In aqueous solution the ionization constants for carbonic acid are K₁=4.2 × 10^{-7} and K₂=4.8 × 10^{-11} Select the correct statement for a saturated 0.034 {M} solution of the carbonic acid. [AIEEE 2010]

Correct Answer is : The concentration of H+{H}^{+}H+and HCO3−{HCO}₃^{-}HCO3−are approximately equal.

The concentration of {CO}₃²-} is 0.034 {M}.

The concentration of {CO}₃²-} is greater than that of {HCO}₃^{-}

The concentration of {H}^{+}and {HCO}₃^{-}are approximately equal.

The concentration of {H}^{+}is double that of {CO}₃^{-}.

Correct Answer is : The concentration of H+{H}^{+}H+and HCO3−{HCO}₃^{-}HCO3−are approximately equal.

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AIEEE 2010 Solved Paper

Section: Chemistry

Question : 9 of 89

Marks: +1, -0

K_1=4.2 \times 10^{-7}

K_2=4.8 \times 10^{-11}

0.034 \text{M}

The concentration of $\mathrm{CO}_3^{2-}$ is $0.034 \text{M}$ .
The concentration of $\mathrm{CO}_3^{2-}$ is greater than that of $\mathrm{HCO}_3^{-}$
The concentration of $\mathrm{H}^{+}$ and $\mathrm{HCO}_3^{-}$ are approximately equal.
The concentration of $\mathrm{H}^{+}$ is double that of $\mathrm{CO}_3^{-}$ .

Solution:

\;\mathrm{H}_2 \underset{0.034-x}{\mathrm{CO}}_3 \text{(aq)} + \mathrm{H}_2\mathrm{O} \text{(l)} \rightleftharpoons \mathrm{HCO}_3^{-} \text{(aq)} + \mathrm{H}_3\mathrm{O}_x^{+} \text{(aq)}

K_1 = \frac{[\mathrm{HCO}_3^{-}] [\mathrm{H}_3\mathrm{O}^{+}]}{[\mathrm{H}_2\mathrm{CO}_3]}

= \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}

\mathrm{H}_2\mathrm{CO}_3

is a weak acid so the concentration of

\mathrm{H}_2\mathrm{CO}_3

will remain

0.034

0.034 \gg x

x = [\mathrm{H}^{+}] = [\mathrm{HCO}_3^{-}]

= 1.195 \times 10^{-4}

Now,

\;\mathrm{HCO}_3^{\underset{x-y}{-}} \text{(aq)} + \mathrm{H}_2\mathrm{O} \text{(l)} \rightleftharpoons \mathrm{CO}_3^{2-} \text{(aq)} + \mathrm{H}_3\mathrm{O}_y^{+} \text{(aq)}

\mathrm{HCO}_3^{-}

is again a weak acid (weaker than

\mathrm{H}_2\mathrm{CO}_3

)

with

x \gg y

K_2 = \frac{[\mathrm{CO}_3^{-}] [\mathrm{H}_3\mathrm{O}^{+}]}{[\mathrm{HCO}_3^{-}]}

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

Note :

[\mathrm{H}_3\mathrm{O}^{+}] = \mathrm{H}^{+}

from first step

(x)

and from second step

(y) = (x+y)

[As

x \gg y

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 = [\mathrm{CO}_3^{2-}]

So the concentration of

[\mathrm{H}^{+}] = [\mathrm{HCO}_3^{-}] =

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

[\mathrm{H}^{+}] = [\mathrm{HCO}_3^{-}] = 1.195 \times 10^{-4}

\& [\mathrm{CO}_3^{2-}] = 4.8 \times 10^{-11}

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