Consider a 0.10 M solution of a weak polyprotic acid (H2A) with the possible values of Ka1 and Ka2 given below. Calculate the contributions to [H3O+] from each ionization step. Ka1=1.0×10−4;Ka2=5.0×10−5

The first dissociation occurred at 0.00311 M and the second at 0.0000484 M.

Explanation:

From the given information:

The ICE table can be computed as follows:

                       H2A         →          HA⁻    +   H⁺        

Initial                0.10                      0              0

Change              -x                        x               x

Equilibrium        0.10 - x                x               x

By solving for x;

x² = (1.0 × 10⁻⁴ × 0.1)

x =

x = [H⁺] =[HA⁻] = 0.00311 M

The acid then further its dissociation again, So;

The ICE table can be computed as follows:

                         HA⁻         →          A⁻    +   H⁺        

Initial                0.00311                  0        0.00311

Change              -x                        x               x

Equilibrium      0.00311 - x             x           0.00311 + x

By solving for x;

x = [H⁺] = 0.0000484 M

Therefore, the first dissociation occurred at 0.00311 M and the second at 0.0000484 M.

it's c. elements are placed in rows based on how may valencies electrons have.