Question

Carbonated beverages must be kept acidic in order to generate the levels of CO₂ (g) desired. The acidity if provided by 0.0070 M H₃PO₄. Calculate the [H+], [OH-], [H₃PO₄], [H₂PO₄^-], [HPO₄^2-], and [PO₄^3-], as well as pH, in a can of cola.

Answer 1

Let the Triprotic acid, H3PO4, be H3A, where H+ are the protons and A-3 is the conjugate

Then, we know that this is a weak acid, so we should expect dissociations.

Since this is triprotic, expect three dissociations

H3A <-> H+ + H2A- Ka1

H2A- <-> H+ + HA-2 Ka2

HA-2 <-> H+ + A-3 Ka3

Each having its Equilibrium constant:

Ka1 = [H+][H2A-]/[H3A]

Ka2 = [H+][HA-2]/[H2A-]

Ka3 = [H+][HA-3]/[HA-2]

Note that typically, Ka1 > Ka2 > Ka3

Then,

for the concentration [H3A] = 0.007 M given

Ka1 = 7.52*10^-3; Ka2 = 6.23*10^-8; Ka3 = 4.8*10^-13

First ionization

Ka1 = [H+][H2A-]/[H3A]

let

[H+] = x

[H2A-] = x

[H3A] = M-x = 0.007-x

substitute

Ka1 = (x)(x)/(M-x)

x^2 + Ka1*x - M*Ka1 = 0

solving for x:

x = (-Ka1 +/- sqrt(Ka1^2 - 4*M*Ka1) ) /(2)

x = 0.00441 M

now,

[H+] = x = 0.00441

[H2A-] = x = 0.00441

[H3A] = M-x = 0.007-0.00441 = 0.00259

Second ionization

Ka2 = [H+][HA-2]/[H2A-]

let y be the new ionization of H+

[H+] = x + y = 0.00441+ y

[HA-2] = y = y

[H2A-] = H3A -x = x-y = 0.00441 - y

substitute in Ka2

Ka2 = [H+][HA-2]/[H2A-]

Ka2 = (x+y)(y)/(x-y)

6.23*10^-8 = (0.00441+y)(y) / (0.00441 - y)

note that ionization is too low, so x-y ; x >>> y, then x-y = x

Ka2 = (x+y)(y)/(x)

y^2+ xy = Ka2*x

y^2 + x*y - Ka2*x = 0

from previous work, x = 0.00441 ; Ka2 = 6.23*10^-8

y^2 + x*y - Ka2*x = 0

y^2 + 0.00441*y - (6.23*10^-8)(0.00441) = 0

y = 6.23*10^-8

substitute in

[H+] = x + y = 0.00441+ 6.23*10^-8

[HA-2] = y = 6.23*10^-8

[H2A-] = H3A -x = x-y = 0.00441 - 6.23*10^-8 = 0.00440993

Third Ionization

Ka3 = [H+][A-3]/[HA-2]

let z be the new ionization of H+

[H+] = x + y + z = 0.00441+ z

[HA-2] = z = z

[H2A-] = HA-2 - z = y-z = 6.23*10^-8 - z

substitute in Ka3

Ka3 = [H+][A-3]/[HA-2]

Ka3 = (x+y+z)(z)/(y-z)

note that ionization is too low, so y >>> z, then y-z = y

Ka3 = (x+y+z)*(z)/(y)

Ka3 = (x+z)*(z)/(y)

z^2 + x*z - Ka3*y = 0

substitute values

z^2 + 0.00441*z - (4.8*10^-13)*(6.23*10^-8) = 0

z = 6.78*10^-18

substitute

[H+] = x + y + z = 0.00441+ 6.78*10^-18 = 0.00441

[A-3] = z = z 6.78*10^-18

[HA-2] = y-z = 6.23*10^-8 - 6.78*10^-18 = 6.23*10^-8

pH = -log(H) = -log(0.00441) = 2.355

notes: there was no need to calculate 2nd and 3rd ionization for pH, since [H+] won't chang eedrastically.