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Chapter 18: Problem 23

Challenge Calculate the number of \(\mathrm{H}^{+}\) ions and the number of \(\mathrm{OH}^{-}\) ions in 300 \(\mathrm{mL}\) of pure water at 298 \(\mathrm{K} .\)

Short Answer

Expert verified
Both \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) ions in 300 mL of pure water are \( 1.81 \times 10^{16} \) ions each.

Step by step solution

01

Determine the Concentrations

Pure water at 298 K has a self-ionization constant given by \[ \mathrm{K}_w = [\mathrm{H}^+] [\mathrm{OH}^-] = 1.0 \times 10^{-14} \].In pure water, the concentrations of \( \mathrm{H}^+ \) ions and \( \mathrm{OH}^- \) ions are equal, \([\mathrm{H}^+] = [\mathrm{OH}^-]\). Hence, \([\mathrm{H}^+]^2 = 1.0 \times 10^{-14}\). We find the concentration of \( \mathrm{H}^+ \) as follows:\[ [\mathrm{H}^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \mathrm{M} \]
02

Calculate the Number of Moles

Now that we have the concentration of \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) both equaling \( 1.0 \times 10^{-7} \mathrm{M} \), we need to find the number of moles in 300 mL of water. Recall that \[ 1 \text{ mole/liter} = 1 \text{ M} \]. Therefore, the number of moles of \( \mathrm{H}^+ \) ions in 300 mL = \[ 1.0 \times 10^{-7} \frac{\text{moles}}{\text{L}} \times 0.300 \text{ L} = 3.0 \times 10^{-8} \text{ moles}.\]
03

Convert Moles to Number of Ions

To find the number of ions, we use Avogadro鈥檚 number, \(6.022 \times 10^{23} \) ions/mole. Thus, the number of \( \mathrm{H}^+ \) ions is: \[ 3.0 \times 10^{-8} \text{ moles} \times 6.022 \times 10^{23} \left( \text{ions/mole} \right) = 1.81 \times 10^{16} \text{ ions}. \]Since the concentrations and volumes are the same for \( \mathrm{OH}^- \), the number of \( \mathrm{OH}^- \) ions is also:\[ 1.81 \times 10^{16} \text{ ions}. \]

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Key Concepts

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

Self-ionization constant
In pure water, a small number of water molecules undergo a process known as self-ionization. This process produces hydrogen ions \( \mathrm{H}^{+} \) and hydroxide ions \( \mathrm{OH}^{-} \). The equilibrium constant for this process is called the self-ionization constant, represented by \( K_w \). At 298 K, this constant is given by \( K_w = [\mathrm{H}^+] [\mathrm{OH}^-] = 1.0 \times 10^{-14} \).
  • The constant indicates the relationship between the concentrations of \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) ions in the water, showing that they are equal in pure water.
  • When presented in the formula \( [\mathrm{H}^+] = [\mathrm{OH}^-] \), it simplifies to \( [\mathrm{H}^+]^2 = 1.0 \times 10^{-14} \).
  • Solving for the concentration of \( \mathrm{H}^+ \) results in \( [\mathrm{H}^+] = 1.0 \times 10^{-7} \text{ M} \), equivalent to that of \( \mathrm{OH}^- \).
Understanding this equilibrium constant is essential because it illustrates how even in neutral water, there are still ions present, resulting from the natural ionization of water.
Avogadro's number
Avogadro's number is a fundamental constant used in chemistry to relate moles of a substance to the number of entities (atoms, ions, molecules, etc.) it contains. It is denoted as \( 6.022 \times 10^{23} \). This number indicates how many particles are in one mole of any substance.
  • When calculating the number of ions in a solution, we make use of Avogadro's number to convert moles to actual quantities of particles.
  • Therefore, knowing the number of moles allows us to determine the number of individual ions by simply multiplying by Avogadro鈥檚 number.
  • This conversion is crucial in chemical calculations when determining how many discrete ions or molecules are present in a given sample.
Practically speaking, Avogadro's number provides a bridge between the macroscopic and microscopic worlds, allowing us to quantify substances in a tangible and meaningful way.
Moles to ions conversion
The conversion from moles to ions is a standard calculation in chemistry that enables us to understand the scale of chemical reactions. Once you have calculated the moles of ions in a solution, the next step is converting that into the actual number of ions using Avogadro's number.
  • For instance, going back to our example, when you have \( 3.0 \times 10^{-8} \) moles of \( \mathrm{H}^+ \), you can calculate the number of ions by \( 3.0 \times 10^{-8} \times 6.022 \times 10^{23} \), resulting in \( 1.81 \times 10^{16} \) ions.
  • This conversion shows us how many individual \( \mathrm{H}^+ \) or \( \mathrm{OH}^- \) ions are present in a sample rather than just a concentration.
  • It's a critical skill in chemistry to visualize and understand the molecular scale of the substances we work with.
Understanding this process can significantly enhance your grasp of chemical concentrations and reactions, enabling more accurate assessments and predictions in lab scenarios and examinations.

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Most popular questions from this chapter

Apply Concepts Use the ion product constant of water at 298 \(\mathrm{K}\) to explain why a solution with a pH of 3.0 must have a poH of \(11.0 .\)

Calculate the molarity of a solution of hydrobromic acid (HBr) if 30.35 \(\mathrm{mL}\) of 0.1000 \(\mathrm{M} \mathrm{NaOH}\) is required to titrate 25.00 \(\mathrm{mL}\) of the acid to the equivalence point.

What is the \(\mathrm{pH}\) of a 0.200\(M\) solution of hypobromous acid \((\mathrm{HBrO}) ? K_{\mathrm{a}}=2.8 \times 10^{-9}\)

Explain why many compounds that contain one or more hydrogen atoms are not classified as Arrhenius acids.

Calculate the \(\mathrm{pH}\) of aqueous solutions with the following \([\mathrm{H}+]\) at 298 \(\mathrm{K}\). \begin{equation} \text { a. }\left[\mathrm{H}^{+}\right]=0.0055 M \quad \text { b. }\left[\mathrm{H}^{+}\right]=0.000084 \mathrm{M} \end{equation}
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