What is the difference between monoprotic and polyprotic acids?

Monoprotic acids contain a single ionizable proton, like hydrofluoric acid, while polyprotic acids contain two or more ionizable protons. Sulfurous acid has two ionizable protons, and phosphoric acid has three. Each proton is lost sequentially, with each reaction having its own acid ionization constant (Ka).

Why does each successive proton become harder to remove from a polyprotic acid?

Removing a proton from a neutral molecule is easier than removing one from a negatively charged ion because negative charge strengthens the bond between the proton and the anion. This is why Ka1 is larger than Ka2, which is larger than Ka3 for polyprotic acids like phosphoric acid.

When can you estimate the pH of a polyprotic acid using only the first ionization?

You can estimate pH using only the first reaction if Ka1 is at least one thousand times larger than subsequent Ka values. For example, ascorbic acid's Ka1 of 8 × 10−5 is much larger than its Ka2, so the first ionization dominates the pH calculation using an ICE table.

How do you calculate the concentration of dianion species in a polyprotic acid solution?

Use an ICE table with the second ionization constant (Ka2) and the monoanion concentration from the first ionization. For ascorbic acid, Ka2 equals 1.6 × 10−12, and the resulting dianion concentration is negligible, confirming that the first ionization determines the solution's pH.

What are examples of diprotic and triprotic acids?

Sulfuric acid and carbonic acid are diprotic acids with two ionizable protons. Phosphoric acid is a triprotic acid with three ionizable protons. Each successive ionization constant decreases significantly, typically by factors of 10^5 to 10^6, allowing simplified calculations for weak acid solutions and dissociation.

How do stepwise ionization constants differ in polyprotic acids?

Successive ionization constants decrease dramatically in polyprotic acids. For phosphoric acid, Ka1 is 7.5 × 10−3, Ka2 is 6.2 × 10−8, and Ka3 is 4.2 × 10−13. These large differences allow use of math-simplifying assumptions when calculating equilibrium concentrations in solutions.

Can polyprotic bases accept multiple protons?

Yes, polyprotic bases can accept more than one hydrogen ion. The carbonate ion is a diprotic base capable of accepting two protons. Like polyprotic acids, equilibrium calculations for polyprotic bases follow similar approaches using stepwise ionization constants and ICE tables.

15.12

Polyprotic Acids

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Monoprotic acids, like hydrofluoric acid, contain a single ionizable proton. In contrast, polyprotic acids contain two or more ionizable protons. For example, sulfurous acid has two ionizable protons, and phosphoric acid has three.

A polyprotic acid loses each of its protons sequentially, and each reaction has its own K_a.

It is easier to remove a proton from a neutral molecule than a negatively charged one because the negative charge increases the strength of the bond between the proton and the anion.

Therefore, for phosphoric acid, the K_a for the removal of the first proton, K_a1, is higher than the second one, K_a2, which is higher than K_a3.

The pH of a polyprotic acid can be estimated using only the first reaction if K_a1 is at least one thousand times larger than its subsequent K_a’s.

For example, the pH of a 0.050 molar ascorbic acid solution can be determined using its K_a1 and an ICE table.

When dissolved in water, ascorbic acid dissociates into hydronium and ascorbate ions. The K_a1 for this reaction is 8 × 10⁻⁵, and it is equal to the concentration of hydronium times the concentration of ascorbate monoanion, divided by the concentration of ascorbic acid.

An ICE table can be prepared for this reaction with the initial and equilibrium concentrations. Due to the small value of x, 0.050 minus x is approximately equal to 0.050.

Substituting these values into the expression for K_a1, the value for x is equal to 0.0020 molar, which is only 4% of the initial concentration of ascorbic acid. Thus, the approximation is valid.

The pH of the solution equals 2.70.

The concentration of ascorbate dianion formed in the second step of the reaction can also be calculated using K_a2 and an ICE table.

K_a2 is equal to 1.6 × 10⁻¹², and it can be expressed as the concentration of hydronium times the concentration of ascorbate dianion divided by the concentration of ascorbate monoanion.

For this reaction, the ICE table is filled in with the initial concentration of ascorbate monoanion and hydronium from the first reaction, 0.002 molar. Due to the small value of x, it can be omitted from the equilibrium concentrations of ascorbate monoanion and hydronium.

After substituting these values in the K_a2 expression and solving, x is 1.6 ×10⁻¹² molar.

Since x is less than 5% of 0.002 molar, the approximation is valid.

As the concentration of hydronium ions that were generated during the second step of ascorbic acid dissociation is negligible, the first ionizable proton determines the pH of the solution.

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Acids are classified by the number of protons per molecule that they can give up in a reaction. Acids such as HCl, HNO₃, and HCN that contain one ioni…