Diprotic Acids and Bases Calculations: Videos & Practice Problems

Sulfuric acid (H₂SO₄) is unique as the only strong diprotic acid, meaning it fully dissociates its first acidic proton in water, forming HSO₄^- and H₃O⁺. The second acidic proton dissociates only partially, making it a weak acid in this step, indicated by reversible arrows in the chemical equation. This two-step dissociation process is crucial for understanding the overall concentration of H⁺ ions in a solution, which is essential for calculating the pH. The constants K₁ and K₂ represent the dissociation constants for the first and second protons, respectively, with K₁ being a measure of the strong acid dissociation and K₂ representing the weaker second dissociation. Understanding these concepts is vital for comprehending acid behavior in aqueous solutions.

pH of Diprotic Acid Dissociation

Diprotic acids are characterized by their ability to donate two protons (H⁺) in solution. Among these, sulfuric acid (H₂SO₄) is unique as it is the only strong diprotic acid, while all other diprotic acids are classified as weak. Understanding the dissociation of these protons is crucial for calculating the concentration of hydrogen ions (H⁺) and, consequently, the pH of the solution.

When sulfuric acid is dissolved in water, the first proton dissociates completely, represented by the reaction:

This reaction is characterized by a strong arrow indicating complete dissociation, and the equilibrium constant for this step is denoted as K_a1.

In contrast, the second proton dissociation from the bisulfate ion (HSO₄^-) is only partial, represented by the reversible reaction:

This reaction is indicated by reversible arrows, reflecting that not all bisulfate ions will dissociate to form sulfate ions (SO₄^2-) and additional hydronium ions (H₃O⁺). The equilibrium constant for this step is denoted as K_a2.

In summary, when analyzing sulfuric acid, it is essential to recognize that the first acidic proton is lost completely, while the second proton is only partially dissociated. This understanding is vital for accurately determining the pH of solutions containing sulfuric acid.

Calculating pH of Diprotic Acids Example

To calculate the pH of a 0.0550 molar solution of sulfuric acid (H₂SO₄), we first recognize that sulfuric acid is a strong diprotic acid. This means it can donate two protons (H⁺) in solution, but the first dissociation is complete while the second is weaker. The dissociation reactions can be represented as follows:

1. H₂SO₄(aq) + H₂O(l) → HSO₄^-(aq) + H₃O⁺(aq)

2. HSO₄^-(aq) + H₂O(l) ⇌ SO₄^2-(aq) + H₃O⁺(aq)

In the first step, the initial concentration of sulfuric acid is 0.0550 M, which means that at the start, the concentration of H₃O⁺ and HSO₄^- is also 0.0550 M due to complete dissociation.

Next, we set up an ICE (Initial, Change, Equilibrium) table for the second dissociation involving HSO₄^-. The initial concentrations are:

• [HSO₄^-] = 0.0550 M
• [H₃O⁺] = 0.0550 M
• [SO₄^2-] = 0

As HSO₄^- donates a proton, we denote the change in concentration as -x for HSO₄^- and +x for both H₃O⁺ and SO₄^2-. The equilibrium concentrations become:

• [HSO₄^-] = 0.0550 - x
• [H₃O⁺] = 0.0550 + x
• [SO₄^2-] = x

Using the second dissociation constant (K_a2), which is 1.2 × 10^-2, we can set up the equilibrium expression:

K_a2 = \(\frac{[SO₄^{2-3O+]}{[HSO₄-]}\)}

Substituting the equilibrium concentrations into the expression gives:

1.2 × 10^-2 = \(\frac{x(0.0550 + x)}{0.0550 - x}\)

To solve for x, we rearrange and simplify, leading to a quadratic equation:

x² + 0.067x - 6.6 × 10^-4 = 0

Applying the quadratic formula, x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where a = 1, b = 0.067, and c = -6.6 × 10^-4, we find two potential solutions for x. Only the positive solution is valid, as concentrations cannot be negative.

