Answer to Question #207112 in General Chemistry for Avi

Question

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 the second ionization step.

Answer 1

Given:

 Ka₁=1.0×10⁻⁴

Ka₂=5.0×10⁻⁵

Explanation:

$\star$ From the given information :

             H₂A    →     HA⁻  + H⁺    

Initial         0.10            0       0

Change        -x              x       x

Equilibrium   0.10 - x        x       x

$Ka_1={[HA^-][H^+]\over[H_2A]}$

$1 \times 10^{-4}={[x][x]\over[0.10-x]}$

By solving for x;

x² = (1.0 × 10⁻⁴ × 0.1)

x = 

x = [H⁺] =[HA⁻] = 0.00311 M

$\star$ The acid then further its disociation again,

So;

              HA⁻     →       A⁻  + H⁺    

Initial         0.00311           0    0.00311

Change        -x             x        x

Equilibrium    0.00311 - x       x     0.00311 + x

$Ka_2={[A^-][H^+]\over[HA]}$

$5 \times 10^{-5}={[ 0.00311 + x][x]\over[ 0.00311 -x]}$

By solving for x;

x = [H⁺] = 0.0000484 M

$\bigstar$ Therefore, the first disociation occured at 0.00311 M and the second at 0.0000484 M.