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Standard XII

Chemistry

Ostwald Dilution Law Mathematical Interpretation

It has been f...

Question

It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.

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Solution

Let the organic acid be HA
$\Rightarrow H A \leftrightarrow H^{+} + A^{-}$
Concentration of HA = 0.01 M pH
= 4.15
$- l o g [H^{+}] = 4.15 [H^{+}] = 7.08 \times 10^{- 5}$
$K_{a} = \frac{[H^{+}] [A^{-}]}{[H A]}$
Now,
$[H^{+}] = [A^{-}] = 7.08 \times 10^{- 5} [H A] = 0.01$
Then,
$K_{a} = \frac{(7.08 \times 10^{- 5}) (7.08 \times 10^{- 5})}{0.01} K_{a} = 5.01 \times 10^{- 7} p K_{a} = - l o g K_{a} = - l o g (5.01 \times 10^{- 7}) p K_{a} = 6.3001$

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Ostwald Dilution Law Mathematical Interpretation

Standard XII Chemistry

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It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.

Let the organic acid be HA ⇒HA↔H++A− Concentration of HA = 0.01 M pH = 4.15 −log[H+]=4.15[H+]=7.08×10−5 Ka=[H+][A−][HA] Now, [H+]=[A−]=7.08×10−5[HA]=0.01 Then, Ka=(7.08×10−5)(7.08×10−5)0.01Ka=5.01×10−7pKa=−log Ka=−log(5.01×10−7)pKa=6.3001