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Chapter 12: Problem 425

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

Expert verified
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

01

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: \[[H^+]_1 = 0.15 \mathrm{M}\]
02

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 \([H^+]_1\).
03

Solve for [H+]_2

Rearrange the equation from Step 2 and solve for \([H^+]_2\): \[[H^+]_2 = \sqrt{1.26 \times 10^{-2} \times 0.15}\] \[[H^+]_2 = \sqrt{1.89 \times 10^{-3}}\] \[[H^+]_2 = 4.35 \times 10^{-2} \mathrm{M}\]
04

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: \[[H^+]_{\text{total}} = [H^+]_1 + [H^+]_2\] \[[H^+]_{\text{total}} = 0.15 \,\text{M} + 4.35 \times 10^{-2} \,\text{M}\] \[[H^+]_{\text{total}} = 0.1935 \,\text{M}\]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid Dissociation
Acid dissociation refers to the process in which an acid releases hydrogen ions (H鈦�) 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鈦� ions into the solution.
  • For example, hydrochloric acid (HCl) dissociates completely into H鈦� and Cl鈦� ions.
  • Weak acids, like acetic acid (CH鈧僀OOH), partially dissociate, releasing fewer H鈦� 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鈦� ions. Understanding acid dissociation helps predict the behavior of acids in chemical reactions.
H2SO4 Dissociation
Sulfuric acid (H鈧係O鈧�) 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鈦� ions as the initial concentration of H鈧係O鈧�.
The second dissociation is weaker and involves the partial dissociation of the hydrogen sulfate ion (HSO鈧勨伝):
  • \[ \mathrm{HSO_4^-} \leftrightarrow \mathrm{H^+} + \mathrm{SO_4^{2-}} \]
The second step significantly contributes to the total H鈦� 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 (HNO鈧�), and sulfuric acid (H鈧係O鈧�).
  • Complete dissociation means that nearly all the acid molecules break down into H鈦� 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鈦� 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 (K鈧�) provides insight into how far a reaction will proceed before reaching equilibrium. For sulfuric acid's second dissociation, K鈧� is particularly important as it indicates the degree to which HSO鈧勨伝 dissociates into H鈦� and SO鈧劼测伝.
K鈧� is defined by the equation:
  • \[ K_2 = \frac{[\mathrm{H^+}][\mathrm{SO_4^{2-}}]}{[\mathrm{HSO_4^-}]} \]
A large K鈧� value suggests that the reaction proceeds significantly towards products, whereas a small value indicates limited dissociation. Understanding K鈧� helps chemists determine how the second dissociation of sulfuric acid affects the concentration of ions in a solution and, subsequently, its pH.

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Most popular questions from this chapter

Find the \(\mathrm{pH}\) of a \(0.2 \mathrm{M}\) solution of formic acid. \(\mathrm{K}_{\mathrm{a}}=1.76 \times 10^{-4}\)

A chemist mixes together \(100 \mathrm{ml}\) of \(0.5 \mathrm{M}\) sodium acetate, \(100 \mathrm{ml}\) of \(0.25 \mathrm{M}\) hydrochloric acid \((\mathrm{HCl})\), and \(100 \mathrm{ml}\) of a \(1.0 \mathrm{M}\) salt solution. She dilutes this to \(1000 \mathrm{ml}\). Determine the concentrations of all the ions present, undissociated acetic acid, and the final \(\mathrm{pH}\) of the solution, using \(\mathrm{K}_{\mathrm{a}}\) of acetic acid \(=1.7 \times 10^{-5}\) and \(\mathrm{K}_{\mathrm{W}}=1.0 \times 10^{-14}\).

Assuming \(\mathrm{K}_{\mathrm{W}}=1.0 \times 10^{-14}\) and \(\mathrm{K}_{\mathrm{a}}=4.0 \times 10^{-10}\), find the \(\mathrm{pH}\) of a \(1.0 \mathrm{M} \mathrm{NaCN}\) solution.

Find \(\mathrm{K}_{\mathrm{b}}\) and \(\mathrm{pK}_{\mathrm{b}}\) for the acetate ion \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), The ionization constant of \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(\mathrm{K}_{\mathrm{a}}=1.75 \times 10^{-5} ; \mathrm{K}_{\mathrm{W}}=1.00 \times 10^{-14}\)

Calculate the \(\mathrm{pH}\) of (a) a \(0.5 \mathrm{M}\) solution with respect to \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa} ;\) (b) the same solution after \(0.1\) mole \(\mathrm{HCl}\) per liter has been added to it. Assume that the volume is unchanged. \(\mathrm{K}_{\mathrm{a}}=1.75 \times 10^{-5}\).
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