Answer to Question #147532 in General Chemistry for gia

Question

Calculate the concentrations of all species in a 0.100 M H3PO4 solution.(Ka1 = 7.1 x 10^-3

, Ka2 = 6.3 x 10^-8 and

Ka3 = 4.5 x 10^-13)

Answer 1

First Dissociation

H₃PO_4(aq)= H⁺_(aq) + H₂PO^-_4(aq)

I          0.100        0         0

C         -x             +x          +x

E         (0.1-x)        x             x

Ka₁= [H⁺] [H₂PO^-₄]/[H₃PO₄]

x²/(0.1-x) =7.1× 10^-3

x²/0.1=7.1×10^-3

Assuming that x˂˂0.1

x²=0.1×7.1×10^-3

x=0.027M

Second Dissociation

H₂PO^-_4(aq)=H⁺_(aq) + HPO^2-_4(aq)

I          0.027       0.027       0

C         -y              +y             +y

E        ( 0.027-y)   (0.027+y)   y

K_a2= [H⁺] [HPO^2-₄]/[H₂PO^-₄] = y(0.19+y)/(0.19-y)

    y=6.3×10^-8M

Assuming that y˂˂0.027

Third Dissociation

HPO^2-_4(aq)= H⁺_(aq)  +     PO^3-_4(aq)

I          6.3×10^-8     0.19          0

C         -z                 +z               +z

E          (6.3×10^-8)-z  (0.027+z)  z

K_a3=[H⁺] [PO^3-₄]/[HPO^2-₄]

   (0.027+z) z/ (6.3×10^-8)-z = 4.5×10^-13

0.027z/ (6.3×10^-8) = 4.5×10^-13

Assuming that [H+] from third dissociation completely negligible

Z= 1.05×10^-18 M