Question

Given the equilibrium \(\mathrm{H}_{2} \mathrm{~S}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HS}^{-}\) and \(\mathrm{HS}^{-}+\) \(\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{S}^{-2}\), find the final concentration of \(\mathrm{S}^{-2}\), if the final concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{H}_{2} \mathrm{~S}\) are \(0.3 \mathrm{M}\) and \(0.1 \mathrm{M}\), respectively. $$ \begin{aligned} &\mathrm{k}_{1}=6.3 \times 10^{-8} \text { for } \mathrm{H}_{2} \mathrm{~S}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HS}^{-} \text {and } \\ &\mathrm{k}_{2}=1 \times 10^{-14} \mathrm{for} \mathrm{HS}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{S}^{-2} \end{aligned} $$

Short answer

The final concentration of $\mathrm{S}^{-2}$ is approximately \(7 \times 10^{-23}\,\mathrm{M}\).

Step-by-step solution

Write the equilibrium expressions for both reactions.

For the first reaction, the equilibrium constant expression is: \[ K_1 = \frac{[H_3O^+][HS^-]}{[H_2S][H_2O]} \] Since the concentration of water is constant, we can combine it with the equilibrium constant: \[ K_1' = \frac{[H_3O^+][HS^-]}{[H_2S]} \] For the second reaction, the equilibrium constant expression is: \[ K_2 = \frac{[H_3O^+][S^{-2}]}{[HS^-][H_2O]} \] Again, combining the constant concentration of water with the equilibrium constant: \[ K_2' = \frac{[H_3O^+][S^{-2}]}{[HS^-]} \]

Substitute the given values into the expressions.

We are given the final concentrations of H3O^+ and H2S, as well as the values of K1 and K2. Let's substitute these values into the modified equilibrium expressions: \[ K_1' = 6.3 \times 10^{-8} = \frac{[0.3\,\mathrm{M}][HS^-(\mathrm{M})]}{[0.1\,\mathrm{M}]} \] \[ K_2' = 1 \times 10^{-14} = \frac{[0.3\,\mathrm{M}][S^{-2}(\mathrm{M})]}{[HS^-(\mathrm{M})]} \]

Solve for the concentration of HS^- and S^(-2).

First, we will solve for the concentration of HS^- from the equation for K1': \[ [HS^-]=\frac{K_1'[H_2S]}{[H_3O^+]} = \frac{6.3 \times 10^{-8} \times 0.1}{0.3} = 2.1 \times 10^{-8}\,\mathrm{M} \] Now, we can substitute this value into the equation for K2': \[ 1 \times 10^{-14} = \frac{[0.3\,\mathrm{M}][S^{-2}(\mathrm{M})]}{[2.1 \times 10^{-8}\,\mathrm{M}]} \] Now, solve for the concentration of S^(-2): \[ [S^{-2}] = \frac{1 \times 10^{-14} \times 2.1 \times 10^{-8}}{0.3} \approx 7 \times 10^{-23}\,\mathrm{M} \]

Report the final concentration of S^(-2).

The final concentration of S^(-2) is approximately 7 × 10^(-23) M.

Key concepts

Equilibrium Constant Expression

Understanding the equilibrium constant expression is crucial for predicting the extent of a chemical reaction under a set of given conditions. In essence, the equilibrium constant (\(K_{eq}\)) is a ration that relates the concentrations of the products to the concentrations of the reactants at equilibrium. This ratio is calculated using the molar concentrations (denoted by square brackets) of the reactants and products, raised to the power of their stoichiometric coefficients in the balanced chemical equation.
For example, if our balanced equation is \(aA + bB \rightleftarrows cC + dD\), the equilibrium constant expression would be written as: \[K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}\], where \(A\) and \(B\) are reactants, \(C\) and \(D\) are products, and \(a\), \(b\), \(c\), and \(d\) are their respective coefficients in the balanced equation.
In the provided exercise, the given equilibrium constants \(K_1'\) and \(K_2'\) are special cases, as the concentration of water (\(H_2O\)) is constant for dilute solutions and can be combined with the equilibrium constants \(K_1\) and \(K_2\) to simplify calculations. This approach streamlines the process, letting us directly relate the concentrations of ions without factoring in the water concentration separately.
When dealing with such calculations, remember that temperature can affect the value of equilibrium constants, as they are temperature-dependent. Not considering changes in temperature could lead to inaccuracies in the calculated concentrations at equilibrium.

Concentration of Ions in Equilibrium

The concentration of ions at equilibrium is a fundamental aspect of understanding how chemical systems behave over time. In a closed system where chemical reactions occur, concentrations of different ions change until equilibrium is reached, meaning the rate of the forward reaction equals the rate of the reverse reaction. From this point on, the concentrations remain constant as long as the external conditions such as temperature and pressure do not change.
In this exercise, we determine the concentrations of \(HS^-\) and \(S^{2-}\) ions at equilibrium using the known concentrations of \(H_3O^+\) and \(H_2S\) and the equilibrium constant expressions \(K_1'\) and \(K_2'\). Using the given final concentration values, we can algebraically solve for the unknowns, which in this case are the concentrations of \(HS^-\) and \(S^{2-}\). This step-by-step approach is essential for solving equilibrium problems.
Key to this process is the initial setup, ensuring that the equations are based on the balanced chemical equations and formulated correctly to include all participating species. Once the equilibrium constants are applied, algebraic manipulation will yield the desired concentrations. It's a precise yet straightforward method, provided that students take care with their units and keep balance between the respective sides of the equilibrium expression.

Acid-Base Equilibria

Acid-base equilibria involve the transfer of protons (\(H^+\)) between acid and base species in solution. In the given exercise, we're dealing with a bicarbonate system that includes weak acids and their conjugate bases, \(H_2S\), \(HS^-\), and \(S^{2-}\). Acids such as \(H_2S\) are molecules that can donate a proton to a water molecule (\(H_2O\)), forming hydronium (\(H_3O^+\)) and the conjugate base of the acid (\(HS^-\)). The conjugate bases can also act as acids in a second step, donating another proton to form \(S^{2-}\).
These sequential proton transfers are represented by the stepwise dissociation constants \(K_1\) and \(K_2\). Each constant represents an individual equilibrium point where a specific acid-base pair exchange protons. When working with acid-base equilibria, particularly involving polyprotic acids, students should consider each proton transfer separately and recognize the dependence of subsequent equilibria on the previous ones.
It's noteworthy that in many acid-base equilibrium problems, the concentration of water (\(H_2O\)) remains virtually unchanged and thus can often be lumped in with the equilibrium constant to simplify the calculation, as seen with \(K_1'\) and \(K_2'\) in our example. Recognizing and understanding each equilibrium allows students to piece together the overall picture of how the system behaves, ultimately determining the concentrations of all species involved at equilibrium.