Hello,
I am trying to calculate the pH for a titration experiment after neutralization, and I can't find anywhere the ionization constant (Kb) for C3H7COO-
pH of 50 mL of KOH added # of moles KOH
X = 0.10 = 5.0 X 10-3 moles
0.050 L 1L
# of moles C3H7COOH
X = 0.05 = 5.0 x 10-3 moles
0.100 L 1L
C3H7COOH KOH ----------> C3H7COO-(K) H20
Initiation 5.00 x 10-3 5.00 X 10-3 O O
Change 5.00 X 10-3 5.00 X 10-3 5.00 X 10-3 O
Final O O 5.00 X 10-3 O
COMPLETELY NEUTRALIZED
Molarity C3H7COO-(K) 5.00 x 10-3 = 0.1 M
0.050 L
C3H7COOH KOH ----------> C3H7COO- H30+
Initiation 0.00 0.00 0.10 O
Change x x (x) O
Equilibrium x x 0.10 -x O
Kb [C3H7COOH] [KOH]
[C3H7COO-]
WHAT IS Kb = (x](x]
0.10 - x
Kb =X2
Kb = X2
Kb = X or OH
pOH = log (X)
pOH =
pH + X = 14
pH =