What volume of a 0.217 M hydrobromic acid solution is required to neutralize 29.1 mL of a 0.194 M potassium hydroxide solution? mL hydrobromic acid
What volume of a 0.217 M hydrobromic acid solution is required to neutralize 29.1 mL of a 0.194 M potassium hydroxide solution? mL hydrobromic acid
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
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![**Acid-Base Titration Calculation**
To calculate the volume of a 0.217 M hydrobromic acid solution necessary to neutralize 29.1 mL of a 0.194 M potassium hydroxide solution:
1. **Start with the balanced chemical equation:**
\[
HBr + KOH \rightarrow KBr + H_2O
\]
This equation shows that 1 mole of HBr reacts with 1 mole of KOH.
2. **Identify the known values:**
- Concentration of HBr = 0.217 M
- Volume of KOH = 29.1 mL
- Concentration of KOH = 0.194 M
3. **Calculate the moles of KOH:**
\[
\text{Moles of KOH} = C_{\text{KOH}} \times V_{\text{KOH}}
\]
Given \( C_{\text{KOH}} = 0.194 \text{ M} \) and \( V_{\text{KOH}} = 29.1 \text{ mL} \) or 0.0291 L:
\[
\text{Moles of KOH} = 0.194 \text{ M} \times 0.0291 \text{ L} = 0.0056454 \text{ moles}
\]
4. **Since the stoichiometric ratio of HBr to KOH is 1:1, moles of HBr required = moles of KOH:**
\[
\text{Moles of HBr} = 0.0056454 \text{ moles}
\]
5. **Calculate the volume of HBr solution needed:**
\[
V_{\text{HBr}} = \frac{\text{Moles of HBr}}{C_{\text{HBr}}}
\]
Given \( C_{\text{HBr}} = 0.217 \text{ M} \):
\[
V_{\text{HBr}} = \frac{0.0056454 \text{ moles}}{0.217 \text{ M}} = 0.02602 \text{ L} \times 1000 = 26.02 \text{ mL}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84435934-315d-4f3f-bf89-3ba66c6d068b%2Fc98b4dd0-7ccd-4739-9d4f-53b05cb89419%2Fr59gxro_processed.png&w=3840&q=75)
Transcribed Image Text:**Acid-Base Titration Calculation**
To calculate the volume of a 0.217 M hydrobromic acid solution necessary to neutralize 29.1 mL of a 0.194 M potassium hydroxide solution:
1. **Start with the balanced chemical equation:**
\[
HBr + KOH \rightarrow KBr + H_2O
\]
This equation shows that 1 mole of HBr reacts with 1 mole of KOH.
2. **Identify the known values:**
- Concentration of HBr = 0.217 M
- Volume of KOH = 29.1 mL
- Concentration of KOH = 0.194 M
3. **Calculate the moles of KOH:**
\[
\text{Moles of KOH} = C_{\text{KOH}} \times V_{\text{KOH}}
\]
Given \( C_{\text{KOH}} = 0.194 \text{ M} \) and \( V_{\text{KOH}} = 29.1 \text{ mL} \) or 0.0291 L:
\[
\text{Moles of KOH} = 0.194 \text{ M} \times 0.0291 \text{ L} = 0.0056454 \text{ moles}
\]
4. **Since the stoichiometric ratio of HBr to KOH is 1:1, moles of HBr required = moles of KOH:**
\[
\text{Moles of HBr} = 0.0056454 \text{ moles}
\]
5. **Calculate the volume of HBr solution needed:**
\[
V_{\text{HBr}} = \frac{\text{Moles of HBr}}{C_{\text{HBr}}}
\]
Given \( C_{\text{HBr}} = 0.217 \text{ M} \):
\[
V_{\text{HBr}} = \frac{0.0056454 \text{ moles}}{0.217 \text{ M}} = 0.02602 \text{ L} \times 1000 = 26.02 \text{ mL}
![### Neutralization Calculation Example
#### Problem Statement:
What volume of a 0.129 M calcium hydroxide solution is required to neutralize 19.3 mL of a 0.130 M hydrochloric acid solution?
**Answer:**
\[ \boxed{} \, \text{mL calcium hydroxide} \]
---
**Detailed Explanation:**
This problem involves a neutralization reaction between calcium hydroxide (Ca(OH)₂) and hydrochloric acid (HCl). To solve it, follow these steps:
1. **Write the balanced chemical equation:**
\[ \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \]
2. **Calculate moles of HCl:**
\[
\text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)}
\]
\[
\text{Moles of HCl} = 0.130 \, \text{M} \times 0.0193 \, \text{L} = 0.002509 \, \text{moles}
\]
3. **Determine moles of Ca(OH)₂ required:**
From the balanced equation, 1 mole of Ca(OH)₂ reacts with 2 moles of HCl. Therefore, moles of Ca(OH)₂ required:
\[
\text{Moles of Ca(OH)}_2 = \frac{0.002509 \, \text{moles}}{2} = 0.001255 \, \text{moles}
\]
4. **Find the volume of Ca(OH)₂ solution required:**
\[
\text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}}
\]
\[
\text{Volume (L)} = \frac{0.001255 \, \text{moles}}{0.129 \, \text{M}} = 0.009728 \, \text{L}
\]
Convert liters to milliliters:
\[
\text{Volume (mL)} = 0.009728 \, \text{L}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84435934-315d-4f3f-bf89-3ba66c6d068b%2Fc98b4dd0-7ccd-4739-9d4f-53b05cb89419%2F5lvsbb_processed.png&w=3840&q=75)
Transcribed Image Text:### Neutralization Calculation Example
#### Problem Statement:
What volume of a 0.129 M calcium hydroxide solution is required to neutralize 19.3 mL of a 0.130 M hydrochloric acid solution?
**Answer:**
\[ \boxed{} \, \text{mL calcium hydroxide} \]
---
**Detailed Explanation:**
This problem involves a neutralization reaction between calcium hydroxide (Ca(OH)₂) and hydrochloric acid (HCl). To solve it, follow these steps:
1. **Write the balanced chemical equation:**
\[ \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \]
2. **Calculate moles of HCl:**
\[
\text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)}
\]
\[
\text{Moles of HCl} = 0.130 \, \text{M} \times 0.0193 \, \text{L} = 0.002509 \, \text{moles}
\]
3. **Determine moles of Ca(OH)₂ required:**
From the balanced equation, 1 mole of Ca(OH)₂ reacts with 2 moles of HCl. Therefore, moles of Ca(OH)₂ required:
\[
\text{Moles of Ca(OH)}_2 = \frac{0.002509 \, \text{moles}}{2} = 0.001255 \, \text{moles}
\]
4. **Find the volume of Ca(OH)₂ solution required:**
\[
\text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}}
\]
\[
\text{Volume (L)} = \frac{0.001255 \, \text{moles}}{0.129 \, \text{M}} = 0.009728 \, \text{L}
\]
Convert liters to milliliters:
\[
\text{Volume (mL)} = 0.009728 \, \text{L}
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