Question

Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a \(0.10-M \mathrm{H}_{2} \mathrm{S}\) solution. Assume \(K_{\mathrm{a}_{\mathrm{t}}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}\).

Short answer

The pH of the 0.10 M H₂S solution is 4, and the concentration of S²⁻ ions is \(1.0\times10^{-19} M\).

Step-by-step solution

Write down the major equilibrium reaction and its dissociation constant.

We begin by analyzing the major reactant, \( H_2S\). It can lose two protons sequentially: \[H_{2}S \rightleftharpoons H^+ + HS^-\] and \[HS^- \rightleftharpoons H^+ + S^{2-}\] However, given the significant difference in magnitudes of the dissociation constants (\(K_{a1} = 1.0\times10^{-7}\) and \(K_{a2} = 1.0\times10^{-19}\)), we will primarily focus on the first dissociation since its contribution to pH will be dominant while estimating \( [S^{2-}]\). So, for the first reaction, \[K_{a1} = \frac{[H^+][HS^-]}{[H_{2}S]}\]

Set up an equilibrium expression for [H⁺] ions.

Initial conc. of H₂S is 0.1 M and of H⁺, HS⁻ is zero in the start. As the reaction reaches equilibrium, some of the H₂S will dissociate to form H⁺ and HS⁻ ions. Since the reaction ratio is 1:1:1, we represent concentrations as follows: [H⁺] = x [HS⁻] = x [H₂S] = 0.1 - x Then, \[ K_{a1} = \frac{x^2}{0.1 - x} \Longrightarrow 1.0\times10^{-7} = \frac{x^2}{0.1 - x}\]

Solve for [H⁺] and find pH.

To solve for x, we can make an approximation since \(K_{a1}\) value is low enough. We can assume that x << 0.1, which means we can approximate the denominator with simply 0.1: \[1.0\times10^{-7} = \frac{x^2}{0.1}\] \[x^2 = 1.0\times10^{-8}\] \[x = 1.0\times10^{-4}\] Therefore, [H⁺] = \(1.0\times10^{-4} M\), and to find the pH, we use the formula: \[pH = -\log([H^+]) = -\log(1.0\times10^{-4}) = 4\] Thus, the pH of the H₂S solution is 4.

Find the concentration of [S²⁻] by considering the second dissociation.

Since the second dissociation is a minor one with a very low dissociation constant (K₂), we can assume that the concentration of HS⁻ ions is approximately equal to the [H⁺] calculated before: [HS⁻] = \(1.0\times10^{-4} M\) Now, for the second equilibrium, we have: \[K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]}\] Substitute the given values into the equation: \[1.0\times10^{-19} = \frac{(1.0\times10^{-4})([S^{2-}])}{(1.0\times10^{-4})}\] Solve for [S²⁻]: \[[S^{2-}] = 1.0\times10^{-19}\] So, the concentration of S²⁻ ions in the 0.1 M H₂S solution is \(1.0\times10^{-19} M\).

Key concepts

Dissociation Constant

To understand the behavior of an acid like \( H_2S \) in water, we look at how it dissociates, or breaks apart, into ions. The dissociation constant \( K_a \) is a measure of the strength of an acid based on its tendency to donate protons to water, forming hydronium ions \([H^+]\) and the conjugate base. The first dissociation of \( H_2S \) can be expressed as:

• \[ H_2S \rightleftharpoons H^+ + HS^- \]

This reaction has a dissociation constant \( K_{a1} = 1.0 \times 10^{-7} \). A higher value of \( K_a \) usually means a stronger acid, as it can donate protons more readily.
The second dissociation is less significant:

• \[ HS^- \rightleftharpoons H^+ + S^{2-} \]

Here, \( K_{a2} = 1.0 \times 10^{-19} \), reflecting a much weaker ability to donate another proton. Therefore, the first dissociation contributes more to the solution's acidity and future calculations.

Equilibrium Expression

In chemical reactions such as the dissociation of acids, the equilibrium expression helps us calculate the concentration of various ions in the solution. For \( H_2S \), which dissociates into \( H^+ \) and \( HS^- \), the equilibrium expression for the first dissociation is:

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

This equation tells us how the ionic species are related at equilibrium. By finding \( x \), the concentration of ions at equilibrium, we can estimate the solution's \( pH \). Since \( K_{a1} \) is quite small, we assume that the change in the \( H_2S \) concentration, \( x \), is minimal, simplifying calculations. This assumption helps balance the accuracy needed and simplifies computational complexity.

H₂S Solution

When working with an \( H_2S \) solution in water, we are primarily concerned with how it behaves as a weak diprotic acid, meaning it can release two protons. The \( 0.1 \, M \) solution offers initial insights into its behavior, where the two-step dissociation will occur:

• The first step, resulting in \( HS^- \), primarily determines the solution's \( pH \).
• The second dissociation, forming \( S^{2-} \), is negligible due to the very small \( K_{a2} \).

The \( pH \) value provides a direct clue into the extent of \( H_2S \) ionization in the solution. With calculated values like the pH of 4, you get an idea of the acid's weak nature on the acidity scale. Additionally, this forms a foundation for understanding the sulfur and hydrogen balance in terms of ionic concentration.

Proton Dissociation

Proton dissociation in acids is a vital concept in understanding the \( pH \) level of solutions. For \( H_2S \), dissociation involves the sequential release of two protons:

• The first dissociation yields a hydrogen ion \( H^+ \) and a bisulfide ion \( HS^- \).
• The second dissociation, which is much weaker due to the lower \( K_{a2} \), forms another \( H^+ \) and a sulfide ion \( S^{2-} \).

Understanding these dissociations helps predict the acidity and \( S^{2-} \) presence in the solution. The first dissociation largely determines the solution's \( pH \) since it releases protons that increase the hydrogen ion concentration. The second dissociation's impact is significantly diminished but important for completeness in understanding the chemical equilibrium of such systems.