Question
Which option correctly shows the consideration of H3O ions in pure water
Answers
Answered by
GPT-4o mini
In pure water, the concentration of hydronium ions (H3O⁺) is equal to the concentration of hydroxide ions (OH⁻). This is due to the self-ionization of water, which can be represented by the following equilibrium reaction:
\[
2 \, \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-
\]
At 25°C, the concentrations of H3O⁺ and OH⁻ in pure water are both \(1.0 \times 10^{-7}\) M. Therefore, in a neutral solution like pure water, the consideration of H3O ions is typically expressed as:
- \([H_3O^+] = [OH^-] = 1.0 \times 10^{-7} \, \text{M}\)
This implies that the pH of pure water is:
\[
\text{pH} = -\log[H_3O^+] = -\log(1.0 \times 10^{-7}) = 7
\]
This indicates that pure water is neutral, with equal concentrations of H3O⁺ and OH⁻ ions.
\[
2 \, \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-
\]
At 25°C, the concentrations of H3O⁺ and OH⁻ in pure water are both \(1.0 \times 10^{-7}\) M. Therefore, in a neutral solution like pure water, the consideration of H3O ions is typically expressed as:
- \([H_3O^+] = [OH^-] = 1.0 \times 10^{-7} \, \text{M}\)
This implies that the pH of pure water is:
\[
\text{pH} = -\log[H_3O^+] = -\log(1.0 \times 10^{-7}) = 7
\]
This indicates that pure water is neutral, with equal concentrations of H3O⁺ and OH⁻ ions.
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