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Chapter 14: Problem 109

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

Expert verified
In a 0.10 M H鈧係 solution with given ionization constants \(K_{a_{1}} = 1.0 \times 10^{-7}\) and \(K_{a_{2}} = 1.0 \times 10^{-19}\), the pH is calculated to be 4 and the concentration of S虏鈦� ions is \(1.0 \times 10^{-15}\) M.

Step by step solution

01

Write Equilibrium Equations

For ionization constants, we need to set up the equilibrium expressions for H鈧係: Equation 1: \(H_{2}S \rightleftharpoons H^{+} + HS^{-}\) with equilibrium constant, Ka鈧� Equation 2: \(HS^{-} \rightleftharpoons H^{+} + S^{2-}\) with equilibrium constant, Ka鈧�
02

Write Expressions for Ka鈧� and Ka鈧�

Now we will write the expressions for the ionization constants Ka鈧� and 碍补鈧�: \(K_{a_{1}} = \frac{[H^{+}][HS^{-}]}{[H_{2}S]}\) and \(K_{a_{2}} = \frac{[H^{+}][S^{2-}]}{[HS^{-}]}\)
03

Approximate the Concentrations

Since Ka鈧� is much smaller than Ka鈧�, we can assume for the time being that the majority of hydrogen ions come from the first ionization: Let x be the concentration of H鈦� ions, so [H鈦篯 = x. For the first equilibrium, [HS鈦籡 鈮� x and [H鈧係] 鈮� 0.10 M. We can ignore the contribution from the second ionization in this step.
04

Solve for x (H鈦� Concentration)

Using the expression for Ka鈧� and the given value of Ka鈧�, we can solve for x: \(1.0 \times 10^{-7} = \frac{x^2}{0.10}\) \(x^2 = 1.0 \times 10^{-8}\) \(x = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4}\) Now we have the H鈦� ion concentration, and we can calculate the pH.
05

Calculate the pH

Using the H鈦� ion concentration, we can determine the pH. pH = -log10([H鈦篯) = -log10(1.0 脳 10鈦烩伌) = 4 Now, we will find the concentration of S虏鈦� ions.
06

Solve for [S虏鈦籡 Concentration

Using the second equilibrium equation and the value of Ka鈧�, we can find the concentration of S虏鈦� ions, [S虏鈦籡, by substituting [H鈦篯=1.0 x 10鈦烩伌 and assuming that [HS鈦籡 鈮� x which we found in the previous steps: \(K_{a_{2}} = \frac{[H^{+}][S^{2-}]}{[HS^{-}]}\) \(1.0 \times 10^{-19} = \frac{(1.0 \times 10^{-4})[S^{2-}]}{1.0 \times 10^{-4}}\) \[S^{2-} = 1.0 \times 10^{-15}\] So, the pH of the 0.10 M H鈧係 solution is 4 and the concentration of S虏鈦� ions is \(1.0 \times 10^{-15}\) M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

H鈧係 ionization
Understanding the ionization process of hydrogen sulfide (H鈧係) is essential for calculating its pH and other relevant concentrations in solution. H鈧係 is a weak diprotic acid, meaning it can donate two protons (H鈦� ions) but does so in two separate stages.
  • First ionization: In the initial step, H鈧係 ionizes to form hydrosulfide ions (HS鈦�) and hydrogen ions (H鈦�). This process is represented by the equation: \[ H_{2}S ightleftharpoons H^{+} + HS^{-} \]
  • Second ionization: The hydrosulfide ion (HS鈦�) can further lose a proton to form sulfide ions (S虏鈦�) and an additional hydrogen ion (H鈦�): \[ HS^{-} ightleftharpoons H^{+} + S^{2-} \]
Each stage of ionization has its own equilibrium constant. The initial ionization (first ionization) is more significant, influencing the solution's pH more profoundly than the second.
Equilibrium Expressions
Equilibrium expressions help us to quantify the extent of ionization. For each ionization step of H鈧係, there's a corresponding equilibrium expression involving the concentrations of the reactants and products.
  • Ka鈧� Expression: For the first ionization, we use: \( K_{a_{1}} = \frac{[H^{+}][HS^{-}]}{[H_{2}S]} \)
  • Ka鈧� Expression: For the second ionization, we have: \( K_{a_{2}} = \frac{[H^{+}][S^{2-}]}{[HS^{-}]} \)
These expressions allow us to calculate the concentrations of ions at equilibrium. Knowing the values for biochemical or chemical systems helps predict or determine the behavior in a solution.
Ka values
Ka values are crucial for understanding how strongly an acid will ionize in a solution. The magnitude of a Ka value indicates the strength of the acid.
  • 碍补鈧�: The first ionization constant for H鈧係 is high relative to Ka鈧�, typically \(1.0 \times 10^{-7}\). This indicates that while H鈧係 does ionize, it is still a weak acid.
  • 碍补鈧�: The second ionization constant, \(1.0 \times 10^{-19}\), is significantly lower, showing that the second proton release is much less likely and far weaker.
The smaller Ka鈧� implies that the formation of S虏鈦� ions is minimal compared to HS鈦� ions. This affects how we simplify calculations, such as assuming the majority of H鈦� ions in solution come from the first ionization step.
Acid-Base Equilibrium
Acid-base equilibria involve reversible reactions where acids donate protons and bases accept them. In the case of H鈧係, both ionization stages establish an equilibrium between reactants and products.
  • First Equilibrium: The first equilibrium is more significant because it governs the solution's pH. Since Ka鈧� is much larger than Ka鈧�, this step accounts for most H鈦� ions.
  • Second Equilibrium: The second ionization provides additional insight but doesn't strongly influence pH due to its much smaller Ka value.
Understanding these equilibria allows for calculating pH and ion concentrations by setting up equations and assumptions like neglecting lesser effects. By considering only major contributors to H鈦� ion concentration, approximations simplify the complex equilibrium calculations.

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Most popular questions from this chapter

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Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{H}-\mathrm{S}, 363 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{kJ} / \mathrm{mol} )\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}^{+}\) d. \(\mathrm{NH}_{4}^{+}, \mathrm{PH}_{4}^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H},\) 322 \(\mathrm{kJ} / \mathrm{mol} )\) Give reasons for the orders you chose.
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