The acid-dissociation constants of sulfurous acid (H₂SO₃)are K_a1= 1.7 × 10^-2 and at 25.0 °C.Calculate the pH of a 0.163 M aqueous solution of sulfurous acid.

Question

Quizplus · Accepted Answer

First, we need to write out the two equilibrium reactions involved with sulfurous acid:

HSO3- ⇌ H+ + SO32-      K1 = 1.7 × 10-2
SO32- ⇌ H+ + HSO3-     K2 = 6.4 × 10-8

Next, we can set up an ICE table to determine the concentrations of the various species at equilibrium.

For the first equilibrium:

HSO3- ⇌ H+ + SO32-
I         0.163 M     0             0
C       -x            +x         +x
E        0.163-x     x          x

For the second equilibrium:

SO32- ⇌ H+ + HSO3-
I           x              0          0
C        -x            +x        +x
E          0            x         x

(Note that we can use the x value from the first equilibrium for the initial concentration of the SO32- ion in the second equilibrium.)

Now, we can use the equilibrium constants to set up two expressions for the concentration of the hydrogen ion (H+) in terms of x:

K1 = [H+][SO32-]/[HSO3-] = (x)(x)/(0.163-x) = 1.7 × 10-2
K2 = [H+][HSO3-]/[SO32-] = (x)(x)/(0.163-x) = 6.4 × 10-8

Solving for x in the first equation gives x = 7.1 × 10-3 M. Substituting this value into the second equation and solving gives [H+] = 1.3 × 10-2 M.

Finally, we can use the definition of pH to find the pH of the solution:

pH = -log[H+] = -log(1.3 × 10-2) = 1.28

Therefore, the best choice is (B).

The Acid-Dissociation Constants of Sulfurous Acid (H2SO3)are Ka1 = 1

The Acid-Dissociation Constants of Sulfurous Acid (H₂SO₃)are K_a1= 1