After calculating, we find x ≈ 0.008715. The total concentration of H₃O⁺ is then:

[H₃O⁺] = 0.0550 + 0.008715 = 0.063715 M

Finally, we can calculate the pH using the formula:

pH = -log[H₃O⁺] = -log(0.063715) ≈ 1.20

Thus, the pH of the 0.0550 molar sulfuric acid solution is approximately 1.20.

pH of Weak Diprotic Acids

To calculate the pH of a weak diprotic acid, such as carbonic acid, it is essential to focus on the first dissociation constant, \( K_{a1} \). For a 0.115 M solution of carbonic acid, we begin by setting up an ICE (Initial, Change, Equilibrium) chart to track the concentrations of the species involved in the reaction with water.

Using the Brønsted-Lowry definition, carbonic acid (\( H_2CO_3 \)) donates a proton to water, forming the bicarbonate ion (\( HCO_3^- \)) and hydronium ion (\( H_3O^+ \)). The reaction can be represented as:

\[ H_2CO_3 + H_2O \rightleftharpoons HCO_3^- + H_3O^+ \]

In the ICE chart, we start with the initial concentration of carbonic acid at 0.115 M, while the initial concentrations of the products are zero. The change in concentration for the reactants is represented as \(-x\) and for the products as \(+x\). Thus, the equilibrium concentrations can be expressed as:

Initial: \( [H_2CO_3] = 0.115 \), \( [HCO_3^-] = 0 \), \( [H_3O^+] = 0 \)

Change: \( [H_2CO_3] = 0.115 - x \), \( [HCO_3^-] = x \), \( [H_3O^+] = x \)

Equilibrium: \( [H_2CO_3] = 0.115 - x \), \( [HCO_3^-] = x \), \( [H_3O^+] = x \)

Next, we set up the equilibrium expression using \( K_{a1} \):

\[ K_{a1} = \frac{[HCO_3^-][H_3O^+]}{[H_2CO_3]} \]

Substituting the equilibrium concentrations into the expression gives:

\[ K_{a1} = \frac{x^2}{0.115 - x} \]

To simplify calculations, we can apply the 500 approximation method. This involves dividing the initial concentration of the weak diprotic acid by \( K_{a1} \). If the result is greater than 500, we can ignore the \(-x\) in the denominator. Given that \( K_{a1} = 4.3 \times 10^{-7} \), we calculate:

\[ \frac{0.115}{4.3 \times 10^{-7}} \approx 267,441.9 \]

Since this value is much greater than 500, we can simplify our expression to:

\[ K_{a1} = \frac{x^2}{0.115} \]

Cross-multiplying gives:

\[ x^2 = K_{a1} \times 0.115 = 4.3 \times 10^{-7} \times 0.115 \approx 4.945 \times 10^{-8} \]

Taking the square root of both sides yields:

\[ x = \sqrt{4.945 \times 10^{-8}} \approx 2.224 \times 10^{-4} \]

Since \( x \) represents the concentration of hydronium ions (\( [H_3O^+] \)), we can now calculate the pH:

\[ pH = -\log[H_3O^+] = -\log(2.224 \times 10^{-4}) \approx 3.65 \]

Thus, the pH of the 0.115 M carbonic acid solution is approximately 3.65, indicating its acidic nature.

Calculate Concentration of the Basic Form

Understanding the concentration of the basic form of a weak diprotic acid is crucial in acid-base chemistry. A weak diprotic acid can donate two protons (H⁺ ions) in a stepwise manner. When given the initial concentration of such an acid, the concentration of its basic form can be directly equated to the second acid dissociation constant, K_a2.

For example, consider a solution of carbonic acid (H₂CO₃) with an initial concentration of 0.115 M. To find the concentration of the carbonate ion (CO₃^2-), we can use the value of K_a2, which is 5.6 × 10^-11 M. Thus, the concentration of the carbonate ion is equal to K_a2, confirming that:

[CO₃^2-] = K_a2 = 5.6 × 10^-11 M.

