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Chapter 14: Problem 111

Calculate the pH of a \(2.0-M \mathrm{H}_{2} \mathrm{SO}_{4}\) solution.

Short Answer

Expert verified
The pH of a \(2.0-M \mathrm{H}_{2} \mathrm{SO}_{4}\) solution can be calculated by first finding the concentration of H鈦� ions, which is \(4.0 M\). Then, we use the pH formula: pH = -log鈧佲個[H鈦篯 to get pH 鈮� -0.60. Though unusual, pH values can go beyond the 0-14 range, indicating a very acidic solution in this case.

Step by step solution

01

Write the dissociation equation for sulfuric acid

In water, sulfuric acid dissociates as follows: H鈧係O鈧� 鈫� 2H鈦� + SO鈧劼测伝 This shows that 1 mole of H鈧係O鈧� will produce 2 moles of H鈦� ions.
02

Find the concentration of H鈦� ions

Since we know that 1 mole of H鈧係O鈧� produces 2 moles of H鈦� ions, we can calculate the concentration of H鈦� ions: Concentration of H鈦� ions = 2 脳 concentration of H鈧係O鈧� = 2 脳 2.0 M = 4.0 M
03

Calculate the pH

The pH is calculated using the formula: pH = -log鈧佲個[H鈦篯 Where [H鈦篯 is the concentration of H鈦� ions. Using the concentration from Step 2: pH = -log鈧佲個(4.0)
04

Evaluate the expression and find the pH

Now, we evaluate the expression to find the pH: pH = -log鈧佲個(4.0) 鈮� -0.60 However, note that the pH scale is typically considered to range from 0 to 14, so reporting a negative pH value may seem unusual. In this case, it indicates that the concentration of H鈦� is very high, and the solution is very acidic. It is vital to understand that the pH can go beyond the range of 0 to 14, even though such cases are uncommon.

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Key Concepts

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

Sulfuric Acid Dissociation
Sulfuric acid (\( \mathrm{H}_2\mathrm{SO}_4 \)) is a strong acid that dissociates completely in water, releasing hydrogen ions (\( \mathrm{H^+} \)) and sulfate ions (\( \mathrm{SO_4^{2-}} \)). The dissociation reaction can be represented as:
  • \[ \mathrm{H}_2\mathrm{SO}_4 \rightarrow 2 \mathrm{H^+} + \mathrm{SO_4^{2-}} \]
This equation tells us that one mole of sulfuric acid yields two moles of \( \mathrm{H^+} \) ions. It is crucial to grasp this stoichiometry because it directly impacts calculations concerning acidity and pH. Knowing how many \( \mathrm{H^+} \) ions are produced allows us to determine the extent of acidity. Therefore, understanding this dissociation process is the first step in calculating the pH of a \(2.0-M\) solution of sulfuric acid.
Concentration of H鈦� Ions
Determining the concentration of hydrogen ions (\( \mathrm{H^+} \)) is essential for assessing the acidity of a solution. From sulfuric acid's dissociation, we know it produces two moles of \( \mathrm{H^+} \) for every mole of \( \mathrm{H}_2\mathrm{SO}_4 \). Therefore, to find the \( \mathrm{H^+} \) concentration:
  • Multiply the concentration of \( \mathrm{H}_2\mathrm{SO}_4 \) by 2.
  • In this exercise where \(2.0-M \mathrm{H}_2\mathrm{SO}_4\) was used, the calculation is:\[ 2 \text{ (moles } \mathrm{H^+}/\text{mole } \mathrm{H}_2\mathrm{SO}_4) \times 2.0 \text{ M} = 4.0 \text{ M} \]
So, there are \(4.0 \text{ M}\) of \( \mathrm{H^+} \) ions in the solution. This high concentration reflects the strong acidic nature of sulfuric acid when dissolved in water.
Acidic Solution
An acidic solution is one that releases hydrogen ions (\( \mathrm{H^+} \)) when dissolved in water, resulting in a low pH. Acids like sulfuric acid are classified based on their ability to release \( \mathrm{H^+} \) ions freely. Given the high concentration of \( \mathrm{H^+} \) ions in such solutions;
  • More \( \mathrm{H^+} \) ions lead to lower pH values.
  • This creates a more acidic environment.
The pH scale typically ranges from 0 (most acidic) to 14 (most basic), with 7 being neutral. In this context, the solution of sulfuric acid, being very acidic, significantly lowers the pH value. The understanding of an acidic solution is vital in chemistry, as it influences reaction conditions and the behavior of substances in the solution.
Negative pH Value
The occurrence of a negative pH might appear strange because we are accustomed to a pH scale between 0 and 14. However, for exceptionally acidic solutions with very high concentrations of hydrogen ions (\( \mathrm{H^+} \)), the pH can transcend the expected scale. Calculating pH involves the formula:
  • \[ \text{pH} = -\log_{10}[\mathrm{H^+}] \]
For the \(4.0 \text{ M}\) concentration of \( \mathrm{H^+} \) in this exercise, solving results in a pH of approximately -0.60. Such a negative value indicates an intensely acidic solution, showcasing the sheer volume of hydrogen ions present. Although rare in everyday scenarios, negative pH values can occur in concentrated strong acids, illustrating the solution's potent acidity.

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Most popular questions from this chapter

One mole of a weak acid HA was dissolved in 2.0 \(\mathrm{L}\) of solution. After the system had come to equilibrium, the concentration of HA was found to be 0.45 \(\mathrm{M} .\) Calculate \(K_{\mathrm{a}}\) for \(\mathrm{HA}\) .

Calculate the concentrations of all species present in a \(0.25-M\) solution of ethylammonium chloride \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\right)\)

How many significant figures are there in the following num- bers: \(10.78,6.78,0.78 ?\) If these were pH values, to how many significant figures can you express the \(\left[\mathrm{H}^{+}\right] ?\) Explain any discrepancies between your answers to the two questions.

a. The principal equilibrium in a solution of \(\mathrm{NaHCO}_{3}\) is $$ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ Calculate the value of the equilibrium constant for this reaction. b. At equilibrium, what is the relationship between \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\) and \(\left[\mathrm{CO}_{3}^{2-}\right] ?\) c. Using the equilibrium $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ derive an expression for the pH of the solution in terms of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) using the result from part b. d. What is the pH of a solution of \(\mathrm{NaHCO}_{3} ?\)

When determining the pH of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solutions, sometimes the \(\mathrm{H}^{+}\) contribution from \(\mathrm{HSO}_{4}^{-}\) can be ignored by the 5\(\%\) rule. At what concentrations of an \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution can the \(\mathrm{H}^{+}\) contribution from \(\mathrm{HSO}_{4}^{-}\) be ignored when determining the pH of the solution?
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