pKb = 3.8.
Kb = -log 3.8
Kb = approx 1.6E-4 but you should get a better answer than that.
.......BNH2 + HOH --> BNH3^+ + OH^-
I......0.01............0.......0
C......-x..............x.......x
E......0.01-x..........x.......x
Kb = 1.6E-4 = (x)(x)/(0.01-x)
Solve for x = (OH^-) and convert to pH.
Note: YOu may need to use the quadratic formula as 0.01-x may NOT be = 0.01. You will need to try it.
What would be pH of a 0.01 M solution of
a very expensive weak base, unobtainamine,
whose pKb is 3.8?
1. 2.9
2. 14
3. 2
4. 7
5. 11.1
1 answer
1 answer