To further understand this relationship, we can analyze the dissociation process of carbonic acid. The first dissociation step involves the formation of bicarbonate (HCO₃^-) and hydronium ions (H₃O⁺), represented by the equilibrium expression for K_a1:

K_a1 = \frac{[HCO₃^-][H₃O^+2CO₃]}.

Using the initial concentration of carbonic acid and applying the 500 percent approximation method, we can simplify calculations. If K_a1 is known (4.3 × 10^-7), we can find the concentration of the intermediate form (bicarbonate) and hydronium ions. After determining these concentrations, we can set up a second ICE (Initial, Change, Equilibrium) chart for the second dissociation step, which leads to the formation of the carbonate ion.

In this second step, we again apply the equilibrium constant expression for K_a2:

K_a2 = \frac{[CO₃^2-][H₃O⁺]}{[HCO₃^-]}.

By substituting the known values and applying the approximation method, we can confirm that the concentration of the basic form (carbonate ion) is indeed equal to K_a2. This relationship simplifies calculations significantly, allowing for quick determination of the basic form's concentration without extensive calculations.

In summary, when given the initial concentration of a weak diprotic acid, one can directly use K_a2 to find the concentration of its basic form, streamlining the process of acid-base equilibrium analysis.

Which of the following diprotic acid would produce the most acidic solution when dissolved?

• A. 0.200 M H₂SO₃ (K_a1 = 1.6 x 10^-2 ; K_a2 = 4.6 x 10^-5)
• B. 0.200 M H­₃PO₄ (K_a1 = 7.5 x 10^-3 ; K_a2 = 6.2 x 10^-8 ; K_a3 = 4.2 x 10^-13)
• C. 0.200 M H­₂CO₃ (K_a1 = 4.3 x 10^-7 ; K_a2 = 5.6 x 10^-11)
• D. 0.200 M H­₂S (K_a1 = 8.9 x 10^-8 ; K_a2 = 1.0 x 10^-19)
• E. 0.200 M H­C₉H₇O₄ (K_a1 = 3.3 x 10^-4)

Answer: (A)

Determine the pH of 0.115 M Na₂S. Hydrosulfuric acid, H₂S, possesses K_a1 = 1.0 × 10⁻⁷ and K_a2 = 9.1 × 10⁻⁸.

• A. 7.04
• B. 10.05
• C. 7.90
• D. 13.06

Answer: (B)

Here’s what students ask on this topic:

How can the concentration of a diprotic acid be found?

To determine the concentration of a diprotic acid, which can donate two protons (H⁺ ions), you can perform a titration using a strong base, such as NaOH. Here's a simplified process:

1. Prepare a standard solution of the strong base with known concentration.
2. Take a measured volume of the diprotic acid solution whose concentration you want to find.
3. Add a few drops of an appropriate indicator to the acid solution, which will change color at the endpoint of the titration.
4. Slowly add the base to the acid while stirring until the indicator shows that neutralization has occurred. This is the endpoint, where moles of H⁺ from the acid equal moles of OH^- from the base.
5. Record the volume of base used to reach the endpoint.

Since the acid is diprotic, remember that one mole of acid will react with two moles of base. Use the stoichiometry of the reaction to calculate the concentration of the acid:

Molarity of acid = (Molarity of base × Volume of base used × 2) / Volume of acid

This equation accounts for the diprotic nature of the acid (hence the multiplication by 2). Ensure that all volumes are measured accurately to get precise results.

How do you calculate the pH of a diprotic acid?

Calculating the pH of a diprotic acid, which can donate two protons (H⁺ ions), involves a few steps. First, you need to write down the two dissociation equations:

1. $H_{2}A\to H+{\mathrm{HA}}_{-}$
2. ${\mathrm{HA}}_{-}\to H+A_2-$

Each step has its own dissociation constant, K_a1 and K_a2. To calculate the pH:

1. Determine the concentration of the acid and the values of K_a1 and K_a2.
2. If K_a1 is much larger than K_a2 (by 10³ or more), you can initially ignore the second dissociation.
3. Calculate the concentration of H⁺ from the first dissociation using the formula: $\left[H\right]$