Berapakah pH dari 50 mL CH3COOH 0,2 M dengan 50 mL NaOH 0,1 M (ka CH3COOH = 2 x 10-5)

Dik : [CH3COOH]= 0,2 M
V CH3COOH=50 mL=5×10^-2 L=0,05 L
[NaOH]= 0,1 M
V NaOH= 50 mL=5×10^-2 L=0,05 L
Dit : pH....?
Peny : mencari Mol
n CH3COOH=[CH3COOH].V CH3COOH
=(0,2) (5×10^-2)
=(2×10^-1) (5×10^-2)
=10×10^-3
=1×10^-2 mol
n NaOH=[NaOH].V NaOH
=(0,1) (5×10^-2)
=(10^-1) (5×10^-2)
=5×10^-3 mol
CH3COOH+NaOH—>CH3COONA+H2O
mula" :10×10^-5 5×10^-5 — —
brx :5×10^-5 5×10^-5 5×10^-5 5×10^-5
sisa. :5×10^-5 — 5×10^-5 5×10^-5

[H+]=Ka×mol asam/mol garam
=2×10^-5 (5×10^-5/5×10^-5)coret smua krna sdah habis,sisa
=2×10^-5
pH= –log [H+]
= – log 2×10^-5
= –(log 2+log 10^-5)
= –(log 2+(-5)(1))
= –(log 2 – 5?
= – log 2+5
= 5–log 2
= 5–0,3
= 4,7

maaf yach klau salah