Question

1.What is the pH after the addition of 20.0 mL of 0.14 M HCl to 80.0 mL of a buffer solution that is 0.25 M in NH and 0.25 M in NHCl, (K=1.8 x 10 for NH)?

2.What is the pH of a buffer solution that is 0.25 M in NH and 0.25 M in NHCl? (K for NH = 1.8 x 10)

Answer 1

2)

NH3= 0.25M

NH4Cl = 0.25M

Kb= 1.8x10^-5

-log(Kb) = -log(1.8x10^-5)

PKb = 4.74

POH= PKb+log[salt/base]

POH = 4.74+log(0.25/0.25)

POH= 4.74

PH+POH=14

PH=14-POH

PH= 14-4.74

PH= 9.26

1)

NH3 = 80 mL of 0.25M

number of moles of NH3 = 0.25M x 0.080L = 0.02 moles

NH4Cl = 80 mL of 0.25M

number of moles of NH4Cl = 0.25M x0.08L = 0.02 moles

HCl = 20.0mL of 0.14M

number of moles of NH4Cl = 0.14M x0.020L = 0.0028 moles

after addition of HCl

number of moles of NH3 = 0.02 - 0.0028 = 0.0172 moles

number of moles of NH4Cl = 0.02+0.0028 = 0.0228 moles

POH = 4.74 + log{0.0228/0.0172}

POH= 4.86

PH= 14-4.86

PH= 9.14