Question

Calculate the pH of a solution prepared by adding 20mL of .1M HCl to 80mL of a buffer that is comprimised of .25M NH3 and .25M NH4Cl. Kb of NH3 = 1.8 * 10E-5.

The answer is 9.17 but I dont know how to get the answer or what equations in need.

Answer 1

Apply buffer equation:

for

NH3 + H2O <-> NH4+ + OH-

apply henderson hasslebach equation

pOH = pKb + log(NH4+/NH3)

initial pOH:

pKb = -log(Kb) = log(1.8*10^-5) = 4.744

so

initially:

pOH = 4.744 + log(0.25/0.25) = 4.744

pH = 14-pOH = 14-4.744 = 9.256

so...

mmol of NH3 = MV = 80*0.25 = 20 mmol of NH3

mmol of NH4+ = MV = 80*0.25 = 20 mmol of NH4+

after addition of HCl:

mmol of acid = MV = 20*0.1 = 2 mmol of H+

then

mmol of NH3 = 20 +2 = 22

mmol of NH4+ =20-2 = 18

so

pOH = 4.744 + log(22/18) = 4.8311

pH = 14-pOH = 14-4.8311= 9.1689 which is pretty near to 9.17