Polyprotic Acids

Quantitative Diprotic and Triprotic Acid Problems

  • No test questions or homework problems on quantitative diprotic and triprotic acid problems.
  • These problems involve pH determination and are considered tedious but not conceptually difficult.
  • An example is provided for informational purposes.

Diprotic Acid Dissociation

  • Generic diprotic acid (H₂A) dissociates in two steps:
    • Step 1: H2A⇌H++HA−H_2A \rightleftharpoons H^+ + HA^-
    • Step 2: HA−⇌H++A2−HA^- \rightleftharpoons H^+ + A^{2-}
  • This is a more accurate representation than the oversimplified version taught in introductory chemistry:
    • Incorrect: H2A→2H++A2−H_2A \rightarrow 2H^+ + A^{2-}
  • Each step has its own acid dissociation constant (Ka):
    • Ka1K_{a1} for the first proton dissociation.
    • Ka2K_{a2} for the second proton dissociation.

Setting Up the Equilibrium

  • Initial concentrations:
    • [H<em>2A]</em>0[H<em>2A]</em>0: Initial concentration of the diprotic acid (given).
    • [H+]0=0[H^+]_0 = 0
    • [HA−]0=0[HA^-]_0 = 0
  • Change:
    • For every XX moles of H+H^+ that dissociates in the first step:
      • Gain XX of H+H^+.
      • Gain XX of HA−HA^-.
  • Ka1K_{a1} expression:
    • K<em>a1=[H+][HA−][H</em>2A]=X2[H<em>2A]</em>0−XK<em>{a1} = \frac{[H^+][HA^-]}{[H</em>2A]} = \frac{X^2}{[H<em>2A]</em>0 - X}

Second Dissociation Step

  • Initial concentrations for the second step:
    • [HA−]0=X[HA^-]_0 = X (from the first step).
    • [H+]0=X[H^+]_0 = X (from the first step).
    • [A2−]0=0[A^{2-}]_0 = 0
  • Define a second variable, YY, to represent the dissociation in the second step:
    • For every −Y-Y lost of HA−HA^-, gain +Y+Y of H+H^+ and +Y+Y of A2−A^{2-}.
  • Ka2K_{a2} expression:
    • Ka2=[H+][A2−][HA−]=(X+Y)(Y)X−YK_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]} = \frac{(X+Y)(Y)}{X-Y}

Solving for pH

  • Simultaneous equations:
    • Two equations with two unknowns (XX and YY).
    • Solve the first equation for XX.
    • Plug the value of XX into the second equation to solve for YY.
  • Concentration of H+H^+ at equilibrium:
    • [H+]eq=X+Y[H^+]_{eq} = X + Y (X from the first step, Y from the second step).
  • Calculate pH:
    • pH=−log⁡([H+]eq)pH = -\log([H^+]_{eq})

Relative Magnitude of Ka Values

  • The dissociation constants generally decrease with each step:
    • K{a1} > K{a2} > K_{a3} (for a triprotic acid).

Exam Relevance

  • Quantitative pH problems with diprotic or triprotic acids will not be on the exam.
  • The focus is on understanding and writing the dissociation steps.