Question

Which of the following is the correct statement(s) regarding $K_{a_1}$ for the first dissociation and $K_{a_2}$ for the second dissociation of a polyprotic acid $H_{2}C{O}_3$ ?

• A. $K_{a_1}$ has less value than $K_{a_2}$
• B. $H_{ion}^{+}$ from $H_{2}C{O}_3$ is more easier to obtain compared to $H_{ion}^{+}$ from $HC{O}_3^{-}$
• C. $K_{a_1}$ has more value than $K_{a_2}$
• D. $\left[H^{+}\right]$ from $H_{2}C{O}_3$ is is more difficult to obtain as compared to $H_{ion}^{+}$ from $HC{O}_3^{-}$

Answer 1

Answer: (B), (C)

$K_{a_1}$ has more value than $K_{a_2}$

$K_{a_1}$ is the first dissociation constant. It describes how the first $H^{+}$ ion dissociates from a neutral polyprotic acid.
$K_{a_1}$ is the second dissociation constant that describes how the second $H^{+}$ dissociates from a negatively charged polyprotic acid.
A neutral molecule has higher tendency to lose a proton than an anionic species because negatively charged molecule resists to give a proton. Thus $K_{a_1}$ value is more than $K_{a_2}$ value for polyprotic acid.

Theory :
Polyprotic acids:
Acids which supply/provide more than one $H^{+}$ per molecule.
Consider $H_{2}A\rightleftharpoons H^{+}+H{A}^{-}$ has another equilibrium of $H{A}^{-}\rightleftharpoons H^{+}+A^{2-}$.
Example: $H_{2}C{O}_3,H_{3}P{O}_4$
For $H_{2}C{O}_3$ in water.
$H_{2}C{O}_3$ ionizes in two steps, each step is in equilibrium.
Step 1 :
$H_{2}C{O}_3\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+HC{O}_3^{-}\left(aq\right)$
$att=0C00$
$att=t_{eq}C(1-{\alpha }_1)(C{\alpha }_1{\alpha }_2+C{\alpha }_1)C{\alpha }_1(1-{\alpha }_2)$
Step 2 :
$HC{O}_3^{-}\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$
$att=t_{eq}C{\alpha }_1(1-{\alpha }_2)(C{\alpha }_1{\alpha }_2+C{\alpha }_1)C{\alpha }_1{\alpha }_2$

$K_{a1}=\frac{\left[H^{+}\right]\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}=\frac{[C{\alpha }_1+C{\alpha }_1{\alpha }_2]\left[C{\alpha }_1\right(1-{\alpha }_2\left)\right]}{C(1-{\alpha }_1)}$
Dissociation constant of second step - $K_{a_2}$
$K_{a2}=\frac{\left[H^{+}\right][C{O}_3^2-]}{\left[HC{O}_3^{-}\right]}=\frac{[C{\alpha }_1+C{\alpha }_1{\alpha }_2]\left[C{\alpha }_1{\alpha }_2\right)]}{C{\alpha }_1(1-{\alpha }_2)}$

Approximations :
In a solution $H_{2}A\rightleftharpoons H^{+}+H{A}^{-}$ and $H{A}^{-}\rightleftharpoons H^{+}+A^{2-}$ are equilibria which occur simultaneously and sequentially with a common ion.

we can also express dissociation in simpler way as:
Step 1 :
$H_{2}C{O}_3\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+HC{O}_3^{-}\left(aq\right)$
$att=0C00$
$att=t_{eq}C(1-x)x+yx-y$
Step 2 :
$HC{O}_3^{-}\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$
$att=t_{eq}(x-y)x+yy$

In many cases with $K_{a1}>>K_{a2}$,
$x>>y$
$\therefore x-y\approx xandx+y\approx x$
Note:- If $C>>x$ so $C-x\approx C$ then $C(1-{\alpha }_1)\approx C$
$K_{a1}\approx \frac{\left(x\right)\left(x\right)}{(C-x)}$ when $C-x=\left[H_{2}C{O}_3\right]$ and $x=\left[H^{+}\right]and\left[HC{O}_3^{-}\right]$
$K_{a2}\approx \frac{\left(y\right)\left(x\right)}{\left(x\right)}\approx y$
Therefore the total $[H^{+}{]}_{total}=[H^{+}{]}_{firststep}+[H^{+}{]}_{secondstep}$
Since $K_{a1}>>K_{a2},[H^{+}{]}_{firststep}>>[H{]}_{secondstep}^{+}$
$\therefore [H^{+}{]}_{total}\approx [H^{+}{]}_{firststep}$
Conclusions :
For diprotic acids, $K_{a1}$ is much larger than $K_{a2}$ (Applicable for any generalised polyprotic acid).
This is because $H^{+}$ is lost more easily from a neutral $H_{2}C{O}_3$ than from $HC{O}_3^{-}$
The stronger attraction of opposite charges inhibits the second ionization.
Approximation for diprotic acids :
In general, for $\left[H^{+}\right]$ calculation, we can treat weak polyprotic acids as monoprotic acids and ignore the second ionization step (when dissociation is very less in further steps).