Let's call morphine BN.
....................BN + HOH ==> BNH^+ + OH^-
I....................0.50.....................0...............0
C.....................-x.......................x................x
E...................0.50-x...................x................x
Kb = (BNH^+)(OH^-)/(BN)
Substitute the E line into the Kb expression and solve for OH^-, then convert to pH. Post your work if you get stuck.
the kb for morphine is 7.4x10^-7. What is the pH of 0.50 M morphine
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