Question

1a)Calculate the change in pH when 8.00mL of 0.100 M HCl is added to 100.0mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

b)Calculate the change in pH when 8.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Answer 1

1a)

pH initially

this is abuffer so

pOH = pKb + log(conjugate/base)

pKb for NH3 = 4.75(from databases)

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

pH = 10-4.75 = 9.25

after addition of HCl

mol of HCl added = MV = 8*0.1 = 0.8 mmol of HCl

mol of NH3 initially = MV = 100*0.1 = 10 mmol

mol of NH4+ initially = MV = 100*0.1 = 10 mmol

after addition

NH3 is neutralized so = 10 - 0.8 = 9.2

NH4+ is formed so = 10 + 0.8 = 10.8

therefore

pOH = 4.75 + log(10.8/9.2) = 4.819635

pH = 14-4.819635 = 9.180

pH change = 9.180 - 9.25 = 0.07 (decreases since acid is being added)

b)

when NAOH is added,

pretty simlar to a

after addition of HCl

mol of NaOHadded = MV = 8*0.1 = 0.8 mmol of NAOH

mol of NH3 initially = MV = 100*0.1 = 10 mmol

mol of NH4+ initially = MV = 100*0.1 = 10 mmol

after addition

NH3 is formed so = 10 + 0.8 = 10.8

NH4+ is reacted so = 10 - 0.8 = 9.2

therefore

pOH = 4.75 + log(9.2/10.8) = 4.680

pH = 14-4.680= 9.32

pH change = 9.32 - 9.25 = +0.07 (increases since base is being added)