Question

The acid-dissociation constants of sulfurous acid (h2so3) are kal = 1.7 × 10-2 and ka2 = 6.4 × 10-8 at 25.0°c. calculate the ph of a 0.163 m aqueous solution of sulfurous acid.

Answer 1

Constructing the ICE table for acid-dissociation  of sulfurous acid

 H₂SO₃ --> H⁺ + HSO₃⁻

I    0.163 M      0        0

C     - x           + x       + x

E   0.163 - x    + x       + x

Writing the acid dissociation expression  of sulfurous acid,

Kₐ = [H⁺ ][ HSO₃⁻]/ [H₂SO₃]

Plugging in the values we get,

\frac{x²}{(0.163 - x)} = 1.7 x 10⁻²

Since 1.7 x 10⁻² is small we can ignore x in the denominator,

\frac{x²}{0.163}  = 1.7 x 10⁻²

x = 0.052640 M  

pH = 1.28  

Thus the pH of a 0.163 M aqueous solution of sulfurous acid is 1.28.

Answer 2

Given:

acid-dissociation constants of sulfurous acid:

Ka1 = 1.7 * 10^(-2)

Ka2 = 6.4 * 10^(-8) at 25.0 °C. 

aqueous solution of sulfurous acid = 0.163 M

 

x² / (0.163 - x) = 1.7 * 10^(-2) 

You simplify it to:

x² / (0.163) = 1.7  *10^(-2) 

x = 0.052640 M 

pH = 1.28 

So, the pH of a 0.163 M aqueous solution of sulfurous acid is 1.28.

 

To add, aqueous solutions of sulfur dioxide purpose are as disinfectants and reductant, as are solutions of sulfite salts and bisulfite. By accepting another oxygen atom, they are oxidised to sulfuric acid or sulfate.