Calculate [ H + ] of a 0.220 M solution of Aniline C6H5NH2 Kb = 7.40e-10?

Answer 1

Kb is the equilibrium constant of the base:

x^2/.220-x=7.4e-10
This is approximatey equal to:

x^2/.220=7.4e-10

Solve for x:
x=square root (7.4e-10*.220)
x=1.276e-5M

This gives us the [OH+], but we want the [H+]

We know that pH + pOH=14

pOH= -log [OH-]=-log[1.276e-5]=4.89

pH=14-4.89
=9.10583 (so we know that this solution is slightly basic, since it is over 7)

We do inverse log (10^x key) of negative(don't forget the negative) 9.10583.

The result is 7.837e-10M for [H+]