Polyprotic Acids and Bases: Videos & Practice Problems

In polyprotic acids, the removal of acidic hydrogens involves distinct acid ionization constants: K1, K2, and K3. The fully protonated form behaves like a monoprotic acid, while the fully deprotonated form acts as a base. Intermediate forms require specific equations to calculate [H⁺] concentrations, utilizing overlapping constants such as K_a and K_b.

Polyprotic Acids

Understanding polyprotic acids builds on the knowledge gained from diprotic acids. Polyprotic acids, represented as H₃A, can donate multiple protons (H⁺) in a stepwise manner. The first step involves the complete dissociation of the first acidic hydrogen, leading to the formation of H₃O⁺ and the anion H₂A^-. This process is characterized by the acid dissociation constant K_a1, which is defined by the equation:

K_a1 = \frac{[H₂A^-][H₃O^+3A]}

In the second step, H₂A^- can further donate another proton to water, producing HA^2- and another H₃O⁺. This reaction is described by the second dissociation constant K_a2:

K_a2 = \frac{[HA^2-][H₃O^+2A-]}

Finally, HA^2- can donate its last acidic hydrogen, resulting in the formation of A^3- and H₃O⁺. This step is represented by the third dissociation constant K_a3:

K_a3 = \frac{[A^3-][H₃O^+2-]}

It is important to note that the removal of the first acidic hydrogen is the easiest, and as each subsequent hydrogen is removed, the process becomes increasingly difficult. Consequently, the concentrations of the products formed decrease with each step, resulting in a smaller amount of H₃O⁺ and the corresponding anions. This trend highlights the fundamental behavior of triprotic and polyprotic acids in solution.

Polyprotic Bases

In the study of polyprotic bases, we explore species capable of accepting multiple protons (H⁺ ions). For instance, a species represented as A^3- can accept up to three protons, transitioning through various forms as it interacts with water. Initially, A^3- accepts one H⁺ from water, resulting in the formation of HA^2- and hydroxide ions (OH^-). This reaction can be expressed using the base dissociation constant (K_b1), which is defined as:

K_b1 = \frac{[OH^-][HA^{2-}]}{[A^{3-}]}

Continuing this process, HA^2- can accept another H⁺ from water, yielding H₂A^- and additional OH^- ions. The equilibrium for this reaction is represented by the second base dissociation constant (K_b2):

K_b2 = \frac{[OH^-][H_2A^-]}{[HA^{2-}]}

Finally, H₂A^- can react with a third water molecule to accept another H⁺, forming H₃A and OH^-. This reaction is characterized by the third base dissociation constant (K_b3):

K_b3 = \frac{[OH^-][H_3A]}{[H_2A^-]}

By examining the transitions between these various forms—A^3-, HA^2-, H₂A^-, and H₃A—we gain insight into the relationships between the different species and their corresponding acid dissociation constants (K_a) and base dissociation constants (K_b). This understanding is crucial for grasping the behavior of polyprotic species in chemical reactions.

Polyprotic Dissociation Constants

In the study of polyprotic acids, understanding the relationship between their various forms and the corresponding equilibrium constants is crucial. A polyprotic acid can lose multiple protons, and each step of deprotonation is associated with a specific acid dissociation constant, denoted as K_a1, K_a2, and K_a3. Conversely, when protons are added back, the base dissociation constants are represented as K_b1, K_b2, and K_b3.

For instance, when the fully protonated form, H₃A, loses its first acidic hydrogen, it involves K_a1. The second loss corresponds to K_a2, and the third to K_a3. The reverse process, where protons are added, utilizes the base constants. Notably, there are overlapping relationships between these constants, leading to the equations: K_a1 × K_b3 = K_w, K_a2 × K_b2 = K_w, and K_a3 × K_b1 = K_w, where K_w is the ion product of water.

