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Chemical Equilibrium question

2010 · Shift 0 · Q5
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  5. /2010 · Shift 0 · Q5

Chemical Equilibrium question

2010 · Shift 0 · Q5

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
In aqueous solution the ionization constants for carbonic acid are K1 = 4.2 x 10–7 and K2 = 4.8 x 10–11 Select the correct statement for a saturated 0.034 M solution of the carbonic acid.
  1. A
    The concentration of CO32−CO_3^{2−}CO32−​ is 0.034 M.
  2. B
    The concentration of CO32−CO_3^{2−}CO32−​ is greater than that of HCO3−HCO_3^{−}HCO3−​
  3. C
    The concentration of H+H^+H+ and HCO3−HCO_3^−HCO3−​ are approximately equal.
  4. D
    The concentration of H+H^+H+ is double that of CO3−CO_3^−CO3−​.
View written solutionFree

Correct answer: C

  1. Write the dissociation steps of carbonic acid

Carbonic acid is a diprotic acid:

H2CO3⇌H++HCO3−K1=4.2×10−7H_2CO_3 \rightleftharpoons H^+ + HCO_3^- \qquad K_1 = 4.2\times 10^{-7}H2​CO3​⇌H++HCO3−​K1​=4.2×10−7

HCO3−⇌H++CO32−K2=4.8×10−11HCO_3^- \rightleftharpoons H^+ + CO_3^{2-} \qquad K_2 = 4.8\times 10^{-11}HCO3−​⇌H++CO32−​K2​=4.8×10−11

Given initial concentration of carbonic acid:

C=0.034 MC = 0.034\,\text{M}C=0.034M

We must identify the correct statement for this solution.


  1. First ionization dominates

Since K1≫K2K_1 \gg K_2K1​≫K2​, the first dissociation is much greater than the second. So we first calculate [H+][H^+][H+] and [HCO3−][HCO_3^-][HCO3−​] from the first ionization approximately as for a weak acid.

Let

[H+]=[HCO3−]=x[H^+] = [HCO_3^-] = x[H+]=[HCO3−​]=x

and

[H2CO3]=0.034−x[H_2CO_3] = 0.034 - x[H2​CO3​]=0.034−x

Then,

K1=x20.034−x=4.2×10−7K_1 = \frac{x^2}{0.034-x} = 4.2\times 10^{-7}K1​=0.034−xx2​=4.2×10−7

Since xxx is small compared to 0.0340.0340.034, take 0.034−x≈0.0340.034-x \approx 0.0340.034−x≈0.034:

x2=(4.2×10−7)(0.034)x^2 = (4.2\times 10^{-7})(0.034)x2=(4.2×10−7)(0.034)

x2=1.428×10−8x^2 = 1.428\times 10^{-8}x2=1.428×10−8

x≈1.2×10−4 Mx \approx 1.2\times 10^{-4}\,\text{M}x≈1.2×10−4M

Thus,

[H+]≈[HCO3−]≈1.2×10−4 M[H^+] \approx [HCO_3^-] \approx 1.2\times 10^{-4}\,\text{M}[H+]≈[HCO3−​]≈1.2×10−4M

So statement C already appears correct.


  1. Check carbonate ion concentration using second dissociation

From

K2=[H+][CO32−][HCO3−]K_2 = \frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}K2​=[HCO3−​][H+][CO32−​]​

Since [H+]≈[HCO3−][H^+] \approx [HCO_3^-][H+]≈[HCO3−​], we get

[CO32−]≈K2=4.8×10−11 M[CO_3^{2-}] \approx K_2 = 4.8\times 10^{-11}\,\text{M}[CO32−​]≈K2​=4.8×10−11M

This is extremely small.

So:

  • it is certainly not 0.034 M0.034\,\text{M}0.034M
  • it is much smaller than [HCO3−][HCO_3^-][HCO3−​]

Therefore A and B are false.


  1. Check statement D

Statement D says: concentration of H+H^+H+ is double that of carbonate ion.

But actually,

[H+]≈1.2×10−4 M[H^+] \approx 1.2\times 10^{-4}\,\text{M}[H+]≈1.2×10−4M

and

[CO32−]≈4.8×10−11 M[CO_3^{2-}] \approx 4.8\times 10^{-11}\,\text{M}[CO32−​]≈4.8×10−11M

Clearly,

[H+]≫2[CO32−][H^+] \gg 2[CO_3^{2-}][H+]≫2[CO32−​]

So statement D is false.

(Also, if interpreted as CO3−CO_3^-CO3−​, that species is not relevant here; carbonate is CO32−CO_3^{2-}CO32−​.)


  1. Evaluate all options
  • A: False, because [CO32−]≈4.8×10−11 M[CO_3^{2-}] \approx 4.8\times 10^{-11}\,\text{M}[CO32−​]≈4.8×10−11M, not 0.034 M0.034\,\text{M}0.034M.
  • B: False, because [CO32−]≪[HCO3−][CO_3^{2-}] \ll [HCO_3^-][CO32−​]≪[HCO3−​].
  • C: True, because for the first dissociation of weak diprotic acid, [H+]≈[HCO3−]≈K1C[H^+] \approx [HCO_3^-] \approx \sqrt{K_1C}[H+]≈[HCO3−​]≈K1​C​
  • D: False, because [H+][H^+][H+] is enormously greater than [CO32−][CO_3^{2-}][CO32−​], not merely double.

  1. Final answer

The correct option is:

C\boxed{\text{C}}C​

This matches the stored correct answer.

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