Verified Solution • 2025–26 Session moderate

QJEEM10Ch6972: JEE Main 2010 Chemistry - Class 11 JEEM10Ch6972 Solution: In aqueous solution the ionization constants for carbonic acid…

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.

Multiple Choice Options

  • A The concentration of CO32−CO_3^{2−} is 0.034 M.
  • B The concentration of CO32−CO_3^{2−} is greater than that of HCO3−HCO_3^{−}
  • C The concentration of H+ and HCO3−HCO_3^− are approximately equal.
  • D The concentration of H+ is double that of CO3−CO_3^− .

Step-by-Step Verified Solution

H2CO3(aq)0.034−x + H2O(l) ⇌ HCO3−(aq)x + H3O+(aq)x\mathop {{H_2}C{O_3}\left( {aq} \right)}\limits_{0.034 - x} \, + \,{H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {HCO_3^ - \left( {aq} \right)}\limits_x \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_x

K1=[HCO3−][H3O+][H2CO3]{K_1} = {{\left[ {HCO_3^ - } \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {{H_2}C{O_3}} \right]}}

=x×x0.034−x= {{x \times x} \over {0.034 - x}}

⇒4.2×10−7=x20.034\Rightarrow 4.2 \times {10^{ - 7}} = {{{x^2}} \over {0.034}}

⇒x=1.195×10−4\Rightarrow x = 1.195 \times {10^{ - 4}}

As H2CO3{H_2}C{O_3} is a weak acid so the concentration of

H2CO3{H_2}C{O_3} will remain 0.034  0.034\,\, as 0.034 > > x.

x=[H+]=[HCO3−]x = \left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right]

=1.195×10−4= 1.195 \times {10^{ - 4}}

Now, HCO3−x−y(aq)+H2O(l) ⇌ CO32−(aq)y + H3O+(aq)y\mathop {HCO_3^ - }\limits_{x - y} \left( {aq} \right) + {H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {CO_3^{2 - }\left( {aq} \right)}\limits_y \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_y

As HCO3− HCO_3^ - \, is again a weak acid (weaker than H2CO3){H_2}C{O_3})

with x > > y.

K2=[CO32−][H3O+][HCO3−]{K_2} = {{\left[ {CO_3^{{2^ - }}} \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {HCO_3^ - } \right]}}

=y×(x+y)(x−y)= {{y \times \left( {x + y} \right)} \over {\left( {x - y} \right)}}

Note : [H3O+]=H+  \left[ {{H_3}{O^ + }} \right] = {H^ + }\,\, from first step (x)(x)

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

[As \,\,\,x > > y\,\, so    x+y≃x   \,\,\,x + y \simeq x\,\,\, and x−y≃xx - y \simeq x ]

So,    K2≃y×xx=y\,\,\,{K_2} \simeq {{y \times x} \over x} = y

⇒K2=4.8×10−11\Rightarrow {K_2} = 4.8 \times {10^{ - 11}}

=y=[CO32−]= y = \left[ {CO_3^{{2^ - }}} \right]

So the concentration of    [H+]=[HCO3−]=   \,\,\,\left[ {{H^ + }} \right] = \left[ {HCO_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

[H+]=[HCO3−]=1.195×10−4\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}}

&   [CO32−]=4.8×10−11\& \,\,\,\left[ {CO_3^{2 - }} \right] = 4.8 \times {10^{ - 11}}

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