$K_{a_{1}}$ for dibasic acid is $X \times 10^{- 5}$ mole/l, if its concentration is $0.05 N$ and $p H$ is $3$ . Find the value of $X$ . (Assume second dissociation constant very less)

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Solution

$p H = 3$
Therefore, $[H^{+}] = 10^{- 3} M$ .

$H_{2} A$
$H A^{-}$
$H^{+}$
Initial conc.
0.05
0
0
Equilibrium conc.
0.05-0.001
0.001
0.001
The equilibrium constant expression is $K = \frac{[H A^{-}] [H^{+}]}{[H A]} = \frac{0.001 \times 0.001}{0.049} = 2 \times 10^{- 5} = X \times 10^{- 5}$ .
Hence, $X = 2$ .

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	$H_{2} A$	$H A^{-}$	$H^{+}$
Initial conc.	0.05	0	0
Equilibrium conc.	0.05-0.001	0.001	0.001

Ka1 for dibasic acid is X×10−5 mole/l, if its concentration is 0.05N and pH is 3. Find the value of X. (Assume second dissociation constant very less)

pH=3Therefore, [H+]=10−3 M.H2AHA−H+Initial conc. 0.0500Equilibrium conc.0.05-0.001 0.001 0.001The equilibrium constant expression is K=[HA−][H+][HA]=0.001×0.0010.049=2×10−5=X×10−5.Hence, X=2.

$K_{a_{1}}$ for dibasic acid is $X \times 10^{- 5}$ mole/l, if its concentration is $0.05 N$ and $p H$ is $3$ . Find the value of $X$ . (Assume second dissociation constant very less)