be .05M.

Good

                      CH3COOH(aq) <=> H+(aq)  +  CH3COO-(aq)
Initial (M):         .05                          0.0           0.0
Change (M):       -X                        +X            +X
Equilibrium(M):    .05-X                      X                 X

Good - in the absence of HCl.

                      HCl(aq) <=> H+(aq)      +      Cl-(aq)              (Cl becomes a spectator ion)

No ICE for HCl - it is strong acid, assumed to be always 100% dissociated.

Not sure what you did later  However, here is a hint as to what you shoul do:

                     CH3COOH(aq) <=> H+(aq)  +  CH3COO-(aq)
Initial (M):         .05                          0.0           0.0

If there is strong, 100% dissociated acid in the solution, what is initial concentration of H⁺?

[J-Mart]

Buffer must contain both weak acid and its conjugated base. Do you have them both in solution?

No, I have a strong acid(HCl). Correct?

If there is strong, 100% dissociated acid in the solution, what is initial concentration of H+?

I believe the H+ initial concentration would be .05M as well. Meaning my ICE equation would look like:

                      CH3COOH(aq) <=> H+(aq)  +  CH3COO-(aq)
Initial (M):         .05                         .05           0.0
Change (M):       -X                        +X            +X
Equilibrium(M):    .05-X                   .05+X           X

And then 1.8E-5=(.05+X)X/(.05-X)
Approximate (.05+X)=.05   and  (.05-X)=(.05)
Giving (.05)X/(.05) which simplifies to X.
Therefore X=1.8E-5
pH=-log(1.8E-5)
Giving me a pH of 4.74

Is that correct? It seems high to me given the addition of a strong acid. But I guess the concentration has decreased, so that could be why?