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

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
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
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

Explanation / Answer

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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