Question

Putin ajotor va rog:)
Sa se calculeze pH-ul urmatoarelor solutii:
a) solutie de C6H5-SO3H (pka=-6,5) de concentrație 2*10^-4 moli/L
b) solutie de HCOOH (ka=17,7*10^-5 mol/L, pka=3,77) ce
contine 230 mg de HCOOH in 100mL solutie
c) solutie CH3-NH-CH3 (kb=5,2*10^-4 mol/L, pkb=3,28) de concentrație 10^-4 M

Answer 1

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×$10^{-5}$×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