Answer to 1E.
35mL
VT
.7143
+.7143
0
[NH4+]
remain
0
1.429
Rxt LR
-.7143
-.7143
Start
(10/35)*2.5
=.7143
10
(25/35)*3
=2.143
25
[H3O+]
V.H3O
[NH3]
V NH3
Buffer remains. Use HH:
pH=-log(5.56(10-10)) + log(1.429/.7143)
pH=5.046
Step back to the problem
Kb=1.8(10-5) for NH3, Thus Ka for NH4+=10-14/1.8(10-5)=5.56(10-10)