Codeine is added to 100ml of water until the solution concentration is 0.800M. What is the pH of this solution?
What are you using for pKb?
Call codeine BN. Then (BN) = 0.800M
...........BN + HOH ==> BNH^+ + OH
initial...0.800.........0.........0
change.....-x...........x.........x
equil....0.800-x........x.........x
Kb = ? = (BNH^+)(OH^-)/(BN)
Substitute from the ICE chart into the Kb expression and solve for x = OH^-, then convert to pH.