Respuesta :
Answer:
ph at first equivalency point= 4.51, ph at second equivalency point= 10
Explanation:
ph at first equivalency point is calculated by finding the mid- point of pka1 and pka2, hence;
ph=(pka1 + pka2)/2 = (1.857 +7.172)/2 = 4.51
At second equivalency point SO₃²⁻ is ionized, as shown:
SO₃²⁻ + H₂O ⇄ HSO₃⁻ + OH⁻
Moles of H₂SO₄ neutralized = Volume x Molarity = 0.05 x 0.194 = 0.0097 moles = Moles of SO₃⁻ produced
Second equivalency point means that 1 mole of acid reacts with 2 moles of base, hence
Moles of NaOH required to neutralize SO₃⁻ = 2 x 0.0097 = 0.0194 moles
Volume of NaOH required = No. of moles / Molarity = 0.0194/0.194= 0.1 L
Total Volume of Solution at Second equivalency point = 0.05 + 0.1= 0.15L
Concentration of SO₃⁻ at second equivalency point = No. of moles/ Volume = 0.0097/0.15= 0.065 M
SO₃⁻ is acting as a weak base, thus
Kb= [HSO₃⁻][OH⁻]/[SO₃⁻]
pKb= 14 - pka2 = 14 - 7.127 = 6.828
Kb = 10^ -6.828 = 1.49 x 10⁻⁷
[HSO₃⁻] = x = [OH⁻]
1.49 x 10^⁻⁷ = x²/0.065
x = 9.84 x 10⁻⁵ = [OH⁻]
p[OH⁻] = -㏒[OH⁻] = 4.007≅4
ph = 14 - p[OH⁻] = 14 - 4 = 10
