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.
The ph of a 0.163 m aqueous solution of sulfurous acid is:
This refers to type of solution which has a high solubility in water.
To find the ph of a 0.163 m aqueous solution of sulfurous acid, we would have to:
First construct the ICE table
Next, we would have to express the acid dissociation of sulfurous acid:
[tex]Kₐ = [H⁺ ][ HSO₃⁻]/ [H₂SO₃][/tex]
Next, we input the values
{x²}{(0.163 - x)} = 1.7 x 10⁻²
Expand further
{x²}{0.163} = 1.7 x 10⁻²
x = 0.052640 M
pH = 1.28 .
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