Q.The first ionization constant of H 2S is 9.1 × 10⁻⁸. Calculate the concentration of HS – ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H 2S is 1.2 × 10⁻¹³, calculate the concentration of S2– under both conditions.

Concept: Weak Acid Ionization – H₂S is a diprotic weak acid; the first ionization dominates in pure water, but a common ion (H⁺ from HCl) suppresses it.

Step 1 – First ionization in pure 0.1 M H₂S

Let $x = [\text{HS}^-] = [\text{H}^+]$. Then $K_{a1} = \frac{x^2}{0.1-x} \approx \frac{x^2}{0.1}$ (since $x \ll 0.1$).

Step 2 – Effect of 0.1 M HCl

Now $[\text{H}^+] \approx 0.1$ M (from HCl). Using $K_{a1}$:

The common ion drastically reduces $[\text{HS}^-]$.

Step 3 – Second ionization for $[\text{S}^{2-}]$

In pure 0.1 M H₂S: $[\text{H}^+] \approx 9.54\times10^{-5}$ M, $[\text{HS}^-] \approx 9.54\times10^{-5}$ M. …

Concept and Intuition

$\ce{H2S}$ is a diprotic weak acid. It dissociates in two steps:

1. $\ce{H2S <=> H+ + HS-}$, $K_{a1} = 9.1 \times 10^{-8}$
2. $\ce{HS- <=> H+ + S^{2-}}$, $K_{a2} = 1.2 \times 10^{-13}$

Because $K_{a1} \gg K_{a2}$, the first dissociation is the main source of $\ce{H+}$ and $\ce{HS-}$ in pure water. The second dissociation is so weak that it hardly affects the $\ce{HS-}$ concentration. When we add 0.1 M HCl, the common ion $\ce{H+}$ suppresses the first dissociation drastically — this is Le Chatelier's principle in action.

The key insight: for a diprotic acid where $K_{a1} \gg K_{a2}$, the concentration of the intermediate ion $\ce{HS-}$ is essentially determined by the first equilibrium alone, and the concentration of the final ion $\ce{S^{2-}}$ is given by $K_{a2}$ times the ratio $[\ce{HS-}]/[\ce{H+}]$.

Step-by-step Solution

1. Pure 0.1 M $\ce{H2S}$ — find $[\ce{HS-}]$

Let the initial concentration of $\ce{H2S}$ be $c = 0.1~\text{M}$. Let $x$ be the concentration of $\ce{H+}$ (and $\ce{HS-}$) from the first dissociation.

The equilibrium for the first step:

The equilibrium expression:

Since $K_{a1}$ is very small, $x \ll 0.1$, so we approximate $0.1 - x \approx 0.1$:

Now $\sqrt{91} \approx 9.539$, so:

Thus, in pure 0.1 M $\ce{H2S}$:

2. Effect of adding 0.1 M HCl

Now the solution is 0.1 M in $\ce{H2S}$ and 0.1 M in HCl. HCl is a strong acid, so it contributes $[\ce{H+}] = 0.1~\text{M}$ initially. The first dissociation of $\ce{H2S}$ will produce an additional tiny amount of $\ce{H+}$, but that is negligible compared to 0.1 M.

Let $y$ be the concentration of $\ce{HS-}$ formed from $\ce{H2S}$ in the presence of HCl. The equilibrium:

The equilibrium expression:

Since $y$ will be tiny compared to 0.1, we approximate $0.1 + y \approx 0.1$ and $0.1 - y \approx 0.1$:

So:

3. Concentration of $\ce{S^{2-}}$ in pure 0.1 M $\ce{H2S}$

The second dissociation is:

From step 1, in pure water: $[\ce{HS-}] \approx 9.5 \times 10^{-5}~\text{M}$ and $[\ce{H+}] \approx 9.5 \times 10^{-5}~\text{M}$ (from the first dissociation). Let $z$ be the concentration of $\ce{S^{2-}}$ formed.

The equilibrium for the second step:

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