d. Calculate the pH at equivalence (after 24.0 mL of base, NaOH has been added).
HF 0H =F
+
+
H28
+
(0.054X
2
0.05
-
1.4706x15
-"
=
b
-
b 0.27410
=
7
e. Calculate the pH after 30.0 mL of base, NaOH has been added.
-
AF OH
-
⑩
0.048MH7
0.06 M
F
-
+
0.012M
Naut
0.048M
H28 Hf
=
F
-
+
-
0.048
1.4706x10+)
=
-
7
0H
-
+
+
0.048
x(x 0.012)
H20
+
↓
⑩23
0
0.012
4