You made a typo in your post. Aniline is C6H5NH2
Let's call aniline a simple ANH2.
........ANH2 + HOH ==> ANH3^+ + OH^-
I.......0.13............0........0
C........-x.............x........x
E.......0.13-x..........x........x
Substitute the E line into the Kb expression for aniline and solve for x and 0.13-x. x = (OH^-), convert that to pOH then to pH.
what is the pH in a 0.13M solution of aniline (Kb=4.3*10^-10).
What is the concentration of C6HNH2 in the same solution?
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