Question

A)      Carbonic acd, H2CO3, can be found in a wide variety of body fluids (from dissolved CO2). Calculate the hydronium ion concentration of 6.38 x 10 ^-4 M, H2CO3 solution. What is the concentration of CO3 ^2-?     Ka (H2CO3)= 4.3 x 10 ^-7      Ka (HCO3)= 4.8 x 10 ^-4

B)     Calculate the base-ionizatoin constant for SO4 ^2- and PO4 ^3-.   Ka (HSO4)= 1.1 x 10 ^-2   Ka (PO4^2-)= 4.8 x 10 -13

Answer 1

(A)         Step 1: Compute x for the first ionization:

H2CO3        + H2O   ---------> HCO3^-       +        H3O⁺

I         6.38 x 10 ^-4 M

C                  - x                                           + x                     +x

E      6.38 x 10 ^-4 M - x       x    x                                         

Ka1 = [HCO3^-]*[H3O+] / [H2CO3] = x*x / (6.38 x 10 ^-4 - x)

4.3 x 10 ^-7 = x² / (6.38 x 10 ^-4 - x)

Solve for x

x = 1.63*10^-5 M = [H3O+] = [HCO3^-]

Step 2: compute the second ionization

                      HCO3^-    +   H2O   ------------->        H3O+       + CO3-2

I                   1.63*10^-5 M                          1.63*10^-5 M

C                       -x                                                 +x                           +x

E             1.63*10^-5 M-x                               1.63*10^-5 M+x                   x

Ka2= [CO3-2][H3O+] / [HCO3^- ]

4.8 x 10 ^-4 = ((1.63*10^-5 +x )*x)/(1.63*10^-5 -x)

Solve for x: (final answer)

x = 1.53*10^-5 M = [CO3^-2]

[H3O+] = 1.63*10^-5 M+x   = 1.63*10^-5 + 1.53*10^-5 M = 3.16*10^-5 M

(B) We know that:

Kw = Ka*Kb = 1*10^-14

So, for SO4 ^2-:

Kb = Kw/Ka = 1*10^-14 / 1.1 x 10 ^-2 = 9.1*10^-13

And for PO4 ^3-:

Kb = Kw/Ka = 1*10^-14 / 4.8 x 10 ^-13 = 0.0208