Question

Calculate the pH of the solution that results from each of the following mixtures.

Part A

140.0 mL of 0.23 M  HF with 230.0 mL of 0.31 M  NaF

Part B

175.0 mL of 0.12 M C2H5NH2 with 270.0 mL of 0.20 M C2H5NH3Cl

What mass of ammonium chloride should be added to 2.65 L of a 0.150 M NH3 in order to obtain a buffer with a pH of 9.45?

Answer 1

part A

pH of acidic buffer = pka + log(salt/acid)

no of mole of acod = 140*0.23 = 32.2 mmole

no of mole of salt = 230*0.31 = 71.3 mmole

pka of HF = 3.17

pH = 3.17 + log((71.3/32.2)

   = 3.51

part B

pH = 14 - (pkb+log(salt/base))

no of mole of base(C2H5NH2 ) = 175*0.12 = 21 mmole

no of mole of salt(C2H5NH3+) = 270*0.2 = 54 mmole

pkb of C2H5NH2 = 3.3

pH = 14 - (3.3 + log(54/21))

    = 10.3

part C

pH = 14 - (pkb+log(salt/base))

pkb of NH3 = 4.74

no of mole of NH3 = 2.65*0.15 = 0.4 mole

no of mole of NH4Cl = x mole

9.45 = 14 - (4.74+log(x/0.4))

x = no of mole of NH4Cl = 0.26 mole

amount of NH4Cl = 0.26*53.5 = 13.91 g