Question

Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a \(0.10-M \mathrm{H}_{2} \mathrm{~S}\) solution. Assume \(K_{\mathrm{a}_{1}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}\).

Short answer

In the 0.10 M \(\mathrm{H}_{2}\mathrm{S}\) solution, the pH is 4, and the concentration of \(\mathrm{S}^{2-}\) ions is approximately \(1.0 \times 10^{-18}\,\mathrm{M}\).

Step-by-step solution

Write the equilibrium reactions

Write the equilibrium expressions for the dissociation of \(\mathrm{H}_{2}\mathrm{S}\) and determine the first and second ionization constants. The dissociation reaction for the first ionization: \[\mathrm{H}_{2} \mathrm{S} \leftrightarrows \mathrm{H}^+ + \mathrm{HS}^{-}\] The equilibrium expression for the first ionization: \[K_{\mathrm{a}_{1}} = \frac{[\mathrm{H}^+][\mathrm{HS}^-]}{[\mathrm{H}_{2}\mathrm{S}]}\] The dissociation reaction for the second ionization: \[\mathrm{HS}^{-} \leftrightarrows \mathrm{H}^+ + \mathrm{S}^{2-}\] The equilibrium expression for the second ionization: \[K_{\mathrm{a}_{2}} = \frac{[\mathrm{H}^+][\mathrm{S}^{2-}]}{[\mathrm{HS}^-]}\]

Set up the equation

Make the assumption that only the first ionization contributes significantly to the hydrogen concentration because \(K_{\mathrm{a}_{1}}\) is much greater than \(K_{\mathrm{a}_{2}}\). Therefore: \[ K_{\mathrm{a}_{1}} = \frac{x^2}{0.10-x} \] where \(x\) is the approximate hydrogen ion concentration.

Solve for the hydrogen ion concentration

Solve the equation for \(x\) to find the hydrogen ion concentration. It is important to notice that \(K_{\mathrm{a}_{1}}\) is quite small, meaning that the dissociation is very minimal. So, we can assume that \(x << 0.10\), and we can simplify the equation: \[ K_{\mathrm{a}_{1}} \approx \frac{x^2}{0.10} \] Solve for \(x\): \[ x = \sqrt{K_{\mathrm{a}_{1}} \times 0.10} = \sqrt{1.0 \times 10^{-7} \times 0.10} = 1.0 \times 10^{-4} \] The hydrogen ion concentration is approximately \(1.0 \times 10^{-4}\,\mathrm{M}\).

Calculate the pH

Use the hydrogen ion concentration to calculate the pH: \[ \mathrm{pH} = -\log([\mathrm{H}^+]) = -\log(1.0 \times 10^{-4})=4 \] The pH of the solution is 4.

Solve for the sulfide ion concentration

Now, find the concentration of \(\mathrm{S}^{2-}\) ions using the second equilibrium expression and the determined hydrogen ion concentration: \[ K_{\mathrm{a}_{2}} = \frac{[\mathrm{H}^+][\mathrm{S}^{2-}]}{[\mathrm{HS}^-]} \] However, \([\mathrm{HS}^{-}]\) can be determined using the first equilibrium expression: \[ [\mathrm{HS}^{-}] = \frac{K_{\mathrm{a}_{1}}[\mathrm{H}_{2}\mathrm{S}]}{[\mathrm{H}^+]} = \frac{1.0 \times 10^{-7} \times 0.10}{1.0 \times 10^{-4}} = 1.0 \times 10^{-3} \] Now plug in the values for \(K_{\mathrm{a}_{2}}\), \([\mathrm{H}^+]\), and \([\mathrm{HS}^{-}]\): \[ 1.0 \times 10^{-19} = \frac{(1.0 \times 10^{-4})([\mathrm{S}^{2-}])}{1.0 \times 10^{-3}} \] Solve for \([\mathrm{S}^{2-}]\): \[ [\mathrm{S}^{2-}] = \frac{1.0 \times 10^{-19} \times 1.0 \times 10^{-3}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-18} \] The concentration of \(\mathrm{S}^{2-}\) ions in the solution is approximately \(1.0 \times 10^{-18}\,\mathrm{M}\). In conclusion, the pH of the 0.10 M \(\mathrm{H}_{2}\mathrm{S}\) solution is 4, and the concentration of \(\mathrm{S}^{2-}\) ions is \(1.0 \times 10^{-18}\,\mathrm{M}\).

Key concepts

Acid Dissociation Constant

Understanding the acid dissociation constant, often represented as Ka, is crucial when studying the behavior of acids in solution. The acid dissociation constant is a numerical value that conveys the extent to which an acid can donate a proton (H+) to the surrounding solution, essentially measuring the strength of the acid. The lower the value of Ka, the weaker the acid, as it implies less dissociation into ions.

For a general acid dissociation reaction, such as:

\[\mathrm{HA} \leftrightarrows \mathrm{H}^+ + \mathrm{A}^-\]

the acid dissociation constant is given by:

\[K_{\mathrm{a}} = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}\]

In the given exercise, we are dealing with a weak polyprotic acid, hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)), which has two dissociation steps, each with its own Ka, labeled \(K_{\mathrm{a}_1}\) and \(K_{\mathrm{a}_2}\), corresponding to the disassociation of the first and second hydrogen ions respectively. These constants are essential for predicting the concentration of hydrogen ions and, subsequently, calculating the pH of the solution.

Chemical Equilibrium

Chemical equilibrium pertains to the condition in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This results in the concentrations of reactants and products remaining constant over time, not necessarily equal. Understanding equilibrium is paramount when dealing with reversible reactions, such as acid dissociation, which can be represented as:

\[\mathrm{HA} \leftrightarrows \mathrm{H}^+ + \mathrm{A}^-\]

At equilibrium, the rate at which the reactants form products is equal to the rate at which products revert to reactants. Equilibrium constants, like Ka for acids, allow us to determine the ratio of these concentrations at equilibrium. In the exercise solution, we see practical application of this concept as the equilibrium expressions for the dissociation of \(\mathrm{H}_2\mathrm{S}\) are set up, leading us to an understanding of how significantly (or insignificantly) products are formed from the reactants.

Hydrogen Ion Concentration

The hydrogen ion concentration, \([\mathrm{H}^+]\), in a solution is a direct measurement of its acidity. It quantifies the number of moles of hydrogen ions per liter of solution and is a fundamental parameter in determining the pH of the solution. pH is calculated as the negative logarithm (base 10) of the hydrogen ion concentration:

\[\mathrm{pH} = -\log([\mathrm{H}^+])\]

In our textbook example, the calculation of \([\mathrm{H}^+]\) is derived from the first dissociation constant of hydrogen sulfide, taking into account the simplifying assumption that the first dissociation predominates due to its relatively larger Ka value. This assumption streamlines the mathematics, allowing for a straightforward calculation of hydrogen ion concentration and, therefore, the pH. A detailed grasp of these underlying principles enhances the understanding of how pH is an indicator of a solution's acidic or basic nature, which is vital across various scientific and industrial applications.