If the ionization constant of ${H_2}C{O_3}$ as an acid in aqueous solution at room temperature is $X$. If the first and second ionisation constants of${H_2}C{O_3}$ are ${X_1}$ and ${X_2}$ respectively, then what will be the equation relating $X$ to ${X_1}$ and ${X_2}$ ?
(A) $X = \dfrac{{{X_1}}}{{{X_2}}}$
(B) $X = \dfrac{{{X_2}}}{{{X_1}}}$
(C) $X = {X_1}{X_2}$
(D) $X = \dfrac{{{X_1}{X_2}}}{2}$
Answer
606.6k+ views
Hint:As we know that the ionisation constant of an acid is the ratio of the concentration of the products and the concentrations of the reactants. It is represented as ${K_a}$ and informs about the strength of an acid. A strong acid is that acid which will completely ionise in the given aqueous solution.
Formula used: \[A + B \to C + D\]
Ionization constant $({K_a})$$ = \dfrac{{[C][D]}}{{[A][B]}}$$\dfrac{{{X_1}{X_2}}}{2}$
Complete step-by-step solution:An acid dissociation constant, ${K_a}$, is a measure of quantitative strength of an acid in solution. When the system achieves equilibrium it acts as the equilibrium constant for chemical reaction known as dissociation in the context of acid-base reaction.
The ionization of ${H_2}C{O_3}$ takes place in the following steps given below:
${H_2}C{O_3}$ + ${H_2}O$ $ \to $ $HC{O_3}^ - $+ ${H_3}{O^ + }$,
We can assume the ionisation constant for this reaction as: ${K_{a1}} = {X_1}$
$HC{O_3}^ - $ + ${H_2}O$ $ \to $ $C{O_3}^{2 - }$ + ${H_3}{O^ + }$
Let us assume the ionisation constant for this reaction as: ${K_{a2}} = {X_2}$
For step of ${H_2}C{O_3}$ ionisation, the ionisation constant is given as $X = \dfrac{{[C{O_3}^{2 - }][{H_3}{O^ + }]}}{{[{H_2}C{O_3}][{H_2}O]}}$
For${X_1}$, the ionisation constant would be: ${X_1} = \dfrac{{[HC{O_3}^ - ][{H_3}{O^ + }]}}{{[{H_2}C{O_3}][{H_2}O]}}$ -(1)
For${X_2}$, the ionisation constant would be: ${X_2} = \dfrac{{[C{O_3}^{2 - }][{H_3}{O^ + }]}}{{[HC{O_3}^ - ][{H_2}O]}}$ -(2)
After computing equations 1 and 2, we will get:
$X = \dfrac{{[HCO_3^ - ][{H_3}{O^ + }]}}{{[{H_2}C{O_3}][{H_2}O]}} \times \dfrac{{[CO_3^{2 - }][{H_3}{O^ + }]}}{{[HCO_3^ - ][{H_2}O]}}$
$X = \dfrac{{[CO_3^{2 - }][{H_3}{O^ + }]}}{{[{H_2}C{O_3}][{H_2}O]}}$
$\dfrac{{[CO_3^{2 - }][{H^ + }]}}{{[{H_2}C{O_3}]}} = \dfrac{{[CO_3^{2 - }][{H^ + }]}}{{[{H_2}C{O_3}]}}$
Both sides are equal so the correct option is $X = {X_1}{X_2}$.
Additional information:Like we can calculate the ionisation constant of the acid, similarly we can identify the ionisation constant of a base which is again the ratio of product to reactant concentrations and is represented by the symbol ${K_b}$ which informs us about the strength of a base in aqueous solution.
Therefore, the correct answer is (C).
Note:Always remember that the strong acids or strong bases are the one which completely ionises in an aqueous solution and a weak acid and a weak base are the one which will partially ionise in the aqueous solution and therefore the ionisation constant will be given as the ratio of product concentrations and reactant concentrations.
Formula used: \[A + B \to C + D\]
Ionization constant $({K_a})$$ = \dfrac{{[C][D]}}{{[A][B]}}$$\dfrac{{{X_1}{X_2}}}{2}$
Complete step-by-step solution:An acid dissociation constant, ${K_a}$, is a measure of quantitative strength of an acid in solution. When the system achieves equilibrium it acts as the equilibrium constant for chemical reaction known as dissociation in the context of acid-base reaction.
The ionization of ${H_2}C{O_3}$ takes place in the following steps given below:
${H_2}C{O_3}$ + ${H_2}O$ $ \to $ $HC{O_3}^ - $+ ${H_3}{O^ + }$,
We can assume the ionisation constant for this reaction as: ${K_{a1}} = {X_1}$
$HC{O_3}^ - $ + ${H_2}O$ $ \to $ $C{O_3}^{2 - }$ + ${H_3}{O^ + }$
Let us assume the ionisation constant for this reaction as: ${K_{a2}} = {X_2}$
For step of ${H_2}C{O_3}$ ionisation, the ionisation constant is given as $X = \dfrac{{[C{O_3}^{2 - }][{H_3}{O^ + }]}}{{[{H_2}C{O_3}][{H_2}O]}}$
For${X_1}$, the ionisation constant would be: ${X_1} = \dfrac{{[HC{O_3}^ - ][{H_3}{O^ + }]}}{{[{H_2}C{O_3}][{H_2}O]}}$ -(1)
For${X_2}$, the ionisation constant would be: ${X_2} = \dfrac{{[C{O_3}^{2 - }][{H_3}{O^ + }]}}{{[HC{O_3}^ - ][{H_2}O]}}$ -(2)
After computing equations 1 and 2, we will get:
$X = \dfrac{{[HCO_3^ - ][{H_3}{O^ + }]}}{{[{H_2}C{O_3}][{H_2}O]}} \times \dfrac{{[CO_3^{2 - }][{H_3}{O^ + }]}}{{[HCO_3^ - ][{H_2}O]}}$
$X = \dfrac{{[CO_3^{2 - }][{H_3}{O^ + }]}}{{[{H_2}C{O_3}][{H_2}O]}}$
$\dfrac{{[CO_3^{2 - }][{H^ + }]}}{{[{H_2}C{O_3}]}} = \dfrac{{[CO_3^{2 - }][{H^ + }]}}{{[{H_2}C{O_3}]}}$
Both sides are equal so the correct option is $X = {X_1}{X_2}$.
Additional information:Like we can calculate the ionisation constant of the acid, similarly we can identify the ionisation constant of a base which is again the ratio of product to reactant concentrations and is represented by the symbol ${K_b}$ which informs us about the strength of a base in aqueous solution.
Therefore, the correct answer is (C).
Note:Always remember that the strong acids or strong bases are the one which completely ionises in an aqueous solution and a weak acid and a weak base are the one which will partially ionise in the aqueous solution and therefore the ionisation constant will be given as the ratio of product concentrations and reactant concentrations.
Recently Updated Pages
The given figure shows two endocrine glands marked class 11 biology NEET_UG

Match columnI with columnII and select the correct class 11 biology NEET

Match column I with column II and select the correct class 11 biology NEET_UG

Which floral family has left 9 right + 1 arrangement class 11 biology NEET_UG

Which is not a variety of sheep A Lohi B Beetal C Nellore class 11 biology NEET_UG

Match column I with column II and select the correct class 11 biology NEET_UG

Trending doubts
Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

10 examples of law on inertia in our daily life

