Question

In an aqueous solution of a certain acid the acid is 0.094% dissociated and the pH is 4.55. Calculate the acid dissociation constant Ka of the acid. Round your answer to 2 significant digits

Answer 1

Answer:

Explanation:

pH = 4.55

[ H⁺ ] = 10⁻⁴°⁵⁵

= 2.82 x 10⁻⁵

Let the acid be HA and its molar concentration be a .

         HA         ⇄       H ⁺       +        A ⁻

a - .094x10⁻³a      .094x10⁻³a    .094x10⁻³a

.094x10⁻³a  = 2.82 x 10⁻⁵

a = 30 x 10⁻²

= .3

a - .094x10⁻³a   = .3 - .094 x 10⁻³ x .3

= .29997 approx

Ka =  2.82 x 10⁻⁵ x  2.82 x 10⁻⁵ /  .29997

= 7.95 x10⁻¹⁰ / .29997

= 26.5 x 10⁻¹⁰

= 27 x 10⁻¹⁰ ( rounding off to two digits )