check my work, ph and Molarity are known, what is Kb?

Question

check work and sig figs to see if i got them right

The pH of a 0.010M solution of a weak base is 11.20. Calculate Kb for the base.

first I take 10^-11.20 to get the [H+] = 6.3x10^-12

Since I need [OH-] not [H+] I devide 10^-14/6.3x10^-12=
.0016 = [OH-]

now that I have OH- I plug in the #'s and get

(.0016)(.0016)/(0.010-.0016) and the answer to this gives me Kb which is 3.0x10^-4

Does this look good, am I forgetting anything? thanks!

Answer 1

pH = 11.2, so (at 25 oC) pOH = 14 - 11.2, pOH = 2.8. Then:

[OH-] = 10^(-2.8), or [OH-] = 1.6*10^(-3) approx.

From the equilibrium:

B + H2O <==> BH+ + OH-

we have that [B+] = [OH-] = 1.6*10^(-3) and [B] = 0.01 - 1.6*10^(-3), [B] = 8.4*10^(-3) approx.

Kb = [BH+]*[OH-]/[B], Kb = 1.6*10^(-3)*1.6*10^(-3)/8.4*10^(-3) so:

Kb = 3.0^10(-4)

So you are correct!!