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Chapter 4: Problem 56

What is the hydronium ion concentration of a \(1.2 \times 10^{-4} \mathrm{M}\) solution of \(\mathrm{HClO}_{4} ?\) What is its \(\mathrm{pH}\) ?

Short Answer

Expert verified
The hydronium ion concentration is \\(1.2 \times 10^{-4} \, \mathrm{M}\\) and the \\(\mathrm{pH}\\) is approximately 3.92.

Step by step solution

01

Understand the acid's dissociation

Perchloric acid \(\mathrm{HClO}_4\) is a strong acid. This means it completely dissociates in water. The equation for its dissociation is: \[\mathrm{HClO}_{4(aq)} \rightarrow \mathrm{H}^{+}_{(aq)} + \mathrm{ClO}_{4}^{-}_{(aq)}\] Since it dissociates completely, the concentration of hydronium ions \(\mathrm{H}^{+}\) will be equal to the initial concentration of the acid.
02

Calculate the hydronium ion concentration

The concentration of hydronium ions \(\left[\mathrm{H}^{+}\right]\) is equal to the concentration of the \(\mathrm{HClO}_4\) solution since it dissociates completely. Thus, \[\left[\mathrm{H}^{+}\right] = 1.2 \times 10^{-4} \, \mathrm{M}\]
03

Use the hydronium concentration to find the pH

The \(\mathrm{pH}\) of a solution is calculated using the negative logarithm (base 10) of the hydronium ion concentration: \[\mathrm{pH} = -\log\left([\mathrm{H}^{+}]\right)\] Substituting in the given value: \[\mathrm{pH} = -\log\left(1.2 \times 10^{-4}\right)\] Using a calculator, this gives a \(\mathrm{pH}\) of approximately 3.92.

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Key Concepts

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

Perchloric Acid
Perchloric acid, commonly represented by the formula \(\mathrm{HClO}_4\), is known for being a strong acid. This means that when it is added to water, it dissociates completely into its constituent ions.
This dissociation can be expressed with the chemical equation:
  • \(\mathrm{HClO}_{4(aq)} \rightarrow \mathrm{H}^{+}_{(aq)} + \mathrm{ClO}_{4}^{-}_{(aq)}\)
Complete dissociation implies that nearly all the \(\mathrm{HClO}_4\) molecules split apart to form hydronium ions and perchlorate ions.
Because of this characteristic, we can easily determine the concentration of hydronium ions in a solution if we know the initial concentration of the acid.
Acid Dissociation
Acid dissociation refers to the process where an acid breaks down in water to form its constituent ions. In strong acids like perchloric acid, this process is complete.
For weak acids, this is not the case, as they only partially dissociate, leading to an equilibrium between the un-dissociated molecules and the ions.
With strong acids, since dissociation is complete, the concentration of hydrogen ions \([\mathrm{H}^+]\) in the solution is equal to the initial concentration of the acid introduced.
  • For example, if you started with a \( 1.2 \times 10^{-4} \mathrm{M} \) solution of \( \mathrm{HClO}_4\), the \([\mathrm{H}^+]\) after dissociation is also \( 1.2 \times 10^{-4} \mathrm{M} \).
This straightforward ratio is a defining feature of strong acids, facilitating easy calculations for chemistry students.
pH Calculation
The pH of a solution is a measure of its acidity or alkalinity. To find the pH of a solution, you use the formula:
  • \(\mathrm{pH} = -\log([\mathrm{H}^+])\)
This formula indicates that pH is the negative logarithm (base 10) of the hydronium ion concentration.
Hence, a lower pH value signifies a higher concentration of \([\mathrm{H}^+]\) ions, which means greater acidity. For example, with a \( 1.2 \times 10^{-4} \mathrm{M} \) \( \mathrm{HClO}_4\) solution, the pH is calculated using the given concentration:
  • \(\mathrm{pH} = -\log(1.2 \times 10^{-4})\)
Using a calculator, this results in a pH of approximately 3.92.
This demonstrates that the solution is acidic, as expected from a strong acid. Understanding this calculation is crucial for students as pH is a fundamental concept in chemistry.

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Most popular questions from this chapter

You can dissolve an aluminum soft drink can in an aqueous base such as potassium hydroxide. \(2 \mathrm{Al}(\mathrm{s})+2 \mathrm{KOH}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ 2 \mathrm{KAl}(\mathrm{OH})_{4}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g}) $$ If you place \(2.05 \mathrm{g}\) of aluminum in a beaker with \(185 \mathrm{mL}\) of \(1.35 \mathrm{M} \mathrm{KOH},\) will any aluminum remain? What mass of \(\mathrm{KAl}(\mathrm{OH})_{4}\) is produced?

If you dilute \(25.0 \mathrm{mL}\) of \(1.50 \mathrm{M}\) hydrochloric acid to \(500 . \mathrm{mL},\) what is the molar concentration of the dilute acid?

What volume of \(2.06 \mathrm{M} \mathrm{KMnO}_{4},\) in liters, contains \(322 \mathrm{g}\) of solute?

To analyze an iron-containing compound, you convert all the iron to \(\mathrm{Fe}^{2+}\) in aqueous solution and then titrate the solution with standardized \(\mathrm{KMnO}_{4} .\) The balanced, net ionic equation is \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+5 \mathrm{Fe}^{2+}(\mathrm{aq})+8 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow\) $$ \mathrm{Mn}^{2+}(\mathrm{aq})+5 \mathrm{Fe}^{3+}(\mathrm{aq})+12 \mathrm{H}_{2} \mathrm{O}(\ell) $$ A \(0.598-\mathrm{g}\) sample of the iron-containing compound requires \(22.25 \mathrm{mL}\) of \(0.0123 \mathrm{M} \mathrm{KMnO}_{4}\) for titration to the equivalence point. What is the mass percent of iron in the sample?

Awo students titrate different samples of the same solution of HCl using \(0.100 \mathrm{M} \mathrm{NaOH}\) solution and phenolphthalein indicator (Figure 4.14). The first student pipets \(20.0 \mathrm{mL}\) of the HCl solution into a flask, adds \(20 \mathrm{mL}\) of distilled water and a few drops of phenolphthalein solution, and titrates until a lasting pink color appears. The second student pipets \(20.0 \mathrm{mL}\) of the HCl solution into a flask, adds \(60 \mathrm{mL}\) of distilled water and a few drops of phenolphthalein solution, and titrates to the first lasting pink color. Each student correctly calculates the molarity of an HCl solution. What will the second student's result be? (a) four times less than the first student's result (b) four times greater than the first student's result (c) two times less than the first student's result (d) two times greater than the first student's result (e) the same as the first student's result
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