When analyzing the fully protonated form, it can be treated similarly to a monoprotic acid using K_a1. In contrast, the fully deprotonated form, A³⁻, behaves as a base, utilizing K_b1. For the first intermediate, where one acidic hydrogen has been lost, the concentration of H⁺ can be calculated using the formula:

[H⁺] = \frac{K_a1 \times K_a2 \times [C₀] + K_a1 \times K_w}{K_a1 + [C₀]}

For the second intermediate, the equation becomes:

[H⁺] = \frac{K_a2 \times K_a3 \times [C₀] + K_a2 \times K_w}{K_a2 + [C₀]}

In summary, when dealing with polyprotic acids, it is essential to recognize the four possible forms: the fully acidic form, the fully basic form, and the two intermediate forms. Each form requires specific approaches to determine pH or H⁺ concentration, often involving ICE (Initial, Change, Equilibrium) tables for the acid and base forms, while the intermediate forms can be calculated directly using the established formulas.

Polyprotic Acid Calculations

To calculate the equilibrium concentrations of phosphoric acid (H₃PO₄), dihydrogen phosphate (H₂PO₄⁻), hydrogen phosphate (HPO₄²⁻), phosphate (PO₄³⁻), and the hydronium ion (H₃O⁺) for a 0.35 M solution of phosphoric acid, we start by recognizing that phosphoric acid is a triprotic acid, meaning it can donate three protons (H⁺ ions) in a stepwise manner. Each step has a corresponding acid dissociation constant (K_a), with K_a1 = 7.2 × 10⁻³, K_a2 = 6.3 × 10⁻⁸, and K_a3 = 4.2 × 10⁻¹³.

Initially, we set up an ICE (Initial, Change, Equilibrium) table for the first dissociation reaction:

H₃PO₄ + H₂O ⇌ H₂PO₄⁻ + H₃O⁺

Starting with 0.35 M of H₃PO₄, we denote the change in concentration as x. The equilibrium concentrations become:

• [H₃PO₄] = 0.35 - x
• [H₂PO₄⁻] = x
• [H₃O⁺] = x

Using the expression for K_a1:

K_a1 = &frac{[H₂PO₄⁻][H₃O⁺]}{[H₃PO₄]} = &frac{x²}{0.35 - x}

Substituting K_a1 into the equation gives:

7.2 × 10⁻³ = &frac{x²}{0.35 - x}

Applying the 5% rule, we find that x is small enough to ignore in the denominator, leading to a quadratic equation:

x² + 7.2 × 10⁻³x - 0.00252 = 0

Using the quadratic formula, we solve for x, yielding x = 0.0467 M (the negative solution is discarded). Thus, at equilibrium:

• [H₃PO₄] = 0.35 - 0.0467 = 0.3033 M
• [H₂PO₄⁻] = [H₃O⁺] = 0.0467 M

Next, we consider the second dissociation:

H₂PO₄⁻ + H₂O ⇌ HPO₄²⁻ + H₃O⁺

For this reaction, we again set up an ICE table. The initial concentration of H₂PO₄⁻ is 0.0467 M, and we denote the change as y:

• [H₂PO₄⁻] = 0.0467 - y
• [HPO₄²⁻] = y
• [H₃O⁺] = 0.0467 + y

Using K_a2:

K_a2 = &frac{[HPO₄²⁻][H₃O⁺]}{[H₂PO₄⁻]} = &frac{y(0.0467 + y)}{0.0467 - y}

Since K_a2 is small, we can approximate y as negligible, leading to:

6.3 × 10⁻⁸ = &frac{y(0.0467)}{0.0467} = y

Thus, [HPO₄²⁻] = 6.3 × 10⁻⁸ M.

Finally, for the third dissociation:

HPO₄²⁻ + H₂O ⇌ PO₄³⁻ + H₃O⁺

Setting up the ICE table again, we find:

• [HPO₄²⁻] = 6.3 × 10⁻⁸
• [PO₄³⁻] = z
• [H₃O⁺] = 0.0467 + z

Using K_a3:

K_a3 = &frac{[PO₄³⁻][H₃O⁺]}{[HPO₄²⁻]} = &frac{z(0.0467 + z)}{6.3 × 10⁻⁸}

Again, since K_a3 is very small, we can ignore z in the equilibrium concentrations:

4.2 × 10⁻¹³ = &frac{z(0.0467)}{6.3 × 10⁻⁸}

Solving for z gives:

z = 5.67 × 10⁻¹⁹ M, which represents the concentration of PO₄³⁻.

