Question

Consider the titration of 0.100 M HCl, a strong acid, with 0.100 M NH3, a weak base: HCL(aq)+NH3(aq)---NH4+(aq)+H2O(l) Calculate the pH at the following points in the titration when 25.0 mL of 0.0100 M NH3 is titrated with 0.100 M HCL. A. 0.0 mL of HCl added B. 10.0 mL of HCl added C. At equivalence point D. 35.0 mL of HCL added

Answer 1

HCL(aq)+NH3(aq)---NH4+(aq)+H2O

(l) Calculate the pH at the following points in the titration when 25.0 mL of 0.0100 M NH3 is titrated with 0.100 M HCL.

A. 0.0 mL of HCl added

moles NH3 = 0.0250 L x 0.0100 M=0.00025 mol

Kb = 1.8 x 10^-5
pKb = 4.74

a)
NH3 + H2O <=> NH4+ + OH-
1.8 x 10^-5 = x^2/ 0.0100-x

Due to small value of Kb 0.0100-x= 0.0100

X^2= 1.8*10^-7

X=4.24*10^-4
x = [OH-]= 4.24*10^-4M
pOH = 3.37
pH = 14 – 3.37

= 10.63

B. 10.0 mL of HCl added of 0.100 M HCl

25.0 mL of 0.100 M NH3

moles NH3 = 0.0250 L x 0.100 M=0.0025 mol

moles HCl = 0.0100 L x 0.100 M= 0.001 mol HCl

this react with NH3 IN 1;1 ratio.
moles NH3 in excess = 0.00250 - 0.001=0.0015
moles NH4+ = 0.001
total volume =10.0 ml+25.0ml= 35 ml= 0.0350 L
[NH3]= 0.0015 / 0.0350=0.043 M
[NH4+]= 0.001/ 0.0350=0.0285 M
pKb = 4.76
pOH = 4.76 + log 0.0285/0.043

pOH = 4.76 + log 0.663

pOH = 4.76 -0.178

=4.582
pH=14-pOH=14-4.582

=9.418

C. At equivalence point

moles HCl required to reach the equivalence point = 0.0025

volume HCl = 0.0025/ 0.100 = 0.0250 L

25+25 = 50 ml or 0.0500 L
total volume = 0.0500 L
moles NH4+ formed = 0.00500
[NH4+]= 0.0250/ 0.0500=0.500 M

NH4+ <=> NH3 + H+
Ka = Kw/Kb = 5.6 x 10^-10 = x^2/ 0.500-x

5.6 x 10^-10 = x^2/ 0.500-x

x = [H+]= 1.67 x 10^-5 M

pH = 4.78

D. 35.0 mL of HCL added

moles HCl = 0.0350 x 0.100=0.0035 HCl mol
moles H+ in excess = 0.0035- 0.0025 = 0.001
total volume = 0.060 L
[H+]= 0.001/0.0600=0.0167 M

pH =- log [H+]= -log 0.0167

pH=1.78