Diprotic Acids and Bases: Videos & Practice Problems

Diprotic acids can donate two H⁺ ions, represented as H₂A. The first ionization constant, K_a1, is larger than K_a2, indicating it's easier to remove the first H⁺. Diprotic bases, represented as A²⁻, can accept two H⁺ ions, with K_b1 and K_b2 as their respective constants. The relationships between these constants are K_a1 × K_b2 = K_w and K_a2 × K_b1 = K_w, where K_w is the ion product constant for water.

Diprotic Acids

Diprotic acids and bases are unique compounds capable of donating or accepting two hydrogen ions (H⁺). The general formula for a diprotic acid can be represented as H₂A. When a diprotic acid is in its fully protonated form, it contains both acidic hydrogens. In this state, it acts as an acid, while water serves as the base. According to acid-base theory, the acid donates an H⁺ ion to the base, resulting in the formation of the conjugate base (A^-) and hydronium ion (H₃O⁺). This reaction can be expressed in terms of the first acid dissociation constant, K_a1, which is defined by the equilibrium expression:

$$ K_{a1} = \frac{[A^-][H_3O^+]}{[H_2A]} $$

After the first dissociation, the conjugate base (A^-) can further lose its second H⁺ ion. In this subsequent reaction, another water molecule acts as the base, and the acid donates another H⁺. This leads to the formation of the second conjugate base (A^2-) and another hydronium ion (H₃O⁺). The equilibrium expression for this second dissociation is represented by the second acid dissociation constant, K_a2:

$$ K_{a2} = \frac{[A^{2-}][H_3O^+]}{[A^-]} $$

It is important to note that K_a1 is always significantly larger than K_a2, indicating that it is easier to remove the first acidic hydrogen than the second. This relationship highlights the decreasing strength of the acid as it loses protons, with K_a1 typically being much greater than K_a2.

Diprotic Bases

In the study of diprotic compounds, understanding both the acid and base forms is crucial. Diprotic bases can be represented by the notation \( A^{2-} \), indicating their ability to accept two protons (\( H^+ \) ions). The interaction of a diprotic base with water, which acts as an acid, leads to the formation of its conjugate acid. In the first reaction, the diprotic base \( A^{2-} \) accepts an \( H^+ \) from water, resulting in the formation of \( HA^{-} \) and hydroxide ions (\( OH^{-} \)). This process is characterized by the equilibrium constant \( K_{b1} \), defined as:

\( K_{b1} = \frac{[HA^{-}][OH^{-}]}{[A^{2-}]} \)

In this equation, only the concentrations of the products and the diprotic base are considered, as solids and liquids are excluded from equilibrium expressions.

Continuing with the reaction, the newly formed \( HA^{-} \) can further accept another \( H^+ \) ion from a second water molecule, producing \( H_2A \) and another \( OH^{-} \). This second reaction is represented by the equilibrium constant \( K_{b2} \), which is expressed as:

\( K_{b2} = \frac{[H_2A][OH^{-}]}{[HA^{-}]} \)

Again, this expression focuses on the products and the reactants, omitting solids and liquids. Through these reactions, we observe the relationships between the diprotic acid forms and their corresponding base forms, as well as the equilibrium constants \( K_{a1} \), \( K_{a2} \), \( K_{b1} \), and \( K_{b2} \). Understanding these connections is essential for grasping the behavior of diprotic compounds in various chemical contexts.

Diprotic Dissociation Constants

In the study of diprotic acids and bases, understanding the relationships between their various forms is crucial. A diprotic species, such as H₂A, can donate protons in a stepwise manner. The first dissociation leads to the formation of HA^-, which is associated with the equilibrium constant K_a1. The second dissociation results in A^2-, linked to K_a2. Conversely, when considering the basic forms, HA^- can accept a proton to revert to H₂A, represented by K_b1, and A^2- can accept another proton to form HA^-, represented by K_b2.

These equilibrium constants are interconnected. Specifically, the relationships can be expressed as K_a1 × K_b2 = K_w and K_a2 × K_b1 = K_w, where K_w is the ion product of water, valued at 1.0 × 10^-14 at 25 degrees Celsius. It is important to note that K_w is temperature-dependent, meaning it will change with temperature variations.

When working with the fully protonated form H₂A, it behaves like a monoprotic acid, and K_a1 is used to calculate pH. The intermediate form HA^- is versatile; it can either donate a proton (acting as an acid, using K_a2) or accept a proton (acting as a base, using K_b2). This dual capability classifies HA^- as an amphoteric species, which can act as both an acid and a base depending on the context.

Common examples of amphoteric species include hydrogen sulfate (HSO₄^-), hydrogen sulfite (HSO₃^-), and dihydrogen phosphate (H₂PO₄^-). These species can donate protons, utilizing K_a2, or accept protons, utilizing K_b2. Similarly, hydrogen carbonate (HCO₃^-), hydrogen sulfide (HS^-), and hydrogen phosphate (HPO₄^2-) are examples of basic amphoteric species.

