Question

What is the pH to 2 decimal places after 46.9 mL of 0.20 M HCl is added to 125 mL of buffer prepared by mixing 0.10 M NH₃ and 0.10 M NH₄⁺? (K_b = 1.76x10^-5)

Answer 1

initially

millimoles of NH3 = 125 x 0.1 = 12.5

millimoles of NH4⁺ = 125 x 0.1 = 12.5

millimoles of HCl added = 46.9 x 0.2 = 9.38

after HCl added

millimoles of NH3 = 12.5 - 9.38 = 3.12

millimoles of NH4⁺   = 12.5 + 9.38 = 21.88

total volume = 125 + 46.9 = 171.9 mL

[NH3] = 3.12 / 171.9 = 0.0181 M

[NH4⁺] = 21.88 / 171.9 = 0.127 M

pOH = pKb + log [NH4⁺] / [NH3]

pKb = - log Kb = - log [1.76 x 10^-5]

pKb = 4.75

pOH = 4.75 + log [0.127] / [0.0181]

pOH = 5.60

pH = 14 - 5.60

pH = 8.40