Calculating the kb

Question

The pH of a 0.016 M aqueous solution of p-toluidine (CH3C6H4NH2) is 8.60. Calculate the kb.

Answer 1

CH₃C₆H₄NH₂ + H₂O ==> CH₃C₆H₄NH₃⁺ + OH^-

Kb = [CH₃C₆H₄NH₃⁺][OH^-]/[CH₃C₆H₄NH₂]

From pH = 8.60, we can find [H⁺] and hence [OH^-]. [H⁺] = 1x10^-8.60 = 2.51x10^-9 M

[OH^-] = 1x10^-14/2.51x10^-9 = 3.98x10^-6 M

(This can also be found from pOH: pOH = 14-pH = 14-8.60 = 5.4 and OH = 10^-5.4 = 3.98x10^-6 M)

Kb = (3.98x10^-6)(3.98x10^-6)/0.016 - 3.98x10^-6 and assuming 3.98x10-6 is small relative to 0.016, ignore it...

Kb = 1.58x10^-11/0.016

Kb = 9.9x10^-10

Answer 2

First, we write out our chemical equation based on what we know. We know that we are looking for Kb, so we are going to be working with a weak base. Because bases are protonated or gain a proton/H+, we see that NH2 on the p-toluidine becomes NH3 and molecule now has a + charge. We also see that OH- is present, because all bases produce OH-. Next lets write out our ICE table:

CH3C6H4NH2(aq) + H20(l) --> CH3C6H4NH3+(aq) + OH-(aq)

I .016M / 0 0

C -x / +x +x

E .016-x / +x +x

We will use the equilibrium equation, Kb = [products]/[reactants] to find our Kb value.

so, Kb = [CH3C6H4NH3+][OH-]/[CH3C6H4NH2]

now using the values from our ice table, we get

Kb = [x][x]/[.106-x]

We now need to find x. We see that in our ICE table, x is equal to [OH-] or the molarity of hydroxide in the reaction, so we need to find molarity of hydroxide.

We are given pH = 8.6, so because pOH+pH = 14, we can find pOH using 14-8.6 = 5.4 pOH.

Using pOH, we will find molarity of OH- using 10^-(pOH) = [OH-]: 10^-5.4 = 3.981x10^-6

now we can plug this into our equilibrium equation, knowing that the molarity of OH- we just calculated is equal to x:

Kb = [3.981x10^-6]^2/ .016-(3.981x10^-6]

= 9.9077x10^-10 (no units of importance)