ABSE_Polyprotic Acids Lesson

Polyprotic Acids

The name "polyprotic" literally means many protons. Polyprotic acids are acids that have more than one ionizable H+ atom per molecule. Protons are lost through several stages (one at each stage), with the first proton being the fastest and most easily lost. We call these stepwise reactions. Many of these steps are equilibrium reactions.

Here are the stepwise reactions for the ionization of phosphoric acid.

H3PO4   +   H2O    ⇌   H2PO4 -   +   H3O+           ka1 = 7.1 x 10-3

H2PO4 - +   H2O   ⇌   HPO4 -2   +   H3O+           ka2 = 6.3 x 10-8

HPO4 -2 +   H2O    ⇌   PO4 -3   +   H3O+             ka3 = 4.5 x 10-13

Notice that each step involving an equilibrium reaction has its own ka value. These values are listed as ka1 and ka2, etc.  

Here are some more examples of polyprotic acids and their respective ka values.

Common Polyprotic Acids: Formula: Sulfuric acid H2SO4 Sulfurous acid H2SO3 Phosphoric acid H3PO4 Carbonic acid H2CO3 H₂S H2C2O4 Hydrosulfuric acid or Hydrogen sulfide Oxalic acid Malonic acid H2C3H2O4 Number of Ionizable Hydrogens: 2 (diprotic) 2 (diprotic) Ka1: Ka2: Very Large 1.1E-2 1.3E-2 6.2E-8 3 (triprotic) 7.1E-3 6.3E-8 4.2E-13 2 (diprotic) 2 (diprotic) 4.7E-11 4.4E-7 1E-19 1.0E-7 5.3E-5 2 (diprotic) 2 (diprotic) 5.4E-2 2.0E-6 Ka3: 1.5E-3

From the table above, we see that sulfuric acid is the strongest. Since it is the strongest acid, it dissociates the fastest and easiest.

Each step contributes to the overall concentration of H+ but because the ka1 is so much larger than ka2, virtually all of the [H+] comes from the ka1 (first) reaction. The contributions to the overall [H+] from the other reactions is negligible. We therefore assume that in a stepwise reaction, [H+]total =   [H+]step 1. Note that this assumption does not always work!

Calculate the concentrations of ALL the species present in a solution of 0.05 M carbonic acid and calculate the pH of the solution. H2CO3H+ HCO3 HCOH + CO₂ 2 Kat = 4.5 x 10-7 Ka2 = 4.7 x 10-11 k_a1=[H][HCO,]/[H,CO,] 4.5 x 107= x= 1.5 x 10 = [H+] = [HCO3] [H2CO3] 0.05-4.5 x 107 = 0.0499 In step 2, some HCO3 will be used up (but very little) and more H+ will be made (also very little). HCO3 1.5 x 104 -X H+ + CO₂ 2 i C 1.5 x 104 + X 0 + X e 1.5 x 104-x 1.5 x 104-x X [H*][CO2] [HCO, ] 4.7 x 10 x (1.5 x 10+x) (1.5 x 10" - x) 4.7 x 10 x (1.5 x 10") (1.5 x 10") 4.7 x 10¹¹¹ = x = [CO₂²]

In a solution that contains a polyprotic acid as the only solute, the [anion] in step 2 = ka2

Salt Solutions of Polyprotic Acids

We determined that if salts contain ions of weak acids or bases, they undergo hydrolysis and can affect the pH of the solution. Salts that contain anions of polyprotic acids, like NaHCO3 or LiHPO4 can undergo hydrolysis to produce H+ or OH-.

Will the salt solution, Na₂HPO4, be acidic, base or neutral? We need to determine which equilibrium reaction is more likely to occur, ka or kb. HPO₄⁻² + H₂O ↔️ PO₄⁻³ + H₂O+ Ka₃=4.5 x 10⁻¹³ HPO₄⁻² + H₂O ↔️ H₂PO₄⁻ + OH⁻ Kb = ? We determine kw above using the equation, Ka = Ka X Kb 1 x 10⁻¹⁴ = Ka₂ X Kb 1 x 10⁻¹⁴=6.3 x 10⁻⁸ x Kb Kb = 1.6 x 10⁻⁷ Compare kь and ka to determine answer. Ka₃ = 4.5 x 10⁻¹³ Kb = 1.6 x 10⁻⁷ Since k > Ka, the solution will be basic.

What is the pH of a 0.15 M solution of Na₂CO₃? CO₂ + H₂O ↔️ HCO₃⁻ + OH⁻ Kb = 1 x 10⁻¹⁴/4.7 x 10⁻¹¹ Kь= 2.13 x 10⁻⁴ kb = [HCO₃⁻ ]/[OH⁻] [CO₃²] 2.13 x 10⁻⁴ = x² / 0.15-x 2.13 x 10⁻⁴ = x²/0.15 x = 5.6 x 10⁻³ = [OH] pOH = 2.25 pH = 11.75

Titrations of Polyprotic Acids

Since neutralization for polyprotic acids occurs stepwise, the titration curve has steps as well. The features are similar to those for monoprotic acids, but two equivalence points are reached. The titration curve shown below is of a diprotic acid, H2A, by a strong base. As each equivalence point is reached, the pH rises sharply. You will not be required to do the math associated with such curves but do need to understand them conceptually.

image of a titration curve with the first equivalence point labeled

Remember to work on the module practice problems as you complete each section of content.

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