Question

In aqueous solution, the ionisation constants for carbonic acid ($\text{ }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$) are $\text{ }{{\text{k}}_{\text{1}}}\text{ = 4}\text{.2 }\times \text{ 1}{{\text{0}}^{-7}}\text{ }$and $\text{ }{{\text{k}}_{2}}\text{ = 4}\text{.8 }\times \text{ 1}{{\text{0}}^{-11}}\text{ }$. Select the correct statement for a saturated $\text{ 0}\text{.034 M }$ solution of the carbonic acid.
A) The concentration of $\text{ CO}_{3}^{2-}\text{ }$ is $\text{ 0}\text{.034 M }$.
B) The concentration of $\text{ CO}_{3}^{2-}\text{ }$ is greater than that of$\text{ HCO}_{3}^{-}\text{ }$.
C) The concentration of $\text{ }{{\text{H}}^{\text{+}}}\text{ }$ and $\text{ HCO}_{3}^{-}\text{ }$ are approximately equal.
D) The concentration of $\text{ }{{\text{H}}^{\text{+}}}\text{ }$ is double that of $\text{ CO}_{3}^{2-}\text{ }$

Answer 1

The polyprotic or polybasic acid contains two or more hydrogen atoms. The carbonic acid $\text{ }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ is dibasic acid. It liberates two hydrogen ions in the two-step. These are as shown below,
$\text{ }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\rightleftharpoons \text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ + HCO}_{3}^{-}\text{ }$ First stag
$\text{ HCO}_{3}^{-}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\rightleftharpoons \text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ + CO}_{3}^{2-}\text{ }$ Second stage
The first dissociation constant is comparably more than the second dissociation constant, which directly affects the hydrogen ion concentration in the solution.

Complete step by step solution:
Polybasic acids contain two or more hydrogens which can get dissociated. They always dissociate in stages. Let's consider an example of carbonic acid $\text{ }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ .The carbonic acid is diprotic, It contains two hydrogen which is lost by the acid in the following steps,
First stage: in this step the carbonic acid loses its one proton and forms a monohydrogen carbonate.
$\text{ }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\rightleftharpoons \text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ + HCO}_{3}^{-}\text{ }$
The dissociation constant $\text{ }{{\text{k}}_{\text{1}}}\text{ }$ is written as,
$\text{ }{{\text{k}}_{\text{1}}}\text{ = }\dfrac{\left[ {{\text{H}}^{\text{+}}} \right]\left[ \text{HCO}_{\text{3}}^{-} \right]}{\left[ {{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}} \right]}\text{ }$
If C is the concretion of the hydrogen ion and $\text{ HCO}_{3}^{-}\text{ }$then hydrogen ion concentration from the dissociation constant is written as,

& \text{ }{{\text{k}}_{\text{1}}}\text{ = }\dfrac{\left[ {{\text{H}}^{\text{+}}} \right]\left[ \text{HCO}_{\text{3}}^{-} \right]}{\left[ {{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}} \right]}\text{ = }\dfrac{C\alpha \text{ }\times C\alpha }{(1-\alpha )C}\text{ } \\\ & \therefore \text{ }\left[ {{\text{H}}^{\text{+}}} \right]\text{ = }\sqrt{C\times {{\text{k}}_{\text{1}}}}\text{ }\because (1-\alpha )\approx 1 \\\ \end{aligned}$$ Therefore, hydrogen ion concretion is the square root of the concentration of carbonic acid .thus concentration of hydrogen is, $\text{ }\left[ {{\text{H}}^{\text{+}}} \right]\text{ = }\sqrt{C\times {{\text{k}}_{\text{1}}}}\text{ = }\sqrt{\left( 0.034 \right)\times \left( 4.2\times {{10}^{-7}} \right)}\text{ = 1}\text{.428}\times \text{1}{{\text{0}}^{\text{-8}}}\text{ M}$ We have given $\text{ }{{\text{k}}_{\text{1}}}\text{ = 4}\text{.2 }\times \text{ 1}{{\text{0}}^{-7}}\text{ }$ Therefore the total hydrogen concentration is $\text{ 1}\text{.428}\times \text{1}{{\text{0}}^{\text{-8}}}\text{ M }$ . Second stage: in this step the monohydrogen carbonate $\text{ HCO}_{3}^{-}\text{ }$ loses its one proton and forms a carbonate $\text{ CO}_{3}^{2-}\text{ }$ . $\text{ HCO}_{3}^{-}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\rightleftharpoons \text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ + CO}_{3}^{2-}\text{ }$ The dissociation constant $\text{ }{{\text{k}}_{2}}\text{ }$ is written as, $\text{ }{{\text{k}}_{2}}\text{ = }\dfrac{\left[ {{\text{H}}^{\text{+}}} \right]\left[ \text{CO}_{\text{3}}^{2-} \right]}{\left[ \text{HCO}_{3}^{-} \right]}\text{ }$ (2) We have given $\text{ }{{\text{k}}_{2}}\text{ = 4}\text{.8 }\times \text{ 1}{{\text{0}}^{-11}}\text{ }$ From given data, the dissociation constant for the first step$\text{ }{{\text{k}}_{\text{1}}}\text{ }$is greater than the$\text{ }{{\text{k}}_{2}}\text{ }$.$\text{ }{{\text{k}}_{\text{1}}}\text{ }\gg {{\text{k}}_{2}}\text{ , 4}\text{.2 }\times \text{ 1}{{\text{0}}^{-7}}\gg \text{ 4}\text{.8 }\times \text{ 1}{{\text{0}}^{-11}}\text{ }$ Thus, the carbonic acid easily loses its proton as compare to the $\text{ HCO}_{3}^{-}\text{ }$.therefore, the total hydrogen ion concentration in the solution is due to the carbonic acid. That is,$\text{ }{{\left[ {{\text{H}}^{\text{+}}} \right]}_{\text{Total}}}\text{ }\approx \text{ }{{\left[ {{\text{H}}^{\text{+}}} \right]}_{{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}(\text{first stage)}}}\text{ }$ Thus, from the dissociation constant equation (1), we can say that the total hydrogen ion concentration is equal to the concentration of monocarbonate hydrogen. That is, $\text{ }{{\left[ \text{HCO}_{3}^{-} \right]}_{{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}(\text{first stage)}}}\approx \text{ }{{\left[ {{\text{H}}^{\text{+}}} \right]}_{{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}(\text{first stage)}}}\text{ }\approx {{\left[ {{\text{H}}^{\text{+}}} \right]}_{\text{Total}}}$ We know that monohydrogen carbonate $\text{ HCO}_{3}^{-}\text{ }$does not readily dissociate into the carbonate ion. Thus, the concentration of carbonate ion in the solution is considered to be negligible when compared with the concentration of $\text{ HCO}_{3}^{-}\text{ }$or total hydrogen concentration. Thus, the concentration of $\text{ }{{\text{H}}^{\text{+}}}\text{ }$ and $\text{ HCO}_{3}^{-}\text{ }$are approximately equal. **Hence, (C) is the correct option.** **Note:** Note that, the second dissociation constant is found to be $\text{ }\dfrac{1}{10000}th\text{ }$ of the first dissociation constant. In most of the acid (polybasic) the $\text{ }{{\text{k}}_{\text{1}}}\text{ }$is greater than the $\text{ }{{\text{k}}_{2}}\text{ }$which shows that second dissociation takes place to a smaller extent than the first dissociation. The decrease in dissociation constant because the first ion comes from the neutral ion while in the second stage the proton is detached from the negatively charged molecule thus makes it difficult to dissociate.