What is the pH of a solution that results when 40mL of 0.1M NH3;?

Question:

a.)is diluted to 20mL with distilled water
b.)when it is mixed with 20mL of 0.2MHcl solution
c.)mixed with 20mL of 0.25M HCl solution
d.)mixed with 20mL of 0.2M NH4Cl ... solution
e.)is mixed with 20mL of 0.1M HCl solution

Answer:

moles NH3 = 0.040 L x 0.1 M = 0.0040

a) total volume = 0.060 L
[NH3]= 0.0040/0.060=0.067 M
Kb = 1.8 x 10^-5 = x^2/ 0.067-x
x = [OH-]= 0.0011 M
pOH = 3.0
pH = 11

b) moles HCl = 0.020 x 0.2 M=0.004
H+ + NH3 = NH4+
NH4+ + H2O <----> NH3 + OH-
total volume = 0.060 L
[NH4+]= 0.004/ 0.060 =0.067 M
Ka = Kw/Kb = 5.6 x 10^-10 = x^2/ 0.067-x
x = [H+]=6.1 x 10^-6 M
pH = 5.2

c) moles HCl = 0.020 x 0.25 =0.0050
moles H+ in excess = 0.0050 - 0.0040 = 0.0010
total volume = 0.060 L
[H+]= 0.0010 / 0.060=0.017 M
pH = 1.8

d) moles NH4+ = 0.020 x 0.2 = 0.0040
total volume = 0.060 L
[NH3]= 0.0040 / 0.060=0.067 M = [NH4+]
pOH = pKb + log 0.067/0.067 = 4.7
pH = 9.3

e) moles HCl = 0.020 x 0.1 = 0.0020
moles NH3 = 0.0040 - 0.0020 = 0.0020 = moles NH4+
pH = 9.3

Category: Science