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Question

$H_{2} C O_{3}(a q)+H_{2} O(l) \rightleftharpoons H C O_{3}(a q)+H_{3} O_{x}^{+}(a q)$

$\quad 0.034-x\quad \quad \quad \quad \q

Vedclass · Accepted Answer

$H^+$ અને $HCO_3^-$ ની સાંદ્રતા લગભગ સમાન છે

Vedclass · Answer

$H_{2} C O_{3}(a q)+H_{2} O(l) ightleftharpoons H C O_{3}(a q)+H_{3} O_{x}^{+}(a q)$

$\quad 0.034-x\quad \quad \quad \quad \quad \quad \quad x\quad \quad \quad  \quad \quad x$

$K_{1}=\frac{\left.\left[H C O_{3}^{-}ight] | H_{3} O^{+}ight]}{\left|H_{2} C O_{3}ight|}$

$=\frac{x 	imes x}{0.034-x}$

$\Rightarrow 4.2 	imes 10^{-7}=\frac{x^{2}}{0.034}$

$\Rightarrow x=1.195 	imes 10^{-4}$

As $H_{2} C O_{3}$ is a weak acid so the concentration of

$H_{2} C O_{3}$ will remain 0.034 as $0.034>>x$

$x=\left[H^{+}ight]=\left[H C O_{3}ight]$

$=1.195 	imes 10^{-4}$

Now, $H C O_{3}(a q)+H_{2} O(l) ightleftharpoons C O_{3_{y}}^{2-}(a q)+H_{3} O_{y}^{+}(a q)$

As $H C O_{3}$ is again a weak acid (weaker than $H_{2} C O_{3}$ )

with $x>>y$

$K_{2}=\frac{\left[c o_{3}^{2}ight]\left|H_{3} O^{+}ight|}{\left[H C o_{3}ight]}$

$=\frac{y 	imes(x+y)}{(x-y)}$

Note : $\left[H_{3} O^{+}ight]=H^{+}$ from first step $(x)$ and from second step $(y)=(x+y)$

$[	ext { As } x>>y 	ext { so } x+y \simeq x 	ext { and } x-y \simeq x]$

So, $K_{2} \simeq \frac{y 	imes x}{x}=y$

$\Rightarrow K_{2}=4.8 	imes 10^{-11}$

$y=y=\left[C O_{3}^{2}ight]$

So the concentration of $\left[H^{+}ight]=\left[H C O_{3}^{-}ight]=$ concentrations obtained from the first step. As the dissociation will be very low in second step so there will be no change in these concentrations. Thus the final concentrations are $\left[H^{+}ight]=\left[H C O_{3}^{-}ight]=1.195 	imes 10^{-4}$

$\left[\mathrm{CO}_{3}^{2-}ight]=4.8 	imes 10^{-11}$

ઍસિડ	$K_a$
$HCN$	$6.2\times 10^{-10}$
$HF$	$7.2\times 10^{-4}$
$HNO_2$	$4.0\times 10^{-4}$

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