Question

A buffer consists of 0.26 M NH4Cl and 0.36 M NH3. What is the pH of the buffer after 0.003 moles of Ca(OH)2 are added to 0.10 L of this buffer solution. Kb for NH3 is 1.8×10‒5.

Answer 1

NH3 = 0.36M

NH4Cl = 0.26M

kb = 1.8x10^-5

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

PKb = 4.74

volume of buffer = 0.10L

number of moles of NH3 = 0.36Mx0.10L = 0.036 moles

number of moles of NH4Cl = 0.26x0.1L = 0.026 moles

number of moles of Ca(OH)2 = 0.003 moles

after additing Ca(OH)2

number of moles of NH3 = 0.036 + 0.003 = 0.039 moles

number of moles of NH4Cl = 0.026 - 0.003 = 0.023 moles

POH = PKb + log( salt/base)

POH = 4.74+log(0.023/0.039)

POH = 4.51

PH+POH= 14

PH= 14-POH

PH = 14-4.51

PH = 9.49.