Question

Calculate the change in pH when 5.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). A list of ionization constants can be found here.

Calculate the change in pH when 5.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Answer 1

V = 100 mL buffer... 0.1 M of each NH3, NH4+

so..

buffer equation

pOH = pKb +log(NH4+/N3)

pKb = 4.75 for NH3

so...

initially

pOH = 4.75 + log(0.1/0.1) = 4.75

pH = 14-4.75 = 9.25

so..

a)

after adding HCL

mmol of acid = MV = 5*0.1 = 0.5 mmol of acid...

NH3 after reaction = MV - 0.5 = 0.1*100 -0.5 = 10-0.5 = 9.5

NH4+ after addition = MV + 0.5 = 0.1*100+0.5 = 10´0.5 = 10.5

so..

pOH after

pOH = 4.75 + log(10.5/9.5) = 4.7934

pH = 14-4.7934 = 9.2066

pH change = 9.2066 - 9.25 = -0.0434

b)

after adding NaOH

mmol of base = MV = 5*0.1 = 0.5 mmol of base...

NH3 after reaction = MV+ 0.5 = 0.1*100+0.5 = 10.5

NH4+ after addition = MV - 0.5 = 0.1*100-0.5 9.5

so..

pOH after

pOH = 4.75 + log(9.5/10.5) = 4.706

pH = 14-4.706= 9.294

pH change = 9.294- 9.25 = 0.044