7.5 Comprehensive Study Notes: Polyprotic Acids and Bases

Introduction to Polyprotic Acids and Bases

• Conceptual Overview: This section extends the study of equilibrium and Brønsted-Lowry acid-base theory beyond substances that only transfer a single proton. It focuses specifically on polyprotic species.

• Primary Learning Objective: To apply existing equilibrium principles to chemical species capable of donating or accepting more than one proton ($H^+$).

Classification of Acids and Bases by Proton Capacity

• Monoprotic Substances:

  • Definition: A substance that can only donate or accept exactly one hydrogen ion (proton) per molecule.

  • Monoprotic Acid Examples:

    • Hydrogen chloride ($HCl$).

    • Hydrogen carbonate ($HCO_3^-$) — Note: In this specific context of acting as a monoprotic acid.

    • Hydrogen cyanide ($HCN$).

    • When dissolved in water, these release only one proton to the solution.

  • Monoprotic Base Examples:

    • Sodium hydroxide ($NaOH$).

    • Potassium hydroxide ($KOH$).

    • Ammonia ($NH_3$).

    • In each instance, these accept exactly one proton, typically resulting in the formation of water ($H_2O$) or ammonium ($NH_4^+$).

• Diprotic Substances:

  • Definition: Substances capable of donating or accepting two protons.

  • Diprotic Acid Examples:

    • Sulfuric Acid ($H_2SO_4$): It first dissociates to release one proton. Its conjugate base can then dissociate further to donate a second proton, eventually resulting in a sulfate ion ($SO_4^{2-}$).

    • Carbonic Acid ($H_2CO_3$): Donates one proton to form hydrogen carbonate ($HCO_3^-$), which can then donate another proton to form the carbonate ion ($CO_3^{2-}$).

  • Diprotic Bases: These are capable of accepting two protons in a stepwise manner.

• Triprotic Substances:

  • Definition: A substance that can donate or accept up to three protons.

  • Triprotic Acid Example: Phosphoric Acid ($H_3PO_4$):

    • Step 1: Donates the first proton to form dihydrogen phosphate ($H_2PO_4^-$).

    • Step 2: Donates the second proton to form hydrogen phosphate ($HPO_4^{2-}$).

    • Step 3: Donates the final proton to form the phosphate ion ($PO_4^{3-}$).

  • Triprotic Base Example: Phosphate Ion ($PO_4^{3-}$): This acts as the reverse of the acid process, accepting up to three protons to eventually reform the initial reactant, phosphoric acid ($H_3PO_4$).

Dynamics of Stepwise Ionization

• Stepwise Nature: The loss or gain of protons occurs in sequential steps rather than all at once.

• The Ionization Constant Trend:

  • In each instance, the equilibrium constant ($Ka$ for acids or $Kb$ for bases) is significantly larger for the first ionization step than for subsequent steps.

  • Relative Magnitude: Ka_1 >> Ka_2 >> Ka_3.

  • Implication: The first ionization occurs to the greatest extent. The second ionization occurs less, and the third occurs the least. This means the majority of the hydronium ion concentration in a solution of a polyprotic acid usually comes from the first dissociation step.

Guided Practice: Calculating Concentrations in a Diprotic Acid Solution

Problem Statement: Determine the unknown concentrations in a saturated aqueous solution of hydrosulfuric acid ($H_2S$) at room temperature with an initial concentration of $0.1\,M$.

Given Data:

• Initial Concentration of $H_2S = 0.1\,M$

• First Acid Ionization Constant ($Ka_1$): $8.9 \times 10^{-8}$

• Second Acid Ionization Constant ($Ka_2$): $1.0 \times 10^{-19}$

Step 1: Solving the First Ionization

• Equation: $H_2S(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HS^-(aq)$

• ICE Table Setup:

  • Initial: $[H_2S] = 0.1\,M$, $[H_3O^+] = 0$, $[HS^-] = 0$

  • Change: $[H_2S] = -x$, $[H_3O^+] = +x$, $[HS^-] = +x$

  • Equilibrium: $[H_2S] = 0.1 - x$, $[H_3O^+] = x$, $[HS^-] = x$

• Assumption: Because $Ka_1$ is small, we assume $x$ is significantly smaller than the initial concentration (x << 0.1). Therefore, $0.1 - x \approx 0.1$.

• Equilibrium Expression:

  • $Ka_1 = \frac{[H_3O^+][HS^-]}{[H_2S]}$

  • $8.9 \times 10^{-8} = \frac{x \times x}{0.1}$

  • $x^2 = (8.9 \times 10^{-8})(0.1) = 8.9 \times 10^{-9}$

• Solving for $x$:

  • $x = \sqrt{8.9 \times 10^{-9}}$

  • $x \approx 9.433981132 \times 10^{-5}\,M$

• Conclusion for Step 1: Applying two significant figures, the concentrations are:

  • $[H_3O^+] = 9.4 \times 10^{-5}\,M$

  • $[HS^-] = 9.4 \times 10^{-5}\,M$

  • This can also be written as $0.000094\,M$.

Step 2: Solving for the Sulfide Ion Concentration ($[S^{2-}]$)

• Equation: $HS^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + S^{2-}(aq)$

• Equilibrium Expression:

  • $Ka_2 = \frac{[H_3O^+][S^{2-}]}{[HS^-]}$

• Calculation Logic:

  • From Step 1, we found that $[H_3O^+]$ and $[HS^-]$ are essentially equal ($9.4 \times 10^{-5}\,M$).

  • In the expression for $Ka_2$, the concentration of hydronium and the concentration of hydrogen sulfide cancel each other out.

  • $Ka_2 = \frac{(9.4 \times 10^{-5})[S^{2-}]}{9.4 \times 10^{-5}}$

• Result:

  • $[S^{2-}] = Ka_2$

  • $[S^{2-}] = 1.0 \times 10^{-19}\,M$

• Significance: The value of $Ka_2$ is extremely small ($10^{-19}$), indicating that almost no sulfide ion is yielded during the second ionization step.

Summary of Findings

• For a $0.1\,M$ solution of $H_2S$:

  1. The primary concentration of acid remains near $0.1\,M$.

  2. The hydronium and hydrogen sulfide concentrations are determined by $Ka_1$ ($9.4 \times 10^{-5}\,M$).

  3. The concentration of the conjugate base from the second dissociation ($S^{2-}$) is equal to $Ka_2$ ($1.0 \times 10^{-19}\,M$).

• This demonstrates that the species produced in later ionization steps exist in radically lower concentrations than those from the first step.