In summary, the equilibrium concentrations are:

• [H₃PO₄] = 0.3033 M
• [H₂PO₄⁻] = 0.0467 M
• [HPO₄²⁻] = 6.3 × 10⁻⁸ M
• [PO₄³⁻] = 5.67 × 10⁻¹⁹ M
• [H₃O⁺] = 0.0467 M

To find the pH, we can use the concentration of H₃O⁺:

pH = -log[H₃O⁺] = -log(0.0467).

For the hydroxide ion concentration, we can use the relationship:

[OH⁻] = &frac{K_w}{[H₃O⁺]} = &frac{1.0 × 10⁻¹⁴}{0.0467}.

Polyprotic Acid Calculations 2

To determine the pH of a 0.250 molar solution of sodium hydrogen phosphate (Na₂HPO₄), it is essential to understand the relationship between this compound and phosphoric acid (H₃PO₄), which is a triprotic acid. The dissociation of phosphoric acid occurs in three steps, resulting in the following species: H₃PO₄ (fully protonated), H₂PO₄^- (first intermediate), HPO₄^2- (second intermediate), and PO₄^3- (fully deprotonated).

In this case, sodium hydrogen phosphate corresponds to the second intermediate, HPO₄^2-. To find the hydronium ion concentration ([H₃O⁺]), we can use the following formula, which incorporates the acid dissociation constants (K_a2 and K_a3) and the initial concentration:

\[[H_3O^+] = \sqrt{\frac{K_{a2} \cdot K_{a3} \cdot C + K_{a2} \cdot K_w}{K_{a2} + C}}\]

Where:

• C is the initial concentration (0.250 M).
• K_w is the dissociation constant of water (1.0 x 10^-14).
• K_a2 and K_a3 are the second and third dissociation constants of phosphoric acid, with values of 6.2 x 10^-8 and 4.2 x 10^-13, respectively.

Substituting the values into the formula gives:

\[[H_3O^+] = \sqrt{\frac{(6.2 \times 10^{-8}) \cdot (4.2 \times 10^{-13}) \cdot (0.250) + (6.2 \times 10^{-8}) \cdot (1.0 \times 10^{-14})}{(6.2 \times 10^{-8}) + (0.250)}}\]

Calculating this results in:

\[[H_3O^+] = \sqrt{2.852 \times 10^{-20}} \approx 1.69 \times 10^{-10} \text{ M}\]

To find the pH, we use the formula:

\[\text{pH} = -\log([H_3O^+])\]

Substituting the hydronium ion concentration yields:

\[\text{pH} = -\log(1.69 \times 10^{-10}) \approx 9.77\]

In summary, when working with polyprotic acids like phosphoric acid, it is crucial to identify the specific form of the acid or its conjugate base you are dealing with. This identification guides the appropriate calculations to determine the pH of the solution accurately. Understanding these relationships and the relevant constants is key to mastering acid-base chemistry.

Polyprotic Acid Calculations 1

To determine the pH of a 0.150 molar solution of citric acid, we start by recognizing that citric acid is a triprotic acid, meaning it can donate three protons (H⁺) due to its three acidic hydrogens. In this case, we focus on the first dissociation constant, K_a1, which is relevant for the fully protonated form of citric acid.

The dissociation of citric acid in water can be represented as follows:

H₃C₆H₅O₇ + H₂O ⇌ H₂C₆H₅O₇^- + H₃O⁺

Initially, the concentration of citric acid is 0.150 M, while the concentrations of the products (H₂C₆H₅O₇^- and H₃O⁺) are both 0. As the reaction proceeds, we denote the change in concentration of H₃O⁺ as x. Thus, at equilibrium, the concentrations will be:

• [H₃C₆H₅O₇] = 0.150 - x
• [H₂C₆H₅O₇^-] = x
• [H₃O⁺] = x

The expression for K_a1 is given by:

K_a1 = \(\frac{x^2}{0.150 - x}\)

