QuestionMay 24, 2025

ncy General Chemistry In any volume of water or the [H_(3)O^+]=square M M and the [OH^-]=square M. Check

ncy General Chemistry In any volume of water or the [H_(3)O^+]=square M M and the [OH^-]=square M. Check
ncy General Chemistry In any volume of water or the [H_(3)O^+]=square M M and the [OH^-]=square M. Check

Solution
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Answer

[H_3O^+] = 1.0 \times 10^{-7} \text{ M}, [OH^-] = 1.0 \times 10^{-7} \text{ M} Explanation 1. Identify the relationship between [H_3O^+] and [OH^-] In pure water at 25°C, the product of the concentrations of hydronium ions and hydroxide ions is constant: **[H_3O^+][OH^-] = 1.0 \times 10^{-14}**. 2. Calculate [H_3O^+] and [OH^-] For neutral water, [H_3O^+] = [OH^-]. Therefore, [H_3O^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \text{ M}.

Explanation

1. Identify the relationship between $[H_3O^+]$ and $[OH^-]$<br /> In pure water at 25°C, the product of the concentrations of hydronium ions and hydroxide ions is constant: **$[H_3O^+][OH^-] = 1.0 \times 10^{-14}$**.<br /><br />2. Calculate $[H_3O^+]$ and $[OH^-]$<br /> For neutral water, $[H_3O^+] = [OH^-]$. Therefore, $[H_3O^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \text{ M}$.
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