Question

Find the \(\mathrm{pH}\) of \(0.15 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), solution, assuming \(\mathrm{K}_{2}=1.26 \times 10^{-2}\)

Short answer

The pH of a 0.15 M H2SO4 solution, assuming \(K_2 = 1.26 \times 10^{-2}\), can be calculated as follows: First, determine the concentration of H+ ions released by the first dissociation (\([H^+]_1 = 0.15 \,\text{M}\)). Next, use K2 to find the concentration of H+ ions in the second dissociation (\([H^+]_2 = 4.35 \times 10^{-2} \,\text{M}\)). Then, add the H+ ions concentrations to get the total concentration (\([H^+]_{\text{total}} = 0.1935 \,\text{M}\)). Lastly, calculate the pH using the total concentration of H+ ions (\(\text{pH} \approx 0.71\)).

Step-by-step solution

Determine the concentration of H+ ions released by the first dissociation of H2SO4

As H2SO4 is a strong acid, its first dissociation is complete and it donates only one proton to the solution: \[H_2SO_4 \rightarrow H^+ + HSO_4^-\] Since the concentration of H2SO4 is 0.15 M, it will produce an equal concentration of H+ ions: \[[H^+]_1 = 0.15 \mathrm{M}\]

Use K2 to find the concentration of H+ ions released by the second dissociation of H2SO4

The second dissociation of H2SO4 is an equilibrium reaction which occurs to a lesser extent and is represented by: \[HSO_4^- \longleftrightarrow H^+ + SO_4^{2-}\] The equilibrium constant, K2, is given as \(1.26 \times 10^{-2}\). Since \(K_2 = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}\), we can use the given K2 value to find the concentration of H+ ions in the second dissociation: \[1.26 \times 10^{-2} = \frac{[H^+]_2[H^+]_2}{[HSO_4^-]} = \frac{([H^+]_2)^2}{0.15}\] We need to solve for \([H^+]_2\). Realize that \([H^+]_2\) is much lower than the first \([H^+]_1\).

Solve for [H+]_2

Rearrange the equation from Step 2 and solve for \([H^+]_2\): \[[H^+]_2 = \sqrt{1.26 \times 10^{-2} \times 0.15}\] \[[H^+]_2 = \sqrt{1.89 \times 10^{-3}}\] \[[H^+]_2 = 4.35 \times 10^{-2} \mathrm{M}\]

Calculate the total concentration of H+ ions

To find the total concentration of H+ ions, add the H+ ions concentrations from Steps 1 and 3: \[[H^+]_{\text{total}} = [H^+]_1 + [H^+]_2\] \[[H^+]_{\text{total}} = 0.15 \,\text{M} + 4.35 \times 10^{-2} \,\text{M}\] \[[H^+]_{\text{total}} = 0.1935 \,\text{M}\]

Calculate the pH of the solution

Now that we have the total concentration of H+ ions in the solution, calculate the pH using the formula: \[\text{pH} = -\log_{10}([H^+])\] \[\text{pH} = -\log_{10}(0.1935)\] \[\text{pH} \approx 0.71\] The pH of the 0.15 M H2SO4 solution is approximately 0.71.

Key concepts

Acid Dissociation

Acid dissociation refers to the process in which an acid releases hydrogen ions (H⁺) when dissolved in water. This is a fundamental concept in understanding how acids behave in solutions. The degree to which an acid dissociates is expressed in terms of its strength. Strong acids tend to dissociate completely, releasing a high concentration of H⁺ ions into the solution.

• For example, hydrochloric acid (HCl) dissociates completely into H⁺ and Cl⁻ ions.
• Weak acids, like acetic acid (CH₃COOH), partially dissociate, releasing fewer H⁺ ions.

This process of dissociation is essential to calculate the pH of a solution, as pH is a direct measure of the concentration of H⁺ ions. Understanding acid dissociation helps predict the behavior of acids in chemical reactions.

H2SO4 Dissociation

Sulfuric acid (H₂SO₄) is a unique acid because it dissociates in two distinct steps. This two-step process is crucial for understanding reactions involving sulfuric acid.
In the first dissociation, sulfuric acid behaves like a strong acid, meaning it dissociates completely:

• \[ \mathrm{H_2SO_4} \rightarrow \mathrm{H^+} + \mathrm{HSO_4^-} \]

This first step results in the same concentration of H⁺ ions as the initial concentration of H₂SO₄.
The second dissociation is weaker and involves the partial dissociation of the hydrogen sulfate ion (HSO₄⁻):

• \[ \mathrm{HSO_4^-} \leftrightarrow \mathrm{H^+} + \mathrm{SO_4^{2-}} \]

The second step significantly contributes to the total H⁺ concentration only when looking at more dilute solutions.

Strong Acids

Strong acids are characterized by their ability to disassociate completely in water. This characteristic is vital because it determines how they will contribute to the acidity of a solution. Common examples of strong acids include hydrochloric acid (HCl), nitric acid (HNO₃), and sulfuric acid (H₂SO₄).

• Complete dissociation means that nearly all the acid molecules break down into H⁺ ions.
• This results in a very low pH value (less than 1) for concentrated solutions.

Understanding strong acids is crucial for calculating pH since it simplifies the task: you can assume that the concentration of H⁺ ions after dissociation is equal to the initial concentration of the acid. This is crucial in predicting the impact of strong acids in chemical processes and calculating pH accurately.

Equilibrium Constant (K2)

In chemical reactions, the equilibrium constant (K₂) provides insight into how far a reaction will proceed before reaching equilibrium. For sulfuric acid's second dissociation, K₂ is particularly important as it indicates the degree to which HSO₄⁻ dissociates into H⁺ and SO₄²⁻.
K₂ is defined by the equation:

• \[ K_2 = \frac{[\mathrm{H^+}][\mathrm{SO_4^{2-}}]}{[\mathrm{HSO_4^-}]} \]

A large K₂ value suggests that the reaction proceeds significantly towards products, whereas a small value indicates limited dissociation. Understanding K₂ helps chemists determine how the second dissociation of sulfuric acid affects the concentration of ions in a solution and, subsequently, its pH.