Question

If a weak acid, HA, is 3% dissociated in a 0.25 M

Answer 1

Answer:

Ka = $2.32 \times 10^{-4}$

pH = 2.12

Explanation:

Calculation of Ka:

% Dissociation = 3% = 0.03

Concentration of solution = 0.25 M

HA dissociated as:

       $HA \rightarrow H^+ + A^{-}$

      C(1 - 0.03)    C×0.03   C×0.03

HA] after dissociation = 0.25×0.97 = 0.2425 M

$[H^+]= 0.25\times0.03 = 0.0075 M$

$[A^{-}]= 0.25 \times 0.03 = 0.0075 M$

$Ka= \frac{[H^+][A^{-}]}{[HA]}$

$Ka = \frac{(0.0075)^2}{0.2475} =2.32 \times 10^{-4}$

Calculation of pH of the solution

$pH = -log [H^+]$

$H^+[\tex] = 0.0075 M$

$pH = -log 0.0075 = 2.12$