Question

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

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

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Homework Answers

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

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