JEE Main 2010 — Chemical Equilibrium Question with Solution

$$\underset{0.034-x}{H_{2}C{O}_3\left(aq\right)}+H_{2}O\left(l\right)\rightleftharpoons \underset{x}{HC{O}_3^{-}\left(aq\right)}+\underset{x}{H_3{O}^{+}\left(aq\right)}$$

$$K_1=\frac{\left[HC{O}_3^{-}\right]\left[H_3{O}^{+}\right]}{\left[H_{2}C{O}_3\right]}$$

$$=\frac{x\times x}{0.034-x}$$

$$\Rightarrow 4.2\times 10^{-7}=\frac{x^2}{0.034}$$

$$\Rightarrow x=1.195\times 10^{-4}$$

As $$H_{2}C{O}_3$$ is a weak acid so the concentration of

$$H_{2}C{O}_3$$ will remain $$0.034$$ as $$0.034>>x.$$

$$x=\left[H^{+}\right]=\left[HC{O}_3^{-}\right]$$

$$=1.195\times 10^{-4}$$

Now, $$\underset{x-y}{HC{O}_3^{-}}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons \underset{y}{C{O}_3^{2-}\left(aq\right)}+\underset{y}{H_3{O}^{+}\left(aq\right)}$$

As $$HC{O}_3^{-}$$ is again a weak acid (weaker than $$H_{2}C{O}_3)$$

with $$x>>y.$$

$$K_2=\frac{\left[C{O}_3^{2^{-}}\right]\left[H_3{O}^{+}\right]}{\left[HC{O}_3^{-}\right]}$$

$$=\frac{y\times \left(x+y\right)}{\left(x-y\right)}$$

Note : $$\left[H_3{O}^{+}\right]=H^{+}$$ from first step $$(x)$$

and from second step $$\left(y\right)=\left(x+y\right)$$

[As $$x>>y$$ so $$x+y\simeq x$$ and $$x-y\simeq x$$]

So, $$K_2\simeq \frac{y\times x}{x}=y$$

$$\Rightarrow K_2=4.8\times 10^{-11}$$

$$=y=\left[C{O}_3^{2^{-}}\right]$$

So the concentration of $$\left[H^{+}\right]=\left[HC{O}_3^{-}\right]=$$ 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

$$\left[H^{+}\right]=\left[HC{O}_3^{-}\right]=1.195\times 10^{-4}$$

$$\&\left[C{O}_3^{2-}\right]=4.8\times 10^{-11}$$