Question

‌ ‌If‌ ‌‌\[{K_{{a_1}}}\]‌,‌ ‌‌\[{K_{{a_2}}}\]‌‌ ‌and‌ ‌‌\[{K_{{a_3}}}\]‌‌ ‌be‌ ‌the‌ ‌first‌ ‌second‌ ‌and‌ ‌third‌ ‌
ionisation‌ ‌constant‌ ‌of‌ ‌‌${H_3}P{O_4}$‌‌ ‌and‌ ‌‌\[{K_{{a_1}}}\]$‌ ‌>‌ ‌>‌ ‌$\[{K_{{a_2}}}\]$‌ ‌>‌ ‌>‌ ‌
$\[{K_{{a_3}}}\]‌,‌ ‌then‌ ‌which‌ ‌of‌ ‌the‌ ‌following‌ ‌is/are‌ ‌correct?‌ ‌
A. $[{H^‌ ‌+‌ ‌}]‌ ‌=‌ \sqrt {{K_{{a_1}}}[{H_3}P{O_4}]}$
B. $[{H^‌ ‌+‌ ‌}]‌ ‌\approx‌ ‌[HP{O_4}^{2-}]$‌ ‌
C. ${K_{{a_2}}}‌ ‌\approx‌ ‌[HP{O_4}^{2‌ ‌-‌ ‌}]$‌ ‌
D. $[{H_3}P{O_4}]‌ ‌=‌ ‌[P{O_4}^{3‌ ‌-‌ ‌}]$‌ ‌

Answer 1

Hint: Ionisation of an acid or a base simply means that the proton of the acid or the hydronium ion of the base has been dissociated from their respective molecules. The ionisation constant signifies the extent of that dissociation of the acid or base in the solution. Ionisation constant of an acid could also be termed as the equilibrium constant of that acid and is calculated in the same manner.

Complete step-by-step answer:
In order to understand the question we must first understand the definition of each of the terms of the symbol which are being used in the question.
The ionisation is the process in which the given acid or base dissociates into its constituent ions, meaning in case of an acid the dissociation of hydrogen ion denoted by $[{H^ + }]$ is termed as the ionisation of that acid similarly the dissociation of $[O{H^ - }]$ ions in case of bases.
Ionisation constant is a measure by which one could know the extent of ionisation of the constituent particles of that acid in the solution. It is denoted by ${K_a}$ where the $K$ signifies that it is an equilibrium constant, and since we are considering an acid so the letter $a$ in the subscript.
Now, the question is telling us about the ionisation of phosphoric acid, so we will simply write the equations of first, second and third ionisation of the phosphoric acid.
${H_3}P{O_4} + {H_2}O \to {H_2}P{O_4}^ - + {H_3}{O^ + }$
${H_2}P{O_4}^ - + {H_2}O \to HP{O_4}^{2 - } + {H_3}{O^ + }$
$HP{O_4}^ - + {H_2}O \to P{O_4}^{3 - } + {H_3}{O^ + }$
Above are the equations whose ionisation constants are denoted as \[{K_{{a_1}}}\], \[{K_{{a_2}}}\] and \[{K_{{a_3}}}\] respectively.
Now if we consider the options given in the question, the first option provides an equation, we need to justify if it is true or not, so we will start with writing the proper expression of first ionisation constant, which is,
\[{K_{{a_1}}} = \dfrac{{[{H^ + }][{H_2}P{O^ - }_4]}}{{{H_3}P{O_4}}}\]
Now we know that $[{H^ + }] \simeq [{H_2}P{O^ - }_4]$, so we can write ${[{H^ + }]^2}$ In the numerator, so we get,
\[{K_{{a_1}}} = \dfrac{{{{[{H^ + }]}^2}}}{{{H_3}P{O_4}}}\]
So if we rearrange the whole equation, we will get $[{H^ + }] = \sqrt {{K_{{a_1}}}[{H_3}P{O_4}]} $, which is the equation given in option A. Hence the option is correct.
Now if we consider the next option it is asked that if $[{H^ + }] \simeq [HP{O^{2 - }}_4]$, but in this equation if you notice the charges are not balancing on both the sides, as in the overall charge remains $ - 1$ on the right side of the equation, while, if we consider the equation $[{H^ + }] \simeq [{H_2}P{O^ - }_4]$, we can see that all the charges gets cancelled out in both the sides. Hence the option B is incorrect.
Now, for the next option, we need to write the equation of second ionisation constant of phosphoric acid, which is
\[{K_{{a_2}}} = \dfrac{{[{H^ + }][HP{O^{2 - }}_4]}}{{[{H_2}P{O^ - }_4]}}\]
Now using the same logic as used in the previous option, we will cancel out the $[{H^ + }]$ and $[{H_2}P{O^ - }_4]$ as $[{H^ + }] \simeq [{H_2}P{O^ - }_4]$. So this leaves us with, \[{K_{{a_2}}} = [HP{O^{2 - }}_4]\], so the third option is correct.
Now, if we consider the last option we know that the \[{K_{{a_2}}}\] $ > > $ \[{K_{{a_3}}}\], hence the corresponding anions would show, $[{H_3}P{O_4}] > > [P{O_4}^{3 - }]$. And the last option says otherwise which is $[{H_3}P{O_4}] = [P{O_4}^{3 - }]$, hence it is not correct.

So the options A and C turned out to be the correct answers.

Note: The value of the second ionisation constant of the phosphoric acid is more than the third ionisation constant of the same acid.
The concentration of hydronium ion is almost equal to the concentration of the diprotic phosphoric acid anion in the solution where the ionisation takes place.