QuestionMay 23, 2025

Calculate the [H_(3)O^+] of an aqueous solution if the [OH^-]=8.57times 10^-4M Answer: square

Calculate the [H_(3)O^+] of an aqueous solution if the [OH^-]=8.57times 10^-4M Answer: square
Calculate the [H_(3)O^+] of an aqueous solution if the [OH^-]=8.57times 10^-4M Answer: square

Solution
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Answer

[H_{3}O^{+}] = 1.17 \times 10^{-11} M Explanation 1. Use the water dissociation constant The product of [H_{3}O^{+}] and [OH^{-}] is equal to the ion product of water, **K_w = 1.0 \times 10^{-14}** at 25°C. 2. Calculate [H_{3}O^{+}] Use the formula [H_{3}O^{+}] = \frac{K_w} OH^{-} . Substitute K_w = 1.0 \times 10^{-14} and [OH^{-}] = 8.57 \times 10^{-4}M. \[ [H_{3}O^{+}] = \frac{1.0 \times 10^{-14}}{8.57 \times 10^{-4}} \]

Explanation

1. Use the water dissociation constant<br /> The product of $[H_{3}O^{+}]$ and $[OH^{-}]$ is equal to the ion product of water, **$K_w = 1.0 \times 10^{-14}$** at 25°C.<br />2. Calculate $[H_{3}O^{+}]$<br /> Use the formula $[H_{3}O^{+}] = \frac{K_w}{[OH^{-}]}$. Substitute $K_w = 1.0 \times 10^{-14}$ and $[OH^{-}] = 8.57 \times 10^{-4}M$.<br /><br />\[ [H_{3}O^{+}] = \frac{1.0 \times 10^{-14}}{8.57 \times 10^{-4}} \]
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