What is the pH of the resulting solution

A buffer solution is formed by the partial neutralization of 600 mL of 0.5 M CH3COOH with 200 mL of 0.5 M NaOH. What is the pH of the resulting solution? Ka for acetic acid is 1.8 × 10−5.
1. 11.33
2. 7.00
3. 9.56
4. 2.67
5. 4.44
6. 5.05
7. 8.95

You REALLY need to know how to find this type of answer on your own. I will show you here.

First you have to know the molarity of the weak acid, CH3COOH, and the conjugate base, CH3COO^-. You do this by calculating the moles of each species and then molarity.

Initial moles CH3COOH = (0.600 L)(0.5 mol/L) = 0.30 mole

CH3COOH(aq) + NaOH(aq) --> CH3COONa(aq) + H2O(l)

Moles CH3COO^- produced = moles CH3COOH reacted with NaOH
Moles CH3COO^- = (0.200 L)(0.5 mol/L) = 0.10 mole

Final moles CH3COOH = 0.30 mole - 0.10 mole = 0.20 mole CH3COOH
Final moles CH3COO^- = 0.10 mole

Final molarity CH3COOH = (0.20 mol)/(0.80 L) = 0.250 M
Final molarity CH3COO^- = (0.10 mol)/(0.80 L) = 0.125 M

Now, use the Henderson-Hasselbalch equation to solve the problem.

pH = pKa + log[CH3COO^-]/[CH3COOH]
pKa = -logKa = -log(1.8 x 10^-5) = 4.74

pH = 4.74 + log(0.125)/(0.250) = 4.74 + (- 0.30) = 4.44

The answer is obviously choice 5.

Hope this is helpful to you. JIL HIR