задача на расчет pH

[klar]

Расчитать pH 0,04% муравьиной кислоты (КА=2.1*10^-4)

[Че Гулливер]

HCOOH = H+ + HCOO-

pH = -lg[H+]

ρ(0,04%HCOOH) ~ 1000 g/l

Cm = 0,04*1000/46 = 0,9

a = (2,1*10^-4/0,9)^1/2 = 1,53*10^-2

[H+] = 1,53*10^-2*0,9 = 1,4*10^-2

pH = -lg1,4*10^-2 = 2 - 0,15 = 1,85

или еще способ через Ka

Ka = [H+]*[HCOO-]/[HCOOH]

[H+] = x

x^2/0,9 = 2,1*10^-4

x^2 = 1,9*10^-4

x = (1,9*10^-4)^1/2

x = 1,4*10^-2

[H+] = 1,4*10^-2

pH = -lg1,4*10^-2 = 2 - 0,15 = 1,85

[klar]

HCOOH = H+ + HCOO-

pH = -lg[H+]

ρ(0,04%HCOOH) ~ 1000 g/l

Cm = 0,04*1000/46 = 0,9

a = (2,1*10^-4/0,9)^1/2 = 1,53*10^-2

[H+] = 1,53*10^-2*0,9 = 1,4*10^-2

pH = -lg1,4*10^-2 = 2 - 0,15 = 1,85

или еще способ через Ka

Ka = [H+]*[HCOO-]/[HCOOH]

[H+] = x

x^2/0,9 = 2,1*10^-4

x^2 = 1,9*10^-4

x = (1,9*10^-4)^1/2

x = 1,4*10^-2

[H+] = 1,4*10^-2

pH = -lg1,4*10^-2 = 2 - 0,15 = 1,85

спасибо огромное