What is the hydronium-ion concentration of a 0.0038 M Ba(OH)2 solution? 1.3 x 10-12 M O 7.6 x 10-3 M O 1.0 x 10-7 M O 3.8 x 10-3 M O 2.6 x 10-12 M
What is the hydronium-ion concentration of a 0.0038 M Ba(OH)2 solution? 1.3 x 10-12 M O 7.6 x 10-3 M O 1.0 x 10-7 M O 3.8 x 10-3 M O 2.6 x 10-12 M
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter13: Acids And Bases
Section: Chapter Questions
Problem 60QAP: Consider a 0.33 M solution of the diprotic acid H2X. H2X H+(aq)+ HX(aq)Ka1=3.3 10 4 HX H+(aq)+...
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![### Question:
What is the hydronium-ion concentration of a 0.0038 M Ba(OH)₂ solution?
### Options:
- A) 1.3 × 10⁻¹² M
- B) 7.6 × 10⁻³ M
- C) 1.0 × 10⁻⁷ M
- D) 3.8 × 10⁻³ M
- E) 2.6 × 10⁻¹² M
### Explanation:
In this problem, we need to determine the concentration of hydronium ions (H₃O⁺) in a given solution of barium hydroxide, Ba(OH)₂. Remember, Ba(OH)₂ is a strong base and completely dissociates in an aqueous solution as follows:
\[ \text{Ba(OH)₂} \rightarrow \text{Ba}^{2+} + 2\text{OH}^- \]
For a 0.0038 M Ba(OH)₂ solution:
1. The concentration of OH⁻ ions will be twice the concentration of Ba(OH)₂ because one Ba(OH)₂ produces two OH⁻ ions upon dissociation.
\[ \text{[OH]⁻} = 2 \times 0.0038 \, M = 0.0076 \, M \]
Using the water dissociation constant:
\[ K_w = [H₃O⁺][OH]⁻ = 1.0 \times 10^{-14} \, M^{2}\]
We can find the concentration of H₃O⁺:
\[ [H₃O⁺] = \frac{1.0 \times 10^{-14}}{0.0076} \approx 1.3 \times 10^{-12} \, M \]
Therefore, the correct answer is:
- **A) 1.3 × 10⁻¹² M**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb99220ad-d702-4943-b904-ec3abe399c73%2Fa270c13a-d129-4e89-84cb-947bc68ae1ce%2F2x8hbu_processed.png&w=3840&q=75)
Transcribed Image Text:### Question:
What is the hydronium-ion concentration of a 0.0038 M Ba(OH)₂ solution?
### Options:
- A) 1.3 × 10⁻¹² M
- B) 7.6 × 10⁻³ M
- C) 1.0 × 10⁻⁷ M
- D) 3.8 × 10⁻³ M
- E) 2.6 × 10⁻¹² M
### Explanation:
In this problem, we need to determine the concentration of hydronium ions (H₃O⁺) in a given solution of barium hydroxide, Ba(OH)₂. Remember, Ba(OH)₂ is a strong base and completely dissociates in an aqueous solution as follows:
\[ \text{Ba(OH)₂} \rightarrow \text{Ba}^{2+} + 2\text{OH}^- \]
For a 0.0038 M Ba(OH)₂ solution:
1. The concentration of OH⁻ ions will be twice the concentration of Ba(OH)₂ because one Ba(OH)₂ produces two OH⁻ ions upon dissociation.
\[ \text{[OH]⁻} = 2 \times 0.0038 \, M = 0.0076 \, M \]
Using the water dissociation constant:
\[ K_w = [H₃O⁺][OH]⁻ = 1.0 \times 10^{-14} \, M^{2}\]
We can find the concentration of H₃O⁺:
\[ [H₃O⁺] = \frac{1.0 \times 10^{-14}}{0.0076} \approx 1.3 \times 10^{-12} \, M \]
Therefore, the correct answer is:
- **A) 1.3 × 10⁻¹² M**
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