An aqueous solution of perchloric acid is standardized by titration with a 0.131 M solution of barium hydroxide. If 21.0 mL of base are required to neutralize 29.5 mL of the acid, what is the molarity of the perchloric acid solution? M perchloric acid
An aqueous solution of perchloric acid is standardized by titration with a 0.131 M solution of barium hydroxide. If 21.0 mL of base are required to neutralize 29.5 mL of the acid, what is the molarity of the perchloric acid solution? M perchloric acid
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![**Titration Problem: Standardizing Perchloric Acid Solution with Barium Hydroxide**
**Problem Statement:**
An aqueous solution of **perchloric acid** is standardized by titration with a **0.131 M** solution of **barium hydroxide**.
If **21.0 mL** of base are required to neutralize **29.5 mL** of the acid, what is the molarity of the **perchloric acid** solution?
**Solution Box:**
\[\_\_\_\_ \text{ M perchloric acid}\]
---
**Detailed Explanation:**
This text and the problem are intended for students learning about titration and chemical concentrations. The problem asks students to determine the molarity of a perchloric acid solution (HClO₄) after it has been neutralized by a barium hydroxide solution (Ba(OH)₂).
In this scenario, students need to use the titration formula and the concept of neutralization to determine the unknown concentration of perchloric acid based on the volumes and concentrations given for barium hydroxide.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a354f1c-cdfd-41e5-a863-9a86a504af1c%2F25edee27-ed08-49d8-af78-50a64a31d187%2Fvoard3p_processed.png&w=3840&q=75)
Transcribed Image Text:**Titration Problem: Standardizing Perchloric Acid Solution with Barium Hydroxide**
**Problem Statement:**
An aqueous solution of **perchloric acid** is standardized by titration with a **0.131 M** solution of **barium hydroxide**.
If **21.0 mL** of base are required to neutralize **29.5 mL** of the acid, what is the molarity of the **perchloric acid** solution?
**Solution Box:**
\[\_\_\_\_ \text{ M perchloric acid}\]
---
**Detailed Explanation:**
This text and the problem are intended for students learning about titration and chemical concentrations. The problem asks students to determine the molarity of a perchloric acid solution (HClO₄) after it has been neutralized by a barium hydroxide solution (Ba(OH)₂).
In this scenario, students need to use the titration formula and the concept of neutralization to determine the unknown concentration of perchloric acid based on the volumes and concentrations given for barium hydroxide.
![### Titration Standardization Procedure
An aqueous solution of **potassium hydroxide** is standardized by titration with a **0.133 M solution of hydrobromic acid**.
**Problem:**
If **15.7 mL** of base are required to neutralize **10.9 mL** of the acid, what is the molarity of the potassium hydroxide solution?
#### Solution:
To solve this, let's use the relationship from the titration formula:
\[ M_1 \times V_1 = M_2 \times V_2 \]
Where:
- \( M_1 \) is the molarity of the potassium hydroxide (\( KOH \)) solution.
- \( V_1 \) is the volume of the potassium hydroxide solution.
- \( M_2 \) is the molarity of the hydrobromic acid (\( HBr \)) solution.
- \( V_2 \) is the volume of the hydrobromic acid solution.
Given values for the problem:
- \( M_2 = 0.133 \) M \( HBr \)
- \( V_2 = 10.9 \) mL
- \( V_1 = 15.7 \) mL
Plug the values into the equation:
\[ M_1 \times 15.7 \, \text{mL} = 0.133 \, \text{M} \times 10.9 \, \text{mL} \]
To solve for \( M_1 \):
\[ M_1 = \frac{0.133 \, \text{M} \times 10.9 \, \text{mL}}{15.7 \, \text{mL}} \]
Calculate the result:
\[ M_1 = \frac{1.4497}{15.7} \]
\[ M_1 \approx 0.0923 \, \text{M} \]
So, the molarity of the potassium hydroxide solution is **0.0923 M**.
_[There is a blank box in the original image that indicates where students should input their answers on the educational website.]_
#### Answer Box:
\[ \boxed{ \,\,\, 0.0923 \,\,\, } \] M potassium hydroxide](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a354f1c-cdfd-41e5-a863-9a86a504af1c%2F25edee27-ed08-49d8-af78-50a64a31d187%2Fd76crlg_processed.png&w=3840&q=75)
Transcribed Image Text:### Titration Standardization Procedure
An aqueous solution of **potassium hydroxide** is standardized by titration with a **0.133 M solution of hydrobromic acid**.
**Problem:**
If **15.7 mL** of base are required to neutralize **10.9 mL** of the acid, what is the molarity of the potassium hydroxide solution?
#### Solution:
To solve this, let's use the relationship from the titration formula:
\[ M_1 \times V_1 = M_2 \times V_2 \]
Where:
- \( M_1 \) is the molarity of the potassium hydroxide (\( KOH \)) solution.
- \( V_1 \) is the volume of the potassium hydroxide solution.
- \( M_2 \) is the molarity of the hydrobromic acid (\( HBr \)) solution.
- \( V_2 \) is the volume of the hydrobromic acid solution.
Given values for the problem:
- \( M_2 = 0.133 \) M \( HBr \)
- \( V_2 = 10.9 \) mL
- \( V_1 = 15.7 \) mL
Plug the values into the equation:
\[ M_1 \times 15.7 \, \text{mL} = 0.133 \, \text{M} \times 10.9 \, \text{mL} \]
To solve for \( M_1 \):
\[ M_1 = \frac{0.133 \, \text{M} \times 10.9 \, \text{mL}}{15.7 \, \text{mL}} \]
Calculate the result:
\[ M_1 = \frac{1.4497}{15.7} \]
\[ M_1 \approx 0.0923 \, \text{M} \]
So, the molarity of the potassium hydroxide solution is **0.0923 M**.
_[There is a blank box in the original image that indicates where students should input their answers on the educational website.]_
#### Answer Box:
\[ \boxed{ \,\,\, 0.0923 \,\,\, } \] M potassium hydroxide
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