Q.Ka1, Ka2 and Ka3 are the respective ionisation constants for the following reactions.
H2S ⇌ H^+ + HS^-
HS^- ⇌ H^+ + S^2-
H2S ⇌ 2H^+ + S^2-
The correct relationship between Ka1, Ka2 and Ka3 is

The key idea here is the combination of equilibrium reactions and their corresponding equilibrium constants. When individual equilibrium reactions are summed to yield an overall reaction, the equilibrium constant for the overall reaction is the product of the equilibrium constants of the individual reactions.

1. The first ionization of $H_2S$ is given by:

   $H_2S \rightleftharpoons H^+ + HS^-$

   $K_{a1} = \frac{[H^+][HS^-]}{[H_2S]}$

2. The second ionization of $H_2S$ (from $HS^-$) is given by:

   $HS^- \rightleftharpoons H^+ + S^{2-}$

   $K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]}$

3. The overall ionization of $H_2S$ to $S^{2-}$ is given by:

   $H_2S \rightleftharpoons 2H^+ + S^{2-}$

   $K_{a3} = \frac{[H^+]^2[S^{2-}]}{[H_2S]}$ …

The problem asks us to find the relationship between the ionization constants ($K_a$) for three related reactions involving hydrogen sulfide ($H_2S$). Understanding how equilibrium constants combine when reactions are added or manipulated is key here.

An ionization constant, $K_a$, quantifies the extent to which an acid dissociates in solution. For a general acid $HA$, the reaction is $HA \rightleftharpoons H^+ + A^-$, and its ionization constant is $K_a = \frac{[H^+][A^-]}{[HA]}$.

The fundamental principle we will use is that if a net chemical reaction can be expressed as the sum of two or more individual reactions, then the equilibrium constant for the net reaction is the product of the equilibrium constants of the individual reactions. This is analogous to Hess's Law for enthalpy changes, but applied to equilibrium constants.

Let's apply this concept step-by-step.

1. Identify the given reactions and their ionization constants:

   We are given three reactions and their respective ionization constants:

   • Reaction 1: The first dissociation of $H_2S$:

     $H_2S \rightleftharpoons H^+ + HS^-$

     The ionization constant for this reaction is $K_{a1} = \frac{[H^+][HS^-]}{[H_2S]}$.

   • Reaction 2: The second dissociation of $H_2S$ (dissociation of the bisulfide ion):

     $HS^- \rightleftharpoons H^+ + S^{2-}$

     The ionization constant for this reaction is $K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]}$.

   • Reaction 3: The overall dissociation of $H_2S$ into $2H^+$ and $S^{2-}$:

     $H_2S \rightleftharpoons 2H^+ + S^{2-}$

     The ionization constant for this reaction is $K_{a3} = \frac{[H^+]^2[S^{2-}]}{[H_2S]}$.

2. Determine how Reaction 3 can be formed from Reaction 1 and Reaction 2: Observe the reactants and products in Reaction 3. We start with $H_2S$ and end with $2H^+$ and $S^{2-}$. Let's try adding Reaction 1 and Reaction 2: $(H_2S \rightleftharpoons H^+ + HS^-)$ $+$ $(HS^- \rightleftharpoons H^+ + S^{2-})$

   $H_2S + HS^- \rightleftharpoons H^+ + HS^- + H^+ + S^{2-}$

3. Simplify the combined reaction:

   Notice that the $HS^-$ ion appears on both sides of the combined reaction. We can cancel it out, just like in algebraic equations:

   $H_2S \rightleftharpoons H^+ + H^+ + S^{2-}$

   $H_2S \rightleftharpoons 2H^+ + S^{2-}$

   This simplified reaction is exactly Reaction 3. …