JEE MAIN - Chemistry (2010 - No. 5)
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.
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.
The concentration of
$$CO_3^{2−}$$ is 0.034 M.
The concentration of
$$CO_3^{2−}$$ 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 $$CO_3^−$$.
Explanation
$$\mathop {{H_2}C{O_3}\left( {aq} \right)}\limits_{0.034 - x} \, + \,{H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {HCO_3^ - \left( {aq} \right)}\limits_x \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_x $$
$${K_1} = {{\left[ {HCO_3^ - } \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {{H_2}C{O_3}} \right]}}$$
$$ = {{x \times x} \over {0.034 - x}}$$
$$ \Rightarrow 4.2 \times {10^{ - 7}} = {{{x^2}} \over {0.034}}$$
$$ \Rightarrow x = 1.195 \times {10^{ - 4}}$$
As $${H_2}C{O_3}$$ is a weak acid so the concentration of
$${H_2}C{O_3}$$ will remain $$0.034\,\,$$ as $$0.034 > > x.$$
$$x = \left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right]$$
$$ = 1.195 \times {10^{ - 4}}$$
Now, $$\mathop {HCO_3^ - }\limits_{x - y} \left( {aq} \right) + {H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {CO_3^{2 - }\left( {aq} \right)}\limits_y \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_y $$
As $$HCO_3^ - \,$$ is again a weak acid (weaker than $${H_2}C{O_3})$$
with $$x > > y.$$
$${K_2} = {{\left[ {CO_3^{{2^ - }}} \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {HCO_3^ - } \right]}}$$
$$ = {{y \times \left( {x + y} \right)} \over {\left( {x - y} \right)}}$$
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, $$\,\,\,{K_2} \simeq {{y \times x} \over x} = y$$
$$ \Rightarrow {K_2} = 4.8 \times {10^{ - 11}}$$
$$ = y = \left[ {CO_3^{{2^ - }}} \right]$$
So the concentration of $$\,\,\,\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = \,\,\,$$ 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
$$\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}}$$
$$\& \,\,\,\left[ {CO_3^{2 - }} \right] = 4.8 \times {10^{ - 11}}$$
$${K_1} = {{\left[ {HCO_3^ - } \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {{H_2}C{O_3}} \right]}}$$
$$ = {{x \times x} \over {0.034 - x}}$$
$$ \Rightarrow 4.2 \times {10^{ - 7}} = {{{x^2}} \over {0.034}}$$
$$ \Rightarrow x = 1.195 \times {10^{ - 4}}$$
As $${H_2}C{O_3}$$ is a weak acid so the concentration of
$${H_2}C{O_3}$$ will remain $$0.034\,\,$$ as $$0.034 > > x.$$
$$x = \left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right]$$
$$ = 1.195 \times {10^{ - 4}}$$
Now, $$\mathop {HCO_3^ - }\limits_{x - y} \left( {aq} \right) + {H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {CO_3^{2 - }\left( {aq} \right)}\limits_y \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_y $$
As $$HCO_3^ - \,$$ is again a weak acid (weaker than $${H_2}C{O_3})$$
with $$x > > y.$$
$${K_2} = {{\left[ {CO_3^{{2^ - }}} \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {HCO_3^ - } \right]}}$$
$$ = {{y \times \left( {x + y} \right)} \over {\left( {x - y} \right)}}$$
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, $$\,\,\,{K_2} \simeq {{y \times x} \over x} = y$$
$$ \Rightarrow {K_2} = 4.8 \times {10^{ - 11}}$$
$$ = y = \left[ {CO_3^{{2^ - }}} \right]$$
So the concentration of $$\,\,\,\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = \,\,\,$$ 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
$$\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}}$$
$$\& \,\,\,\left[ {CO_3^{2 - }} \right] = 4.8 \times {10^{ - 11}}$$
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