Calculate the pH at 25°C of 199.0 mL of a buffer solution that is 0.210 M NH4Cl and 0.210 M NH3 before and after the addition of 1.80 mL of 6.0 M HNO3. (The pKa for NH4+ = 9.75)

Question

Calculate the pH at 25°C of 199.0 mL of a buffer solution that is 0.210 M  NH₄Cl and 0.210 M  NH₃ before and after the addition of 1.80 mL of 6.0 M  HNO₃. (The pK_a for NH₄⁺ = 9.75)

pH before =

pH after =

Answer 1

Before any additions...

pH = pKa + log [conj.base]/[acid]

pH = 9.75 + log [0.210] / [0.210]

pH = 9.75

or

pOH = pKb + log [conj.acid] / [base] and pKb = 14 - 9.75 = 4.25

pOH = 4.25 + log [0.210/0.210]

pOH = 4.25

pH = 14 - 4.25

pH = 9.75

After addition of 1.80 ml of 6.0 M HNO₃ (assuming volumes are additive)

HNO₃ will react with NH₃ to produce NH₄⁺

moles NH₃ initially present = 190 ml x 1 L / 1000 ml x 0.210 mol/L = 0.0399 mols

moles NH₄⁺ initially present = 190 ml x 1 L/ 1000 ml x 0.210 mol/L = 0.0399 mols

moles H⁺ added = 1.80 ml x 1 L / 1000 ml x 6.0 mol/L = 0.0108 mols H⁺

NH₃ + H⁺ ==> NH₄⁺

0.0399...0.0108....0.0399........Initial

-0.0108..-0.0108....+0.0108....Change

0.0291.......0..........0.0507.......Equilibrium

pH = pKa + log [0.0291] / [0.0507]

pH = 9.75 + log 0.5740

pH = 9.75 - 0.241

pH = 9.51

or

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

pOH = 4.25 + log [0.0507] / [0.0291]

pOH = 4.25 + 0.241

pOH = 4.49

pH = 14 - 4.49

pH = 9.51