It has been found that the $p H$ of a $0.01 M$ solution of an organic acid is $4.15$ . Calculate the concentration of the anion, the ionization constant of the acid and its ${p K}_{a}$ .

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Solution

$p H = - l o g [H^{+}] = 4.15$
$[H^{+}] = a n t i l o g (- 4.15) = 7.08 \times 10^{- 5}$
$[A^{-}] = [H^{+}] = 7.08 \times 10^{- 5}$
The concentration of undissociated acid is $0.01 - 0.000071 = 0.009929 M$ .
$H A + H_{2} O ⇌ H_{3} O^{+} + A^{-}$
$K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[H A]} = \frac{(7.08 \times 10^{- 5}) (7.08 \times 10^{- 5})}{0.009929} = 5.05 \times 10^{- 7}$
$p K_{a} = - l o g K_{a} = - l o g 5.05 \times 10^{- 7} \approx 6.3$

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

pH=−log[H+]=4.15[H+]=antilog(−4.15)=7.08×10−5[A−]=[H+]=7.08×10−5The concentration of undissociated acid is 0.01−0.000071=0.009929M.HA+H2O⇌H3O++A−Ka=[H3O+][A−][HA]=(7.08×10−5)(7.08×10−5)0.009929=5.05×10−7pKa=−logKa=−log5.05×10−7≈6.3

It has been found that the $p H$ of a $0.01 M$ solution of an organic acid is $4.15$ . Calculate the concentration of the anion, the ionization constant of the acid and its ${p K}_{a}$ .