J.R. S. answered 10/18/22
Ph.D. University Professor with 10+ years Tutoring Experience
We can use the Henderson Hasselbalch equation:
pH = pKa + log [salt] / [acid]
Reaction:
CH3COOH + NaOH ==> CH3COONa + H2O
30 ml x 5 g CH3COOH / 100 ml = 1.5 g CH3COOH x 1 mol CH3COOH / 60.05 g = 0.0250 mols CH3COOH
10 ml x 1 L / 1000 ml x 0.5 mol NaOH / L = 0.005 mols NaOH
Final volume = 30 ml + 10 ml = 40 ml = 0.040 L
CH3COOH + NaOH ==> CH3COONa + H2O
0.025...........0.005................0................Initial
-0.005.......-0.005.............+0.005...........Change
0.02..............0...................0.005..........Equilibrium
You can use moles of convert to M by dividing moles by 0.04 L
pH = pKa + log [salt] / [acid] and pKa for acetic acid = ~4.75 (your tabular value may be different)
pH = 4.75 + log (0.005 / 0.02) = 4.75 + log 0.25
pH = 4.15