With K_a1 for citric acid being approximately 7.4 × 10^-4, we can set up the equation:

7.4 × 10^-4 = \(\frac{x^2}{0.150 - x}\)

To simplify the calculation, we can apply the 5% rule to determine if we can ignore x in the denominator. We calculate:

\(\frac{0.150}{7.4 × 10^{-4}} \approx 202.7\)

Since this value is not greater than 500, we must keep the -x term in our calculations and solve the quadratic equation:

x² + 7.4 × 10^-4x - 1.11 × 10^-4 = 0

Using the quadratic formula, where a = 1, b = 7.4 × 10^-4, and c = -1.11 × 10^-4, we find:

x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Calculating the discriminant and solving gives us two potential values for x. However, only the positive value is physically meaningful, as concentrations cannot be negative. The valid solution for x is approximately 0.010172 M.

Finally, to find the pH, we use the relationship:

pH = -log[H₃O⁺]

Substituting the concentration of H₃O⁺ we found:

pH = -log(0.010172) ≈ 1.99

In summary, understanding the specific form of citric acid being analyzed is crucial, as it dictates the appropriate dissociation constant to use and the method for calculating pH.

Here’s what students ask on this topic:

What are polyprotic acids and how do they differ from monoprotic acids?

Polyprotic acids are acids that can donate more than one proton (H⁺) per molecule. They have multiple acidic hydrogens, each with its own dissociation constant (K_a1, K_a2, K_a3, etc.). In contrast, monoprotic acids can donate only one proton per molecule and have a single dissociation constant (K_a). The dissociation of polyprotic acids occurs in steps, with each step having a distinct equilibrium constant. For example, H₃A dissociates to H₂A^-, then to HA^2-, and finally to A^3-, each step involving a different K_a value.

How do you calculate the pH of a solution containing a polyprotic acid?

To calculate the pH of a solution containing a polyprotic acid, you need to consider the stepwise dissociation of the acid. For the fully protonated form (e.g., H₃A), use the first dissociation constant (K_a1) and set up an ICE (Initial, Change, Equilibrium) chart to find [H⁺]. For intermediate forms, use specific equations that involve overlapping constants. For example, for the first intermediate (H₂A^-), use the equation involving K_a1 and K_a2. For the fully deprotonated form (A^3-), treat it as a base and use K_b1 to find [OH^-], then convert to pH.

What are the equilibrium constants K_a and K_b in the context of polyprotic acids?

In the context of polyprotic acids, K_a (acid dissociation constant) represents the equilibrium constant for the dissociation of each acidic hydrogen. For example, K_a1 is for the first dissociation, K_a2 for the second, and K_a3 for the third. K_b (base dissociation constant) represents the equilibrium constant for the acceptance of protons by the deprotonated forms. For instance, K_b1 is for the first proton acceptance by the fully deprotonated form (A^3-), K_b2 for the second, and K_b3 for the third. These constants are related by the equation K_a * K_b = K_w, where K_w is the ion-product constant of water.

How do you determine the concentration of H⁺ for intermediate forms of polyprotic acids?

To determine the concentration of H⁺ for intermediate forms of polyprotic acids, use specific equations that involve the dissociation constants and the initial concentration of the acid. For the first intermediate (H₂A^-), the equation is:

Ka1Ka2C + Ka1KwKa1 + C

where C is the initial concentration. For the second intermediate (HA^2-), the equation is:

Ka2Ka3C + Ka2KwKa2 + C

These equations help calculate [H⁺] for the intermediate forms.

What is the significance of overlapping equilibrium constants in polyprotic acids?

Overlapping equilibrium constants in polyprotic acids are significant because they simplify the calculations for certain forms of the acid. For example, K_a1 and K_b3 overlap, meaning K_a1 * K_b3 = K_w. This relationship helps in understanding the behavior of the acid and its conjugate base forms. When dealing with the fully protonated form (H₃A), you use K_a1 as if it were a monoprotic acid. For the fully deprotonated form (A^3-), you use K_b1 to treat it as a base. These overlaps allow for more straightforward calculations of pH and [H⁺] for different forms of the polyprotic acid.