Chem final- have to pull up my grade

Question

Any help on this answer and how set up to solve.

Kb for ammonia is 1.8 x 10-5 at 25oC. Calculate the pH of the solution at the equivalence point at 25oC when 74.4 mL of 0.568 M ammonia is titrated with 0.281 M HCl

Answer 1

At the equivalence point, all that remains is NH4Cl in solution.

NH3 + HCl => NH4Cl

Let's calculate how much NH4Cl was produced and what the volume of the solution is.

moles NH3 = M NH3 x L NH3 = (0.568)(0.0744) = 0.0423 moles NH3

moles HCl = moles NH3 since they react 1:1.

0.0423 moles HCl x (1 L HCl / 0.281 moles) = 0.150 L HCl = 150 mL HCl.

So the final volume of the solution at the endpoint is 74.4 + 150 = 224 mL.

[NH4+] = moles NH4+ / L = 0.0423 moles / 0.224 L = 0.189 M

To calculate the pH, this is an acid hydrolysis problem (NH4+ is the conjugate acid of NH3).

Molarity . .NH4+ + H2O  <==>  H3O+ + NH3

. initial . . .0.189 . . . . . . . . . . . . 0 . . . .0

change . . .-x . . . . . . . . . . . . . .+x . . .+x

final . . . .0.189 - x . . . . . . . . . . .x . . .x . .

Now Ka NH4+ = Kw / Kb NH3 =

(1 x 10^-14) / (1.8 x 10^-5) = 5.6 x 10^-10

Ka = [H3O+][NH3] / [NH4+] = (x)(x) / 0.189 - x

Because Ka is so small, the -x term after 0.189 can be ignored.

5.6 x 10^-10 = x^2 / 0.189

1.1 x 10^-10 = x^2

1.0 x 10^-5 = x = [H3O+]

pH = -log [H3O+] = -log (1.0 x 10^-5) = 5.0