A 50.0 ml sample of 0.0240 M NH(aq) is titrated with aqueous hydrochloric acid. What is the pH after the addition of 15.0 mL of 0.0600M HCl(aq)? (Kb of NH3= 1.8 x 10^-5)

Question

Initial amount of NH3 (0.0240)(0.05) = .0012 mol

Amount of HCl added (0.0600)(.015) = .009 mol

Amount of NH3 after reaction = .0012 - .0009 =.003 mol

Amount of NH4^+ after reaction = .0009

pOH = pKb + log (concentration of BH+/concentration of B)

pOH = 1.8 X 10^-5 + log [.0009]/[.0003]

1.85x10^-5 + log 3

pOH = 1.85x10^-5 + .47712 = .47713

Something is wrong. I think that I have a problem with the amount or concentration of NH4^+ after reaction.

Answer 1

I think you have two problems but neither is earth shaking.
First, you wrote pKb but typed in and calculated with Kb, not pKb.
pKb is -log Kb = 4.74 according to my figures.
Then pOH = pKb + log acid/base
I get 5.22 for pOH.
Second, the problem asks for pH so you need to subtract this from 14 to get 8.77. I think, however, that you have done it the hard way.
I find trying to remember two different formulas, one for pH and one for pOH is difficult. So I remember only one.
pH = pKa + log base/acid.
pH = 9.25 + log (0.0003/0.0009) = 8.77 directly without the subtraction step. Check my work.

Answer 2

Amount of ammonia initially = 0.050 x 0.0240 =0.0012 mol.

Amount of HCl added =0.015 x 0.0600 = 0.0009 mol.

Amount of ammonia left = 0.0012-0.0009 =0.0003 mol.

Amount of NH4+ ion = Amount of HCl added = 0.0009 mol.

pOH= pKb + log(NH4+/NH3)

pOH= -log(1.8 x 10-5) + log(0.0009/0.0003)

pOH= 4.74 + 0.477 =5.22.

pH =14-pOH.

pH = 14-5.22= 8.78.

Hence, option A) is correct.