Question

Calculate the change in pH to 0.01 pH units caused by adding 10. mL of 2.25-M NaOH is added to 670. mL of each of the following solutions.

a) water

pH before mixing =
pH after mixing=
pH change =

b) 0.130 M NH₄¹⁺

pH before mixing =
pH after mixing=
pH change =

c) 0.130 M NH₃

pH before mixing =
pH after mixing=
pH change =

d) a buffer solution that is 0.130 M in each NH₄¹⁺ and NH₃

pH before mixing =
pH after mixing=
pH change =

Answer 1

a) molarity of diluted NaOH = 2.25 x 10/680 = 0.033 M

pH before mixing = 14 - (-log(2.25)) = 14.35

pH after mixing = 14 - pOH = 14 - (-log(0.033)) = 12.52

chnnge in pH = 12.52 - 14.35 = -1.83

b) pH before mixing = 14.35

molarity of [NH4+] in total volume = 0.130 x 0.67/(0.68) = 0.13 M

molarity of NaOH in total solution = 2.25 x 0.01/0.68 = 0.033 M

excess [NH4+] = 0.13 - 0.097 = 0.097 M

NH4+ hydrolyzes in water to give H3O+

Ka = 5.55 x 10^-10 = x^2/0.097

x = [H+] = 7.34 x 10^-6

pH = 5.13

pH change = 5.13 - 14.35 = -9.22

c) initial pH = 14.35

initial moles NH4+ = 0.130 x 0.67 = 0.0871 mols

initial moles NH3 = 0.13 x 0.67 = 0.0871 mols

Added moles NaOH = 2.25 x 0.01 =  0.0225 mols

final moles NH4+ = 0.0871 - 0.0225 = 0.0646 mols

final moles NH3 = 0.0871 + 0.0225 = 0.1096 mols

molarity of NH4+ = 0.0646/0.68 = 0.095 M

molarity of NH3 = 0.1096/0.68 = 0.161 M

Using Hendersen-Hasselbalck equation,

pH = pKa + log[base/acid]

pKa = 9.25

pH after mixing = 9.25 + log(0.161/0.095) = 9.48

change in pH = 9.48 - 14.35 = -4.87