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Vitek1552 [10]
3 years ago
12

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

Chemistry
1 answer:
masha68 [24]3 years ago
4 0

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

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