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 $pK_a$.
Let the organic acid be $HA$.
$HA \longleftrightarrow H^{+} + A^{-}$
Concentration of $HA = 0.01 \ M$
$pH = 4.15$
$-\log [H^{+}] = 4.15$
$[H^{+}] = 10^{-4.15} = 7.08 \times 10^{-5} \ M$
Since $[H^{+}] = [A^{-}]$,the concentration of the anion $[A^{-}] = 7.08 \times 10^{-5} \ M$.
Now,$K_a = \frac{[H^{+}][A^{-}]}{[HA]} = \frac{(7.08 \times 10^{-5})^2}{0.01} = 5.01 \times 10^{-7}$.
$pK_a = -\log K_a = -\log (5.01 \times 10^{-7}) = 6.30$.
$HA \longleftrightarrow H^{+} + A^{-}$
Concentration of $HA = 0.01 \ M$
$pH = 4.15$
$-\log [H^{+}] = 4.15$
$[H^{+}] = 10^{-4.15} = 7.08 \times 10^{-5} \ M$
Since $[H^{+}] = [A^{-}]$,the concentration of the anion $[A^{-}] = 7.08 \times 10^{-5} \ M$.
Now,$K_a = \frac{[H^{+}][A^{-}]}{[HA]} = \frac{(7.08 \times 10^{-5})^2}{0.01} = 5.01 \times 10^{-7}$.
$pK_a = -\log K_a = -\log (5.01 \times 10^{-7}) = 6.30$.
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| Triethylamine,$(C_{2}H_{5})_{3}N$ | $6.45 \times 10^{-5}$ |
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| Aniline,$C_{6}H_{5}NH_{2}$ | $4.27 \times 10^{-10}$ |
| Urea,$CO(NH_{2})_{2}$ | $1.3 \times 10^{-14}$ |
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