Question

4. Calculate the pH after 0.010 mol of gaseous HCl is added to 250.0 mL of each of the

following buffered solutions. (Kb for NH3 = 1.76 x 10^5)

a. 0.050 M NH3 and 0.15 M NH4Cl

b. 0.50 M NH3 and 1.50 M NH4Cl

Answer 1

a)

250 ml of 0.050 M NH3 = 0.250 * 0.050 = 0.0125 mole.

250 ml of 0.15 M NH4Cl = 0.250 * 0.15 = 0.0375 mole.

PKb (NH3) = - log (1.76 * 10^-5) = 4.75.

Henderson equation is

POH = PKb + log [salt] / [base]

POH = 4.75 + log [(0.0375 + 0.010) / (0.0125 - 0.010)]

POH = 6.03

PH = 14 - 6.03 = 7.97

PH = 7.97

b)

250 ml of 0.50 M NH3 = 0.250 * 0.50 = 0.125 mole.

250 ml of 1.50 M NH4Cl = 0.250 * 1.50 = 0.375 mole.

POH = PKb + log [salt] / [base]

POH = 4.75 + log [(0.375 + 0.010) / (0.125 - 0.010)]

POH = 5.21

PH = 14 - 5.21 = 8.79

PH = 8.79