In summary, the behavior of diprotic species is characterized by their ability to donate or accept protons, governed by their respective equilibrium constants. Understanding these relationships is essential for calculating pH and concentrations in various chemical contexts.

Diprotic Acids and Bases Calculations 1

Sulfuric acid, represented as H₂SO₃, is classified as a diprotic acid, meaning it can donate two protons (H⁺) in solution. The acid dissociation constants for sulfuric acid are given as K_a1 = 1.6 × 10⁻² and K_a2 = 4.6 × 10⁻⁵. To calculate the pH and concentrations of sulfuric acid, hydrogen sulfide (HSO₃⁻), and sulfate ion (SO₃²⁻), we start with an initial concentration of 0.200 M sulfuric acid.

When H₂SO₃ dissociates in water, it donates a proton to form HSO₃⁻ and hydronium ions (H₃O⁺). The initial concentrations for the ICE (Initial, Change, Equilibrium) table are set up as follows: initial concentration of H₂SO₃ is 0.200 M, while HSO₃⁻ and H₃O⁺ start at 0. The change in concentration for H₂SO₃ is -x, and for the products, it is +x. The equilibrium expression for K_a1 is given by:

$$ K_{a1} = \frac{[HSO_3^-][H_3O^+]}{[H_2SO_3]} $$

Substituting the equilibrium concentrations into the expression yields:

$$ 1.6 \times 10^{-2} = \frac{x^2}{0.200 - x} $$

To simplify the calculation, we apply the 5% approximation method. We find that:

$$ \frac{0.200}{1.6 \times 10^{-2}} = 12.5 $$

Since this value is not greater than 500, we cannot ignore x, and thus we must use the quadratic formula to solve for x. Rearranging the equation leads to:

$$ x^2 + 1.6 \times 10^{-2}x - 0.0032 = 0 $$

Using the quadratic formula:

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

where a = 1, b = 1.6 × 10⁻², and c = -0.0032. Solving this gives two potential values for x, but only the positive value is valid for concentration:

$$ x \approx 0.0491 $$

At equilibrium, the concentration of H₂SO₃ is:

$$ [H_2SO_3] = 0.200 - 0.0491 \approx 0.151 M $$

The concentration of HSO₃⁻ and H₃O⁺ is both equal to x, which is approximately 0.0491 M. The pH can then be calculated using the formula:

$$ pH = -\log[H_3O^+] $$

Substituting the value gives:

$$ pH \approx 1.31 $$

Next, to find the concentration of the sulfate ion (SO₃²⁻), we consider the second dissociation step of HSO₃⁻ reacting with water. The equilibrium expression for K_a2 is:

$$ K_{a2} = \frac{[SO_3^{2-}][H_3O^+]}{[HSO_3^-]} $$

Since the concentration of HSO₃⁻ is approximately 0.0491 M and the change in concentration due to the second dissociation is negligible, we can simplify the calculation. Thus, we find:

$$ K_{a2} \approx [SO_3^{2-}] \cdot 0.0491 $$

Solving for [SO₃²⁻] gives:

$$ [SO_3^{2-}] \approx 4.6 \times 10^{-5} M $$

In summary, the calculated concentrations and pH for the diprotic acid H₂SO₃ are as follows: the pH is approximately 1.31, the concentration of H₂SO₃ at equilibrium is 0.151 M, the concentration of HSO₃⁻ is approximately 0.0491 M, and the concentration of SO₃²⁻ is approximately 4.6 × 10⁻⁵ M.

Diprotic Acids and Bases Calculations 1

To determine the pH of a 0.080 molar sodium sulfide solution, we first recognize that sodium sulfide (Na₂S) is the basic form of the diprotic acid hydrosulfuric acid (H₂S). The relevant acid dissociation constants for H₂S are K_a1 = 1.0 × 10^-7 and K_a2 = 9.1 × 10^-8. Since sodium ions (Na⁺) are spectator ions, we focus on the sulfide ion (S^2-), which will react with water in a hydrolysis reaction.

The reaction can be represented as follows:

S^2- + H₂O ⇌ HS^- + OH^-

Using an ICE (Initial, Change, Equilibrium) table, we set up the initial concentrations. The initial concentration of S^2- is 0.080 M, while the concentrations of HS^- and OH^- are initially 0. We denote the change in concentration as x:

Initial: 0.080, 0, 0

Change: -x, +x, +x

Equilibrium: 0.080 - x, x, x

Since we are dealing with the first dissociation of the sulfide ion, we need to find K_b1. The relationship between K_b1 and K_a2 is given by:

K_b1 = K_w / K_a2

Where K_w = 1.0 × 10^-14. Thus:

K_b1 = (1.0 × 10^-14) / (9.1 × 10^-8) = 1.1 × 10^-7

Next, we set up the expression for K_b1:

K_b1 = (OH^-)(HS^-) / (S^2-)

Substituting the equilibrium concentrations, we have:

1.1 × 10^-7 = (x)(x) / (0.080 - x)

To simplify, we check if we can ignore x in the denominator by using the 5% rule:

0.080 / 1.1 × 10^-7 = 727, which is greater than 500. Thus, we can ignore x.

