An aqueous solution of barium hydroxide is standardized by titration with a 0.179 M solution of hydrochloric acid. If 28.4 mL of base are required to neutralize 24.5 mL of the acid, what is the molarity of the barium hydroxide solution? M barium hydroxide Submit Answer Retry Entire Group 9 more group attempts remaining
An aqueous solution of barium hydroxide is standardized by titration with a 0.179 M solution of hydrochloric acid. If 28.4 mL of base are required to neutralize 24.5 mL of the acid, what is the molarity of the barium hydroxide solution? M barium hydroxide Submit Answer Retry Entire Group 9 more group attempts remaining
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Titration Problem: Determining the Molarity of Barium Hydroxide Solution**
An aqueous solution of **barium hydroxide** is standardized by titration with a **0.179 M solution of hydrochloric acid**.
**Question:**
If **28.4 mL of base** are required to neutralize **24.5 mL of the acid**, what is the molarity of the **barium hydroxide** solution?
[Input field for molarity]
**Buttons:**
- **Submit Answer**
- **Retry Entire Group**
(Note: You have **9 more group attempts remaining**)
---
### Explanation:
To solve this titration problem, follow these steps:
1. **Write the balanced chemical equation** for the reaction between barium hydroxide (Ba(OH)₂) and hydrochloric acid (HCl):
\[
\text{Ba(OH)}_{\text{2}}(aq) + 2\text{HCl}(aq) \rightarrow \text{BaCl}_{\text{2}}(aq) + 2\text{H}_{\text{2}}\text{O}(l)
\]
2. **Determine the moles of HCl used**:
\[
\text{Moles of HCl} = \text{Molarity of HCl} \times \text{Volume of HCl in liters}
\]
\[
\text{Moles of HCl} = 0.179 \, \text{M} \times 0.0245 \, \text{L} = 4.3855 \times 10^{-3} \, \text{mol}
\]
3. **Use the stoichiometry of the reaction** to find the moles of Ba(OH)₂. According to the balanced equation, 1 mole of Ba(OH)₂ reacts with 2 moles of HCl:
\[
\text{Moles of Ba(OH)}_{\text{2}} = \frac{\text{Moles of HCl}}{2} = \frac{4.3855 \times 10^{-3}}{2} = 2.19275 \times 10^{-3} \, \text{mol}
\]
4. **Determine the molarity of the barium hydroxide solution**:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb9bc6dc7-2424-4f69-956a-285f3752d785%2F2495f4f5-02fd-4883-995f-08ba3a1796dc%2Fpf7eg0a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Titration Problem: Determining the Molarity of Barium Hydroxide Solution**
An aqueous solution of **barium hydroxide** is standardized by titration with a **0.179 M solution of hydrochloric acid**.
**Question:**
If **28.4 mL of base** are required to neutralize **24.5 mL of the acid**, what is the molarity of the **barium hydroxide** solution?
[Input field for molarity]
**Buttons:**
- **Submit Answer**
- **Retry Entire Group**
(Note: You have **9 more group attempts remaining**)
---
### Explanation:
To solve this titration problem, follow these steps:
1. **Write the balanced chemical equation** for the reaction between barium hydroxide (Ba(OH)₂) and hydrochloric acid (HCl):
\[
\text{Ba(OH)}_{\text{2}}(aq) + 2\text{HCl}(aq) \rightarrow \text{BaCl}_{\text{2}}(aq) + 2\text{H}_{\text{2}}\text{O}(l)
\]
2. **Determine the moles of HCl used**:
\[
\text{Moles of HCl} = \text{Molarity of HCl} \times \text{Volume of HCl in liters}
\]
\[
\text{Moles of HCl} = 0.179 \, \text{M} \times 0.0245 \, \text{L} = 4.3855 \times 10^{-3} \, \text{mol}
\]
3. **Use the stoichiometry of the reaction** to find the moles of Ba(OH)₂. According to the balanced equation, 1 mole of Ba(OH)₂ reacts with 2 moles of HCl:
\[
\text{Moles of Ba(OH)}_{\text{2}} = \frac{\text{Moles of HCl}}{2} = \frac{4.3855 \times 10^{-3}}{2} = 2.19275 \times 10^{-3} \, \text{mol}
\]
4. **Determine the molarity of the barium hydroxide solution**:
\[
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