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 17
How many milliliters of 0.0259 M Ca(OH)₂ would be required to exactly neutralize 106 mL of 0.419 M HCl?
### Explanation
To solve this problem, we start with the balanced chemical equation for the reaction between calcium hydroxide, Ca(OH)₂, and hydrochloric acid, HCl:
\[ \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \]
From the equation, 1 mole of Ca(OH)₂ neutralizes 2 moles of HCl.
#### Step-by-Step Solution:
1. **Calculate moles of HCl:**
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)} \]
\[ = 0.419 \, \text{mol/L} \times 0.106 \, \text{L} \]
\[ = 0.044414 \, \text{moles HCl} \]
2. **Calculate moles of Ca(OH)₂ needed:**
- From the balanced equation, 1 mole of Ca(OH)₂ reacts with 2 moles of HCl.
- Therefore, moles of Ca(OH)₂ needed = 0.044414 moles HCl / 2
\[ = 0.022207 \, \text{moles Ca(OH)}_2 \]
3. **Calculate the volume of Ca(OH)₂ solution needed:**
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]
\[ = \frac{0.022207 \, \text{moles}}{0.0259 \, \text{mol/L}} \]
\[ = 0.857 \, \text{L} \]
4. **Convert the volume to milliliters:**
\[ \text{Volume (mL)} = 0.857 \, \text{L} \times 1000 \, \text{mL/L} \]
\[ = 857 \, \text{mL} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf942647-17d2-4969-8d53-0bdf20ef3a25%2Fa5eb6ab0-d2c5-48ed-9aa4-f0f3f0b8335b%2Fbf4sp4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 17
How many milliliters of 0.0259 M Ca(OH)₂ would be required to exactly neutralize 106 mL of 0.419 M HCl?
### Explanation
To solve this problem, we start with the balanced chemical equation for the reaction between calcium hydroxide, Ca(OH)₂, and hydrochloric acid, HCl:
\[ \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \]
From the equation, 1 mole of Ca(OH)₂ neutralizes 2 moles of HCl.
#### Step-by-Step Solution:
1. **Calculate moles of HCl:**
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)} \]
\[ = 0.419 \, \text{mol/L} \times 0.106 \, \text{L} \]
\[ = 0.044414 \, \text{moles HCl} \]
2. **Calculate moles of Ca(OH)₂ needed:**
- From the balanced equation, 1 mole of Ca(OH)₂ reacts with 2 moles of HCl.
- Therefore, moles of Ca(OH)₂ needed = 0.044414 moles HCl / 2
\[ = 0.022207 \, \text{moles Ca(OH)}_2 \]
3. **Calculate the volume of Ca(OH)₂ solution needed:**
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]
\[ = \frac{0.022207 \, \text{moles}}{0.0259 \, \text{mol/L}} \]
\[ = 0.857 \, \text{L} \]
4. **Convert the volume to milliliters:**
\[ \text{Volume (mL)} = 0.857 \, \text{L} \times 1000 \, \text{mL/L} \]
\[ = 857 \, \text{mL} \]
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