A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 96. mg of oxalic acid (H,C,04), a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in 250. mL of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 48.1 mL of sodium hydroxide solution. Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 2 significant digits.
A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 96. mg of oxalic acid (H,C,04), a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in 250. mL of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 48.1 mL of sodium hydroxide solution. Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 2 significant digits.
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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2.4
![**Standardizing a Sodium Hydroxide Solution**
A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 96. mg of oxalic acid \((H_2C_2O_4)\), a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in 250. mL of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 48.1 mL of sodium hydroxide solution.
**Problem:** Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 2 significant digits.
#### Answer Entry Field
\[ \text{M} \]
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##### Calculation:
1. **Determine moles of oxalic acid (\(H_2C_2O_4\))**:
- Formula mass of \(H_2C_2O_4\): \(2(1) + 2(12) + 4(16) = 2 + 24 + 64 = 90 \text{ g/mol}\)
- Mass of \(H_2C_2O_4\): 96 mg = 0.096 g
- Moles of \(H_2C_2O_4\): \[\frac{0.096 \text{ g}}{90 \text{ g/mol}} = 0.001067 \text{ moles}\]
2. **Calculate moles of \(OH^-\)**:
- As it is a diprotic acid, 1 mole of \(H_2C_2O_4\) reacts with 2 moles of \(NaOH\).
- Moles of \(NaOH\) = 2 \(\times 0.001067 \approx 0.002134 \text{ moles}\)
3. **Calculate molarity of \(NaOH\)**:
- Volume of \(NaOH\) solution = 48.1 mL = 0.0481 L
- Molarity (M) of \(NaOH\): \[\frac{0.002134 \text{ m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F47bace65-e928-444e-9199-b108470afcd9%2F1861f15b-39b5-4954-9276-ba9a6b5b3d3c%2Fqrl28ls_processed.png&w=3840&q=75)
Transcribed Image Text:**Standardizing a Sodium Hydroxide Solution**
A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 96. mg of oxalic acid \((H_2C_2O_4)\), a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in 250. mL of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 48.1 mL of sodium hydroxide solution.
**Problem:** Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 2 significant digits.
#### Answer Entry Field
\[ \text{M} \]
#### Tools
- **Scientific Calculator** (icon representing \(\times 10\) functionality)
- Other icons for reset, assistance, etc.
##### Calculation:
1. **Determine moles of oxalic acid (\(H_2C_2O_4\))**:
- Formula mass of \(H_2C_2O_4\): \(2(1) + 2(12) + 4(16) = 2 + 24 + 64 = 90 \text{ g/mol}\)
- Mass of \(H_2C_2O_4\): 96 mg = 0.096 g
- Moles of \(H_2C_2O_4\): \[\frac{0.096 \text{ g}}{90 \text{ g/mol}} = 0.001067 \text{ moles}\]
2. **Calculate moles of \(OH^-\)**:
- As it is a diprotic acid, 1 mole of \(H_2C_2O_4\) reacts with 2 moles of \(NaOH\).
- Moles of \(NaOH\) = 2 \(\times 0.001067 \approx 0.002134 \text{ moles}\)
3. **Calculate molarity of \(NaOH\)**:
- Volume of \(NaOH\) solution = 48.1 mL = 0.0481 L
- Molarity (M) of \(NaOH\): \[\frac{0.002134 \text{ m
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