Now, we solve for x:

1.1 × 10^-7 = x² / 0.080

x² = 8.8 × 10^-9

x = √(8.8 × 10^-9) = 9.4 × 10^-5

This value of x represents the concentration of OH^-. To find pOH, we use the formula:

pOH = -log(OH^-) = -log(9.4 × 10^-5) = 4.03

Finally, we can calculate the pH using the relationship:

pH = 14 - pOH = 14 - 4.03 = 9.97

In summary, the pH of a 0.080 molar sodium sulfide solution is approximately 9.97. Understanding the relationship between the acid and base forms of diprotic acids is crucial for solving similar problems involving pH determination.

Diprotic Acids and Bases Calculations 2

To determine the pH of a 0.15 molar solution of sodium bicarbonate (NaHCO₃), we first recognize that sodium bicarbonate is the intermediate form of the diprotic species carbonic acid (H₂CO₃). When carbonic acid loses its first acidic hydrogen, it forms bicarbonate (HCO₃^-), which can further lose its second hydrogen to become carbonate (CO₃^2-). In this case, we are dealing with the bicarbonate ion, which can act as both an acid and a base, making it an amphoteric species.

To find the concentration of hydrogen ions (H⁺), we can use the following formula for the intermediate form of a diprotic species:

[H⁺] = √(K_a1 × K_a2 × C + K_a1 × K_w / (K_a1 + C))

Where:

• K_a1 = 4.46 × 10^-7
• K_a2 = 4.69 × 10^-11
• K_w = 1.0 × 10^-14 (the dissociation constant for water)
• C = 0.15 M (the formal concentration of sodium bicarbonate)

Plugging in the values, we calculate:

[H⁺] = √(4.46 × 10^-7 × 4.69 × 10^-11 × 0.15 + 4.46 × 10^-7 × 1.0 × 10^-14 / (4.46 × 10^-7 + 0.15))

After performing the calculations, we find:

[H⁺] = √(2.09471 × 10^-17)

Taking the square root gives us:

[H⁺] ≈ 4.6 × 10^-9 M

Now, to find the pH, we use the formula:

pH = -log[H⁺]

Substituting the value we calculated:

pH ≈ -log(4.6 × 10^-9) ≈ 8.34

This pH value indicates that sodium bicarbonate is a basic solution, which aligns with its role as an amphoteric species. The presence of bicarbonate allows it to act as a buffer, maintaining a pH above 7, which is characteristic of basic solutions. Understanding this relationship between diprotic species and their intermediate forms is crucial for solving similar problems in acid-base chemistry.

An unknown diprotic acid has an initial concentration of 0.025 M. What is the pH of the solution if pka₁ is  3.25 and pKa₂ is 6.82?

• A. 4.21
• B. 2.46
• C. 1.91
• D. 3.25
• E. 2.79

Answer: (B)

Here’s what students ask on this topic:

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

Diprotic acids are acids that can donate two H⁺ ions, represented as H₂A. They undergo two ionization steps, each with its own ionization constant (K_a1 and K_a2). The first ionization constant, K_a1, is always larger than K_a2, indicating it is easier to remove the first H⁺ ion. In contrast, monoprotic acids can only donate one H⁺ ion and have a single ionization constant (K_a).

How do you calculate the pH of a diprotic acid solution?

To calculate the pH of a diprotic acid solution, you need to consider both ionization steps. First, use the first ionization constant (K_a1) to find the concentration of H₃O⁺ produced in the first step. Then, use the second ionization constant (K_a2) to find the additional H₃O⁺ produced in the second step. The total [H₃O⁺] is the sum of the contributions from both steps. Finally, calculate the pH using the formula:

$\mathrm{pH}=-\log \left(H_{+}\right)$

What is the relationship between K_a and K_b for diprotic acids and bases?

The relationship between the ionization constants of diprotic acids and bases is given by the equations:

$K_1\times K_2=K_{\mathrm{w}}$

and

$K_2\times K_1=K_{\mathrm{w}}$

where K_w is the ion product constant for water (1.0 × 10^-14 at 25°C). These relationships help in understanding the equilibrium between the different forms of diprotic species.

What are the common examples of diprotic acids and bases?

Common examples of diprotic acids include sulfuric acid (H₂SO₄), carbonic acid (H₂CO₃), and oxalic acid (H₂C₂O₄). For diprotic bases, examples include carbonate ion (CO₃^2-) and oxalate ion (C₂O₄^2-). These compounds can donate or accept two H⁺ ions, respectively, and have distinct ionization constants for each step.

How do you determine the equilibrium concentrations in a diprotic acid solution?

To determine the equilibrium concentrations in a diprotic acid solution, follow these steps:

1. Write the ionization equations for both steps.
2. Set up the equilibrium expressions for K_a1 and K_a2.
3. Use the initial concentration of the diprotic acid and the ionization constants to solve for the equilibrium concentrations of all species involved.
4. Apply the ICE (Initial, Change, Equilibrium) table method to keep track of concentration changes.

This approach helps in systematically finding the concentrations of H₃O⁺, the conjugate base, and the remaining diprotic acid at equilibrium.