Question
In aqueous solution the ionization constants for carbonic acid are $${K_1} = 4.2 \times {10^{ - 7}}$$ and $${K_2} = 4.8 \times {10^{ - 11}}.$$
Select the correct statement for a saturated 0.034 M solution of the carbonic acid.
A.
The concentration of $$CO_3^{2 - }$$ is $$0.034\,M.$$
B.
The concentration of $$CO_3^{2 - }$$ is greater than that of $$HCO_3^ - .$$
C.
The concentrations of $${H^ + }$$ and $$HCO_3^ - $$ are
approximately equal.
D.
The concentration of $${H^ + }$$ is double that of $$CO_3^{2 - }.$$
Answer :
The concentrations of $${H^ + }$$ and $$HCO_3^ - $$ are
approximately equal.
Solution :
$$\eqalign{
& \mathop {{H_2}C{O_3}\left( {aq} \right)}\limits_{0.034 - x} + {H_2}O\left( l \right) \rightleftharpoons \mathop {HCO_3^ - \left( {aq} \right)}\limits_x + \mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_x \cr
& {K_1} = \frac{{\left[ {HCO_3^ - } \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {{H_2}C{O_3}} \right]}} = \frac{{x \times x}}{{0.034 - x}} \cr
& \Rightarrow 4.2 \times {10^{ - 7}} \simeq \frac{{{x^2}}}{{0.034}} \Rightarrow x = 1.195 \times {10^{ - 4}} \cr} $$
As $${H_2}C{O_3}$$ is a weak acid so the concentration of $${H_2}C{O_3}$$ will remain $$0.034$$ as $$0.034 > > x.$$
$$\eqalign{
& x = \left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}} \cr
& {\text{Now,}}\,\,\mathop {HCO_3^ - }\limits_{x - y} \left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons \mathop {CO_3^{2 - }}\limits_y \left( {aq} \right) + \mathop {{H_3}{O^ + }}\limits_y \left( {aq} \right) \cr} $$
As $$HCO_3^ - $$ is again a weak acid ( weaker than $${H_2}C{O_3}$$ ) with x >> y.
$${K_2} = \frac{{\left[ {CO_3^{2 - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {HCO_3^ - } \right]}} = \frac{{y \times \left( {x + y} \right)}}{{\left( {x - y} \right)}}$$
Note : $$\left[ {{H_3}{O^ + }} \right] = {H^ + }$$ from first step $$(x)$$ and from second step $$\left( y \right) = \left( {x + y} \right)$$
$$\eqalign{
& \left[ {{\text{As}}\,\,x > > y\,\,\,so\,\,\,x + y \simeq x\,\,\,{\text{and}}\,\,\,x - y \simeq x} \right] \cr
& {\text{So}},\,{K_2} \simeq \frac{{y \times x}}{x} = y \cr
& \Rightarrow {K_2} = 4.8 \times {10^{ - 11}} = y = \left[ {CO_3^{2 - }} \right] \cr
& {\text{So the concentration of}}\,\left[ {{H^ + }} \right] \simeq \left[ {HCO_3^ - } \right] = \cr} $$
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
$$\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}}\& \,\left[ {CO_3^{2 - }} \right] = 4.8 \times {10^{ - 11}}$$