Question

Ka1,Ka2,Ka3 values for \[{H_3}P{O_4}\] are \[{10^{ - 3}}\], \[{10^{ - 8}}\], \[{10^{ - 12}}\] respectively. If \[{K_W}\left( {{H_2}O} \right) = {10^{ - 14}}\], then which is correct?
A.Dissociation constant for \[{H_3}P{O_4}^{2 - }\] is \[{10^{ - 12}}\]
B.\[{K_b}\] of \[{H_3}P{O_4}^{2 - }\] \[{10^{ - 6}}\]
C.\[{K_b}\] of \[{H_3}P{O_4}^{2 - }\]\[{10^{ - 11}}\]
D.All are correct

Answer 1

Hint:Further the conjugate base is we can say the leftover after an acid has donated a proton during a chemical reaction.
From this we can say that
Conjugate acid=H+ + conjugate base
Further we can demonstrate it as:
\[{K_W} = {K_a}.{K_b}\]
And from here we can calculate conjugate bases.

Complete step-by-step answer:We will firstly discuss the dissociation constant so we come to know that the dissociation constant is an equilibrium constant that further measures the propensity of a larger object to separate into its component ions or the particles. Further we must know about the dissociation constant of water which is denoted by Kw. We must know that the value of the water dissociation constant varies with the temperature.
Here in the question we can see that the dissociation constant of phosphoric acid is given which can be shown by Ka1,Ka2,Ka3 values
\[\begin{gathered}
  {K_{a1}} = {10^{ - 3}} \\
  {K_{a2}} = {10^{ - 8}} \\
  {K_{a3}} = {10^{ - {{12}^{}}}} \\
\end{gathered} \]
As we can see that here it is given \[{K_a}\] which simply means the conjugate acid . Now we must know what conjugate acid is so it is basically formed when an acid donates a proton to a base. Further the conjugate base is we can say the leftover after an acid has donated a proton during a chemical reaction.
From this we can say that
Conjugate acid=H+ + conjugate base
So from here we can say that \[{K_a}\] is the conjugate acid and \[{K_b}\] is the conjugate base.
Further we can demonstrate it as:
\[{K_W} = {K_a}.{K_b}\]
So we can say that conjugate base for the first reaction is
\[\dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 3}}}} = {10^{ - 11}}\]
For the second reaction the conjugate base can be calculated as
\[\dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 8}}}} = {10^{ - 6}}\]
And lastly for the third reaction the conjugate base can be calculated as:
\[\dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 12}}}} = {10^{ - 2}}\]
 we must remember that conjugate acids and conjugate bases are interconnected.also, we conclude from the above solution that option second is correct.

Note:The conjugate base is able to absorb protons in a chemical reaction. The further use of the conjugate acid and base lies in the buffer system which further includes buffer solution.