The acid ionization constant (ka) of an unknown acid if a 0.030M solution of it has a pH of 2.946 is 4.8×10⁻⁵.
Relation between the pH and [H⁺] will be shown as:
[H⁺] = [tex]10^{-pH}[/tex]
Given that pH of solution = 2.946
[H⁺] = [tex]10^{-2.946}[/tex] = 0.00113 = 0.0012
Let the ICE table for the unknown acid HA will be:
HA ⇄ H⁺ + A⁻
Initial: 0.03 0 0
Change: -0.0012 0.0012 0.0012
Equilibrium: 0.03-0.0012 0.0012 0.0012
Acid dissociation constant will be calculated as:
Ka = [0.0012][0.0012] / [0.03-0.0012]
Value of 0.0012 is very small as compared to the 0.03, so the equation becomes
Ka = (0.0012)² / 0.03 = 0.000048 = 4.8×10⁻⁵
Hence value of Ka is 4.8×10⁻⁵.
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