QuestionAugust 12, 2025

What is the pH of a solution that is 0.40 M HClO and 1.0 M HCOOH? K_(a(HClO))=3.5times 10^-8 K_(a(HCOOH))=1.77times 10^-4 pH=[?] Be careful to note which acid dissociates more. You may assume you can neglect change in X.

What is the pH of a solution that is 0.40 M HClO and 1.0 M HCOOH? K_(a(HClO))=3.5times 10^-8 K_(a(HCOOH))=1.77times 10^-4 pH=[?] Be careful to note which acid dissociates more. You may assume you can neglect change in X.
What is the pH of a solution that is 0.40 M HClO and 1.0 M HCOOH? K_(a(HClO))=3.5times 10^-8 K_(a(HCOOH))=1.77times 10^-4 pH=[?] Be careful to note which acid dissociates more. You may assume you can neglect change in X.

Solution
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Answer

pH \approx 2.37 Explanation 1. Identify the Dominant Acid Compare K_a values. K_{a(HCOOH)} = 1.77 \times 10^{-4} is greater than K_{a(HClO)} = 3.5 \times 10^{-8}. Thus, HCOOH dissociates more. 2. Calculate [H⁺] from HCOOH Use the formula for weak acid dissociation: **[H^+] = \sqrt{K_a \cdot C}** where C is the concentration of the acid. For HCOOH: [H^+] = \sqrt{1.77 \times 10^{-4} \times 1.0} = \sqrt{1.77 \times 10^{-4}}. 3. Calculate pH **pH = -\log[H^+]**. Substitute [H^+] = \sqrt{1.77 \times 10^{-4}} into the formula to find pH.

Explanation

1. Identify the Dominant Acid<br /> Compare $K_a$ values. $K_{a(HCOOH)} = 1.77 \times 10^{-4}$ is greater than $K_{a(HClO)} = 3.5 \times 10^{-8}$. Thus, HCOOH dissociates more.<br /><br />2. Calculate [H⁺] from HCOOH<br /> Use the formula for weak acid dissociation: **$[H^+] = \sqrt{K_a \cdot C}$** where $C$ is the concentration of the acid.<br /> For HCOOH: $[H^+] = \sqrt{1.77 \times 10^{-4} \times 1.0} = \sqrt{1.77 \times 10^{-4}}$.<br /><br />3. Calculate pH<br /> **$pH = -\log[H^+]$**.<br /> Substitute $[H^+] = \sqrt{1.77 \times 10^{-4}}$ into the formula to find pH.
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