Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![**Problem Statement:**
What volume of a **0.172 M sodium hydroxide** solution is required to neutralize **12.5 mL** of a **0.388 M hydrobromic acid** solution?
**Answer Box:**
____ mL sodium hydroxide
**Explanation:**
To solve this problem, we need to use the concept of stoichiometry and the neutralization reaction between sodium hydroxide (NaOH) and hydrobromic acid (HBr). The balanced chemical equation for the reaction is:
\[ \text{NaOH (aq) + HBr (aq) } \rightarrow \text{ NaBr (aq) + } \text{H}_2\text{O (l)} \]
From the equation, we see that 1 mole of NaOH neutralizes 1 mole of HBr.
To find the volume of NaOH needed, we can use the formula for dilution and neutralization:
\[ \text{M}_1 \text{V}_1 = \text{M}_2 \text{V}_2 \]
Where:
- \(\text{M}_1\) is the molarity of NaOH (0.172 M).
- \(\text{V}_1\) is the volume of NaOH (what we are solving for).
- \(\text{M}_2\) is the molarity of HBr (0.388 M).
- \(\text{V}_2\) is the volume of HBr (12.5 mL).
Rearranging the equation to solve for \(\text{V}_1\):
\[ \text{V}_1 = \frac{\text{M}_2 \times \text{V}_2}{\text{M}_1} \]
Substitute the known values:
\[ \text{V}_1 = \frac{0.388 \text{ M} \times 12.5 \text{ mL}}{0.172 \text{ M}} \]
Calculate to find the volume of sodium hydroxide needed.](https://content.bartleby.com/qna-images/question/84435934-315d-4f3f-bf89-3ba66c6d068b/ae45105f-7cd6-42f1-91b6-76e90ba5905d/8ek7u3c_processed.png)
Transcribed Image Text:**Problem Statement:**
What volume of a **0.172 M sodium hydroxide** solution is required to neutralize **12.5 mL** of a **0.388 M hydrobromic acid** solution?
**Answer Box:**
____ mL sodium hydroxide
**Explanation:**
To solve this problem, we need to use the concept of stoichiometry and the neutralization reaction between sodium hydroxide (NaOH) and hydrobromic acid (HBr). The balanced chemical equation for the reaction is:
\[ \text{NaOH (aq) + HBr (aq) } \rightarrow \text{ NaBr (aq) + } \text{H}_2\text{O (l)} \]
From the equation, we see that 1 mole of NaOH neutralizes 1 mole of HBr.
To find the volume of NaOH needed, we can use the formula for dilution and neutralization:
\[ \text{M}_1 \text{V}_1 = \text{M}_2 \text{V}_2 \]
Where:
- \(\text{M}_1\) is the molarity of NaOH (0.172 M).
- \(\text{V}_1\) is the volume of NaOH (what we are solving for).
- \(\text{M}_2\) is the molarity of HBr (0.388 M).
- \(\text{V}_2\) is the volume of HBr (12.5 mL).
Rearranging the equation to solve for \(\text{V}_1\):
\[ \text{V}_1 = \frac{\text{M}_2 \times \text{V}_2}{\text{M}_1} \]
Substitute the known values:
\[ \text{V}_1 = \frac{0.388 \text{ M} \times 12.5 \text{ mL}}{0.172 \text{ M}} \]
Calculate to find the volume of sodium hydroxide needed.
![**Precipitation Calculation for Lead(II) Bromide (PbBr₂)**
---
**Problem Statement:**
How many grams of PbBr₂ will precipitate when excess FeBr₂ solution is added to 71.0 mL of 0.698 M Pb(NO₃)₂ solution?
**Chemical Reaction:**
\[ \text{Pb(NO}_3\text{)}_2 (\text{aq}) + \text{FeBr}_2 (\text{aq}) \rightarrow \text{PbBr}_2 (\text{s}) + \text{Fe(NO}_3\text{)}_2 (\text{aq}) \]
---
**Calculation:**
To find the mass of PbBr₂ precipitate, follow these steps:
1. **Determine the moles of Pb(NO₃)₂:**
- Volume of Pb(NO₃)₂ solution: 71.0 mL = 0.0710 L
- Concentration of Pb(NO₃)₂ solution: 0.698 M
\[ \text{Moles of Pb(NO}_3\text{)}_2 = \text{Concentration} \times \text{Volume} \]
\[ \text{Moles of Pb(NO}_3\text{)}_2 = 0.698 \, \text{mol/L} \times 0.0710 \, \text{L} \]
\[ \text{Moles of Pb(NO}_3\text{)}_2 = 0.049558 \, \text{mol} \]
2. **Stoichiometry of the reaction:**
- According to the balanced chemical equation, 1 mole of Pb(NO₃)₂ produces 1 mole of PbBr₂.
\[ \text{Moles of PbBr}_2 = 0.049558 \, \text{mol} \]
3. **Calculate the mass of PbBr₂:**
- Molar mass of PbBr₂ = (207.2 g/mol for Pb) + (2 × 79.904 g/mol for Br)
\[ \text{Molar mass of PbBr}_2 = 207.2 \, \text{g/mol} + 159.808 \, \text{g/mol} \]
\[ \text{Molar mass of PbBr}_2 =](https://content.bartleby.com/qna-images/question/84435934-315d-4f3f-bf89-3ba66c6d068b/ae45105f-7cd6-42f1-91b6-76e90ba5905d/v5y2gub_processed.png)
Transcribed Image Text:**Precipitation Calculation for Lead(II) Bromide (PbBr₂)**
---
**Problem Statement:**
How many grams of PbBr₂ will precipitate when excess FeBr₂ solution is added to 71.0 mL of 0.698 M Pb(NO₃)₂ solution?
**Chemical Reaction:**
\[ \text{Pb(NO}_3\text{)}_2 (\text{aq}) + \text{FeBr}_2 (\text{aq}) \rightarrow \text{PbBr}_2 (\text{s}) + \text{Fe(NO}_3\text{)}_2 (\text{aq}) \]
---
**Calculation:**
To find the mass of PbBr₂ precipitate, follow these steps:
1. **Determine the moles of Pb(NO₃)₂:**
- Volume of Pb(NO₃)₂ solution: 71.0 mL = 0.0710 L
- Concentration of Pb(NO₃)₂ solution: 0.698 M
\[ \text{Moles of Pb(NO}_3\text{)}_2 = \text{Concentration} \times \text{Volume} \]
\[ \text{Moles of Pb(NO}_3\text{)}_2 = 0.698 \, \text{mol/L} \times 0.0710 \, \text{L} \]
\[ \text{Moles of Pb(NO}_3\text{)}_2 = 0.049558 \, \text{mol} \]
2. **Stoichiometry of the reaction:**
- According to the balanced chemical equation, 1 mole of Pb(NO₃)₂ produces 1 mole of PbBr₂.
\[ \text{Moles of PbBr}_2 = 0.049558 \, \text{mol} \]
3. **Calculate the mass of PbBr₂:**
- Molar mass of PbBr₂ = (207.2 g/mol for Pb) + (2 × 79.904 g/mol for Br)
\[ \text{Molar mass of PbBr}_2 = 207.2 \, \text{g/mol} + 159.808 \, \text{g/mol} \]
\[ \text{Molar mass of PbBr}_2 =
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