# Chemistry

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Find the pH of mixture of acids. 0.185 M in HCHO2 and 0.225 M in HC2H3O2

Im using an ice chart of weak acid and putting in strong acid in H+ initiAL concentration. I've done the problems many different ways but cannot seem to get the right answer help please?Answer is pH of 2.19.

You said...
You must recognize that this is a mixture of two weak acids; i.e., formic acid and acetic acid. I looked up Ka for both and used 1.77E-4 for Ka HCOOH and 1.8E-5 for Ka CH3COOH.

Calculate the H^+ from the strong acid, then add the H^+ from the weak acid. Formic acid first since it is the stronger. .........HCOOH ==> H^+ + HCOO^-initial..0.185.....0......0 change...-x........x......x equil..0.185-x.....x......x

Ka = (H^+)(HCOO^-)/(HCOOH) Solve for H^+. This is what acts as the common ion (remember Le Chatelier's Principle). This causes the acetic acid to ionize less than it would if were just acetic acid solution.

........CH3COOH ==> H^+ + CH3COO^-initial..0.225.......0......0 change....-x........x........x equil...0.225-x......x........x

Ka acet acid = (H^+)(CH3COO^-)/(CH3COOH) Substitute TOTAL H^+ into the Ka expression for CH3COOH. That will be about 0.00572 from HCOOH from the above calculation plus xfrom this ionization. Solve for x,add this (H^+) to the 0.00572 from HCOOH, then convert to pH. I obtained pH = 2.19

I STILL CAN'T GET 2.19???

I got concentration of HCOOH rxn to be 5.77X10^-3. And the CH3COOH rxn to have concentration of 2.01X10^-3. I added those together to get 7.78X10-3. What do you mean add to ka expression? I keep getting 1.35 now.

• Chemistry - ,

You should have told me what Ka values you are using. Not all texts have the same values although they are close. I'm using 1.77E-4 for formic acid and I will call that HF. I know that isn't formic acid and you know that, too, but it saves some space on the line. HAc is acetic acid and I'm using Ka for HAc of 1.8E-5.
............HF ==> H^+ + F^-
initial....0.185...0......0
change......-x......x.....x
equil.....0.185-x...x.....x

1.77E-4 = (x)(x)/(0.185-x)
I'm not going to do this step by step but this is the way you set it up. You should get an answer for x = 0.00572 if you assume 0.185-x = 0.185.

.............HAc ==> H^+ + Ac^-
initial......0.225....0.....0
change.......-x.......x......x
equil.....0.225-x.....x.......x

Ka = 1.8E-5 = (0.00572+x)(x)/(0.225-x) and solve for x
(Note:I suspect you didn't substitute the 0.00572 here. That's a common error.)

If I assume 0.00572+x = 0.00572 and 0.225-x = 0.225, then x = 7.2E-4
Then 0.00572 + 7.2E-4 = ? and -log of that is 2.19. Voila!.

That's all I did earlier in the day when I first responded to your post. You may ask what happens if we don't make those assumptions so here is what you get.
For formic acid, x = 0.00563 instead of 0.00572 (hardly worth talking about).
For acetic acid, x = 0.000644 (again, not much difference)
So 0.00563 + 0.000644 = 0.00627 and the pH =2.20
Let me know if you don't understand what I did. The only thing I've omitted is the algebra.

• Chemistry - ,

Oh okay yeah I didn't add the 5.77x10^-3 to the weaker acid. So I determine concentration of stronger acid then put that H concentration into the weaker acid then I determine the concentration of the H of CH3COOH and get 7.02x10^-4 and add the H+ concentration of 5.77x10^-3 and get 6.47X10^-3 and take pH of it to get 2.19!!! Ah I got it now!!!
But one question, why do we add the 5.77x10^-3 again in the end if we added it to the equilibrium expression? Why do we have to do that?

• Chemistry - ,

Since the HAc is the weaker acid, the formic acid acts, according to Le Chatelier's Principle, to shift the weaker acid to the left.
HAc ==>H^+ + Ac^-
Adding H^+ from the other acid make HAc ionize to a smaller extent. You can work out how much it would ionize on its own and that is about 0.002 so you can see that it ionizes in the presence of formic acid much less. Back to the point, so that is done to calculate the amount acid contributed by HAc. Then you add the amount contributed by HAc to the amount contributed by the formic acid and calculate pH from the total H^+.