pH calculation of a mixure containing 2 strong acids and a buffer mixture

Question

Calculate the pH of a mixtureof 10 mL 0,1 M HCl ; 5 mL 0,1 M HNO3 ; 10 mL 0,2 M HAc ; 15 mL 0,4 M NaAc.

My understanding was that I have to find the [H+] by:

1. finding the concentration of the products by: a) finding the amount of moles (n) = c*V n(HCl) = 0.1M * 0.010L = 0.001 moles n(HNO3)= 0.1M * 0.005L = 0.0005 moles n(HAc) = 0.2M * 0.010L = 0.002 moles n(NaAc) = 0.4M * 0.015L = 0.006 moles

b) Total volume = (0.01 + 0.005 + 0.01 +0.015)L = 0.04L

c) the concentration of the products are: c = n/V(tot) c(HCl) = 0.001n/0.04L = 0.025M c(HNO3) = 0.0005n/0.04L = 0.0125M c(HAc) = 0.002n/0.04L = 0.05M c(HANa) = 0.006n/0.04L = 0.15M

2. Because strong acids wil react first I added the c (= [H+]) of the acids = 0.025M + 0.0125M = 0.0375M

3. pH = -log (cA) = 1.4259...

4. Because HAc and HANa are a weak acid and it's conjugate base I used the buffer formula to find the pH: pH = pKa + log (cb/ca) pKa of CH3COOH = 4.75 ca = 0.05M cb = 0.15M pH = 4.75 + log (0.15/0.05) = 4.75 + 0.47712 = 5.22 [H+] = 10^-pH = 5.928*10^-6

5. [H+] = 0.0375M + 5.928*10^-6 = 0.037505928 pH = -log(0.037505928) = 1.4

The correction indicated that the answer should have been a pH = 4.88. I don't know where it went wrong, but I have the feeling I terribly misunderstood a concept.

Answer 1

First, you have to allow the strong acids to react with the weak base.

$$\ce{HCl + NaOAc -> HOAc + Na+ + Cl-}$$ $$\ce{HNO3 + NaOAc -> HOAc + Na+ + NO3-}$$

This goes to completion. As a result, you make some acetic acid and use up some, but not all of the acetate.

Now you should be able to figure out the final concentrations of the major species (acetic acid, acetate) and plug it into the buffer equation.

To turn this into a rule: you will never have a strong acid and a weak base in the same solution (same goes for a strong base and a weak acid). They always react with each other until at least one of them is gone.