Let's work through the question step-by-step:
1. Understanding Self-Ionization of Water:
Water (H₂O) can undergo self-ionization, which means that a small fraction of water molecules dissociate into hydrogen ions (H⁺) and hydroxide ions (OH⁻). This process is represented by the following chemical equation:
[tex]\[ 2H₂O (l) ⇌ H₃O⁺ (aq) + OH⁻ (aq) \][/tex]
However, for simplicity, we often write it as:
[tex]\[ H₂O (l) ⇌ H⁺ (aq) + OH⁻ (aq) \][/tex]
2. Concentration of H⁺ and OH⁻ ions in pure water:
In pure water at 25°C, the concentration of hydrogen ions [tex]\([H⁺]\)[/tex] and hydroxide ions [tex]\([OH⁻]\)[/tex] is [tex]\(1 \times 10^{-7}\)[/tex] moles per liter. This means that in one liter of pure water, there are [tex]\(1 \times 10^{-7}\)[/tex] moles of H⁺ ions and [tex]\(1 \times 10^{-7}\)[/tex] moles of OH⁻ ions.
3. Finding the Number of Moles of Self-Ionized Water Molecules:
Since each self-ionization event of a water molecule produces one H⁺ ion and one OH⁻ ion, the number of moles of self-ionized H₂O molecules is equal to the concentration of either H⁺ or OH⁻ ions. Therefore, in one liter of pure water, the number of moles of water molecules that self-ionize is also [tex]\(1 \times 10^{-7}\)[/tex] moles (or 0.0000001 moles).
4. Conclusion:
From the question, the concentration of self-ionized water molecules in one liter of water is indeed [tex]\(1 \times 10^{-7}\)[/tex] moles.
Therefore, the answer is:
c) 0.0000001 moles
