Question

For each solution, calculate the initial and final pH after adding 0.012 mol of NaOH

(c) 250.0 mL of a buffer solution that is 0.259 M in CH₃CH₂NH₂ and 0.226 M in CH₃CH₂NH₃Cl

initial pH
final pH

Answer 1

(c)
K_b of ethylammine = 4.3 x 10^-4
pKb = 3.25
pOH = 3.25 + log 0.226 / 0.259 = 3.190
pH = 14 - 3.190 = 10.81 ( initial pH)
moles CH₃CH₂NH₃⁺ = 0.226 x 0.250 L = 0.0565
moles CH₃CH₂NH₂ = 0.259 x 0.250 L = 0.06475
CH₃CH₂NH₃⁺ + OH^- = CH₃CH₂NH₂ + H₂O
moles CH₃CH₂NH₃⁺ = 0.0565 - 0.012 = 0.0445
moles CH₃CH₂NH₂ = 0.06475 + 0.012 = 0.07675
concentration CH₃CH₂NH₃⁺ = 0.0445 / 0.250 =0.178 M
concentration CH₃CH₂NH₂ = 0.07675/ 0.250 = 0.307 M
pOH = 3.25 + log 0.178 / 0.307 = 3.013
pH = 14 - 3.013 = 10.987 (final pH)