Question

A 0.00300 M solution of an acid is 12% ionized. Calculate the acid ionization constant Ka?

Answer 1

`4.32 × 10^-5`

Explanation:

Acid is `12%` ionised. So, degree of dissociation (`α`) is

`α = 12/100 = 0.12`

For weak acid, acid dissociation constant (`"K"_"a"`) is approximately given by

`"K"_"a" = "c"α^2 = 0.00300 × (0.12)^2 = 4.32 × 10^-5`

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