Question

What pH would be obtained by adding 5mL of 0.100M NaOH to 20 mL of a 0.100M NH3/NH4Cl buffer with a pH of 9.26?

Answer 1

Number of moles of NaOH = 0.1 X 5/1000 = 5*10^-4

The reaction will be as-
NH4Cl + NaOH -------> NH3 + NaCl + H2O
Some of NH4Cl is consumed and NH3 is formed
New no.of moles of NH4Cl = 0.002– 5*10^-4 = 0.0015
And no.of moles of NH3 = 0.002+ 5*10^-4 = 0.0025
New [NH4Cl] = 0.0015/0.025 = 0.06 M
[NH3] =0.0025/0.025 = 0.1 M
pOH = pKb + log [NH4Cl]/[NH3]
pKb for NH3 = 4.745
so pOH = 4.745 + log 0.06/0.1
pOH = 4.745 + log 0.6
pOH = 4.745 -0.22 = 4.525
and the new pH = 14 - pOH = 14 - 4.525 = 9.475
change = 9.475 - 9.26 = 0.215