You are carrying out the titration of 100.0 mL of 1.000M acetic acid with 1.000 M sodium hydroxide. Ka = 1.76x10^-5 of acetic acid.
(a) Calculate the initial pH of your acetic acid sample.
(b) Calculate the pH of the solution after the addition of 25.0mL of NaOH.
For (a) should I just take the -log of the given Ka value, which is 4.75?
or should I do this:
CH3COOH <--> CH3COO- + H+
1.0 0 , 0
-x +x , +x
1.0-x x , x
1.0-x can be approximated to 1.0
so, x^2/1.0=1.76x10^-5
x = 4.20x10^-3, and x = [H+]
-log [H+] = pH = 2.38
I'm not sure if should just be 2.38 or 4.75
then part B, I set it up like this:
so convert both CH3COOH and NaOH to moles
.1 CH3COOH and .025 NaOH
CH3COOH + NaOH <--> CH3COONa + H2O
I subtracted moles of NaOH from CH3COOH on the reactants side, and added moles of NaOH to CH3COONa (which starts at zero) on the products side
.1-0.025 = 0.075 of CH3COOH, and 0+0.025 = 0.025 of CH3COO-
because these are conjugates, I can use Henderson-Hasselbalch:
pH = pKa + log 0.025/0.075
pH = 4.27
Is this correct? Can the pH go down when I'm adding a strong base (even if it's a smaller amount) to a weak acid?
2 answers
For b, I came out with 4.277 which I would round to 4.28 but check my arithmetic. Yes, the pH can do down because you are neutralizing 1/4 of the acid and the extra conjugate base that's formed suppresses the ionization of acetic acid more, also. Good work.