Question

Calculate the change in pH when 9.00 mL of 0.100 M HCl(aq) is added to a 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). Calculate the change in pH when 9.00 mL of 0.100 M NaOH(aq) is added to the original buffer system.

Answer 1

pOH = pKb + log [NH4+]/ [NH3]

pKb of ammonia = 4.74

initial pOH = 4.74 + log 0.100/0.100 = 4.74
pH = 14 - 4.74=9.26

moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100
moles H+ added = 9.00 x 10^-3 L x 0.100 M=0.0009

NH3 + H+ = NH4+
moles NH3 = 0.0100 - 0.0009=0.0091
moles NH4+ = 0.0100 + 0.0009=0.0109

total volume = 9 + 100 = 104 ml = 0.109 L

new [NH3] = 0.0091/0.109 = 0.0834 M
[NH4Cl] = 0.0109/0.109= 0.1 M

new pOH = 4.745 + log 0.1/0.0834

pOH = 4.745 + 0.788

pOH = 5.533

new pH = 14 - 5.533 = 8.467

change in pH = 9.26 - 8.467 = 0.793



no.of moles of NaOH = 0.1 X 9/1000 = 0.0009

this time this reaction will take place :-

NH4Cl + NaOH -------> NH3 + NaCl + H2O

so this time NH4Cl is consumed and NH3 is formed

new no.of moles of NH4Cl = 0.01 - 9 X 10^-4 = 0.0091
and no.of moles of NH3 = 0.01 + 9 X 10^-4 = 0.0109

new [NH4Cl] = 0.0091/0.109 = 0.0834 M
[NH3] =0.0109/0.109 = 0.1 M

pOH = 4.745 + log 0.0834/0.1

pOH = 4.745 -0.788

pOH =3.957

so new pH = 14 - 3.957 = 10.043

change = 9.26 - 10.043 = 0.783