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:
- 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
Because strong acids wil react first I added the c (= [H+]) of the acids = 0.025M + 0.0125M = 0.0375M
pH = -log (cA) = 1.4259...
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
[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.