Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
Problem 121AP: Calcium carbonate, CaCO3, can be obtained in a very pure state. Standard solutions of calcium ion...
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![## Question 1 of 10
### What volume in milliliters of 0.0130 M Ca(OH)₂ is required to neutralize 70.0 mL of 0.0300 M HBr?
**Explanation:**
To solve this problem, we need to use the concept of neutralization reactions, where an acid and a base react to form water and a salt. The balanced chemical equation for the reaction between calcium hydroxide (Ca(OH)₂) and hydrobromic acid (HBr) is:
\[ \text{Ca(OH)}_2 + 2 \text{HBr} \rightarrow \text{CaBr}_2 + 2 \text{H}_2\text{O} \]
This equation indicates that one mole of Ca(OH)₂ neutralizes two moles of HBr.
We use the formula:
\[ M_1V_1 = M_2V_2 \]
where:
- \( M_1 \) and \( V_1 \) are the molarity and volume of the acid (HBr)
- \( M_2 \) and \( V_2 \) are the molarity and volume of the base (Ca(OH)₂)
However, because the mole ratio is 1:2, we need to account for this by modifying the formula:
\[ M_1V_1 = 2(M_2V_2) \]
Given:
- \( M_1 = 0.0300 \) M (HBr)
- \( V_1 = 70.0 \) mL (HBr)
- \( M_2 = 0.0130 \) M (Ca(OH)₂)
We need to find \( V_2 \):
\[ 0.0300 \, \text{M} \times 70.0 \, \text{mL} = 2 \times 0.0130 \, \text{M} \times V_2 \]
\[ 2.1 \, \text{M}\cdot\text{mL} = 0.0260 \, \text{M} \times V_2 \]
\[ V_2 = \frac{2.1 \, \text{M}\cdot\text{mL}}{0.0260 \, \text{M}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57980898-f1ea-4baf-a646-fad54b56b73d%2Fb9855a0d-92ab-409c-acf8-f92db0513ad1%2Fejzhah8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Question 1 of 10
### What volume in milliliters of 0.0130 M Ca(OH)₂ is required to neutralize 70.0 mL of 0.0300 M HBr?
**Explanation:**
To solve this problem, we need to use the concept of neutralization reactions, where an acid and a base react to form water and a salt. The balanced chemical equation for the reaction between calcium hydroxide (Ca(OH)₂) and hydrobromic acid (HBr) is:
\[ \text{Ca(OH)}_2 + 2 \text{HBr} \rightarrow \text{CaBr}_2 + 2 \text{H}_2\text{O} \]
This equation indicates that one mole of Ca(OH)₂ neutralizes two moles of HBr.
We use the formula:
\[ M_1V_1 = M_2V_2 \]
where:
- \( M_1 \) and \( V_1 \) are the molarity and volume of the acid (HBr)
- \( M_2 \) and \( V_2 \) are the molarity and volume of the base (Ca(OH)₂)
However, because the mole ratio is 1:2, we need to account for this by modifying the formula:
\[ M_1V_1 = 2(M_2V_2) \]
Given:
- \( M_1 = 0.0300 \) M (HBr)
- \( V_1 = 70.0 \) mL (HBr)
- \( M_2 = 0.0130 \) M (Ca(OH)₂)
We need to find \( V_2 \):
\[ 0.0300 \, \text{M} \times 70.0 \, \text{mL} = 2 \times 0.0130 \, \text{M} \times V_2 \]
\[ 2.1 \, \text{M}\cdot\text{mL} = 0.0260 \, \text{M} \times V_2 \]
\[ V_2 = \frac{2.1 \, \text{M}\cdot\text{mL}}{0.0260 \, \text{M}} \]
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