Odpowiedź:
Wyjaśnienie:
Zad.1
HCN=[H2O]=H+ + CN-
K = [H+]*[CN-]/[HCN]
[H+]=[CN-]
[HCN]= 0,25M
pKa= 8,52 to Ka = 10⁻⁸'⁵²
Ka = 3,02*10⁻⁹
Ka = [H+]²/025
[H+] = √Ka*025
[H+] = √3,02*10⁻⁹ * 0,25
[H+] = 2,75*10⁻⁵
pH = -log (2,75*10⁻⁵)
pH = 4,56
Zad.2
C = 0,45M
pKb = 4,33
Kb = 10⁻⁴'³³
Kb = 4,68*10⁻⁵
NH3*H2O =[H2O]= NH4+ + OH-
Kb = [NH₄⁺]*[OH⁻]/[NH3*H2O]
Kb = [OH⁻]²/[NH₄OH]
[OH-]=[NH₄⁺]
[NH₃*H₂O] = 0,45M
[OH-] = √Kb* 0,45
[OH-] = √4,68*10⁻⁵*0,45
[OH-]= 4,59*10⁻³
pOH = -log(4,59*10⁻³)
pOH = 2,34
pH = pKw - pOH
pH = 13,97 - 2,34
pH = 11,63
α = Czdys/C
α = 4,59*10⁻³/0,45
α = 0,0102 ≅ 0,01 = 1%
