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6-2.Equilibrium-II (Ionic Equilibrium)
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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 $p{K_a}$
Option A
Option B
Option C
Option D
Solution
Let the organic acid be $HA$.
$\Rightarrow HA \longleftrightarrow H ^{+}+ A$
Concentration of $HA =0.01 \,M \,pH$
$=4.15$
$-\log \left[ H ^{+}\right]=4.15$
$\left[ H ^{+}\right]=7.08 \times 10^{-5}$
Now, $K_{a}=\frac{\left[ H ^{+}\right]\left[ A ^{-}\right]}{[ HA ]}$
$\left[ H ^{+}\right]=\left[ A ^{-}\right]=7.08 \times 10^{-5}$
$[ HA ]=0.01$
Then, $K_{a}=\frac{\left(7.08 \times 10^{-5}\right)\left(7.08 \times 10^{-5}\right)}{0.01}$
$K_{a}=5.01 \times 10^{-7}$
$p K_{a}=-\log K_{a}$
$=-\log \left(5.01 \times 10^{-7}\right)$
$p K_{a}=6.3001$
Standard 11
Chemistry
