Question

what is the pH of a solution prepared by mixing 55 mL of .183 M KOH and 40 mL of .145 M HC2H3O2?

Answer 1

50 ml of .145 M H2H3O2"

What's "H2H3O2" supposed to be? I don't have a clue, so I am sorry that I can't help you with that part.

"Suppose you just added 200 ml of a solution containing .5000 moles of acetic acid per liter to 100 ml of .5000 M NaOH. What is the final pH? The Ka of acetic acid is 1.77 * 10^-5"

(200 mL)*(0.5000 M CH3COOH) = 0.1000 mol CH3COOH
(100 mL)*(0.5000 M NaOH) = 0.0500 mol NaOH

So 0.0500 mol of the CH3COOH will react with the 0.0500 mol NaOH to form 0.0500 mol CH3COO-
There will be (0.1000 mol CH3COOH - 0.0500 mol =) 0.0500 mol CH3COOH left

New volume = 200 mL + 100 mL = 300 mL l= 0.3 L
(it turns out that we don't really need it in the calculation below)

Now we use the Henderson-Hasselbalch equation:

pH = pKa + log(A^-]/[HA])
pH = pKa + log([CH3COO-]/[CH3COOH])
pH = (-log(1.77 * 10^-5)) + log((0.5000/0.3)/(0.5000/0.3))
pH = 4.752 + 0 = 4.752