Calculate [ OH - ] of a 0.380 M solution of Aniline C6H5NH2 Kb = 7.40e-10?

Answer 1

I think I answered a very similar question here.

Set up your equation:

[x][x]/.380-x=7.40e-10=Kb

Solve for x by assuming the term (.380M-x) is approximately
.380

x^2=7.40e-10*.380
Your Answer:

x=1.6769e-5M=[OH-]

If you want the pOH

-log[OH-]=-log[1.6769e-10]=4.775
the pH= 14-pOH
pH=14-4.775=9.2245

At a pH of 9.22, the solution is slightly basic.

Answer 2

Kb = [C6H5NH3+][OH-]/[C6H5NH2]
7.40 * 10^-10 = x^2/0.380-x

Since 0.380 >>> 7.40*10^-10, x will be insignificant compared to 0.380.

7.40*10^-10 = x^2/0.380
2.81*10^-10 = x^2
x = [OH-] = 1.68 * 10^-5 M