Problem 23
Question
Challenge Calculate the number of \(\mathrm{H}^{+}\) ions and the number of \(\mathrm{OH}^{-}\) ions in 300 \(\mathrm{mL}\) of pure water at 298 \(\mathrm{K} .\)
Step-by-Step Solution
Verified Answer
Both \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) ions in 300 mL of pure water are \( 1.81 \times 10^{16} \) ions each.
1Step 1: 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:\[ [\mathrm{H}^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \mathrm{M} \]
2Step 2: 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 = \[ 1.0 \times 10^{-7} \frac{\text{moles}}{\text{L}} \times 0.300 \text{ L} = 3.0 \times 10^{-8} \text{ moles}.\]
3Step 3: Convert Moles to Number of Ions
To find the number of ions, we use Avogadro’s 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:\[ 1.81 \times 10^{16} \text{ ions}. \]
Key Concepts
Self-ionization constantAvogadro's numberMoles to ions conversion
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 \( \mathrm{OH}^- \).
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’s number.
- This conversion is crucial in chemical calculations when determining how many discrete ions or molecules are present in a given sample.
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.
Other exercises in this chapter
Problem 20
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