Calculate the pH at 25°C of 200.0 mL of a buffer solution that is 0.460 M NH4Cl and 0.460 M NH3 before and after the addition of 2.70 mL of 6.0 M HNO3. (The pKa for NH4+ = 9.75)

Question

A. pH before =

B. pH after =

Answer 1

The Henderson Hasselbalch equation for a basic buffer such as this is...

pOH = pKb + log [conj.acid] [base]

pKa + pKb = 14

pKb = 14 - 9.75 = 4.25

pOH = 4.25 + log [0.460] / [0.460] = 4.25 + log 1

pOH = 4.25

pH = 14 - pOH = 14 - 4.25

pH = 9.75 before addition

When HNO₃ is added, it is an acid so will react with the base (NH₃) to form NH₄⁺ and NO₃^-

NH₃ + H+ ==> NH₄⁺

We can use an ICE table to see how the concentrations of each species is affected:

Initial mols NH₃ = 200.0 ml x 1 L / 1000 ml x 0.460 mol / L = 0.0920 mols NH₃

Initial mols NH₄⁺ = 200.0 ml x 1 L / 1000 ml x 0.460 mol / L = 0.0920 mols NH₄⁺

Initial mols H⁺ = 2.70 ml x 1 L / 1000 ml x 6.0 mol / L = 0.0162 mols H⁺

NH₃ + H⁺ ==> NH₄⁺

0.0920...0.0162....0.0920....Initial

-0.0162...-0.0162...+0.0162...Change

0.0758.......0............0.1082....Equilibrium

Taking the final volume into account, we have 200.0 ml + 2.70 ml = 202.7 ml = 0.2027 L

Final concentrations for the buffer are...

[NH₃] = 0.0758 mols / 0.2027 L =0.374 M

[NH₄⁺] = 0.1082 mols / 0.2027 L = 0.534 M

pOH = pKb + log [conj.acid] / [base]

pOH = 4.25 + log [0.1082] / [ 0.0758]

pOH = 4.25 + log 1.43

pOH = 4.25 + 0.16

pOH = 4.41

pH = 14 - pOH = 14 - 4.41

pH = 9.59 after addition