a) pKa = -6,5
C = 2×10⁻⁴ M C₆H₅-SO₃H
C₆H₅-SO₃H - acid tare
[C₆H₅-SO₃H] = C = [H₃O⁺] = 2×10⁻⁴ M
pH = -lg 2×10⁻⁴ = 3,7
b) Ka = 17,7×10⁻⁵ M
pKa = 3,77
md = 230mg = 0,23g
Vs = 100mL = 0,1L
C = md/M×Vs(L) = 0,23/46×0,1 = 0,05 M
[H₃O⁺] = √17,7×[tex] 10^{-5} [/tex]×5×10⁻² = 2,974×10⁻³ M
pH = -lg[H₃O⁺] = -lg 2,974×10⁻³ = 2,52
c) Kb = 5,2×10⁻⁴
pKb = 3,28
C = 10⁻⁴ M H₃C-NH-CH₃
[OH⁻] = √5,2×10⁻⁴×10⁻⁴ = 2,28×10⁻⁴ M
[H₃O⁺] = 10⁻¹⁴/2,28×10⁻⁴ = 4,385×10⁻¹¹
pH = -lg 4,385×10⁻¹¹ = 10,35
