Respuesta :
H3PO4 = 0.30 M
H2PO4- = 4.59 x 10^-2M
HPO2^4- = 6.17 x 10^-8 M
PO3^4- = 6.44 x 10^-19 M
H+ = 4.59 x 10^-2 M
Phosphoric acid can be classified as a weak acid which dissciates to give three acid where the products of one dissociation equilibrium become components of the subsequent dissociation equilibrium. Below are their equilibrium equations:
H3PO4(aq) ⇌ H2PO4-(aq) + H+(aq) H2PO4-(aq) ⇌ HPO2^4-(aq) + H+(aq)
HPO2^4-(aq) ⇌ PO3^4-(aq) + H+(aq)
The pH for the concentrations are:
Ka = 10^-pKa,
Ka1 = 10−2.16
= 6.17 x 10^-3
Ka2 = 10−7.21
= 6.17 × 10^-8
Ka3 = 10−12.32
= 4.79 × 10
The following concentrations are:
Solving for [H2PO4^-] and [H+],
Ka1 = {[H2PO4^-]*[H+]}/[H3PO4]
6.92 x 10^-3 = ((x)*(x))/0.35 - x
2.42 x 10^-3 - 6.92 x 10^-3 = x2
Solving the above using general quadratic formula,
x = 4.59 x 10^-2M
x = [H2PO4^-] = [H+]
Solving for [HPO4^2-],
Ka2 = {[HPO4^2-]*[H+]}/[H2PO4^-]
6.17 x 10^8- = ((x)*(4.59 x 10-2))/4.59 x 10^-2
x = 6.17 x 10^-7
x = [HPO4^2-]
Solving for [PO4^3-],
Ka3 = [PO4^3-]*[H+]}/[HPO4^2-]
4.79 x 10^-13 = ((x)*(4.5 x 10^-×))/6.17 x 10^-8
x = 6.44 x 10^-19M
x = [PO4^3-]
