A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 231. 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 solutic with his sodium hydroxide solution. When the titration reaches the equivalence point, the student finds he has used 130.9 mL of sodium hydroxide solution. Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 3 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...
icon
Related questions
Question

2.3

**Titration Problem: Standardizing a Sodium Hydroxide Solution**

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 231. mg of oxalic acid (H₂C₂O₄), 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 his sodium hydroxide solution. When the titration reaches the equivalence point, the student finds he has used 130.9 mL of sodium hydroxide solution.

**Objective:**
Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 3 significant digits.

**Calculation Steps:**

1. **Determine the Moles of Oxalic Acid**:
   - Given mass of oxalic acid = 231 mg = 0.231 g
   - Molar mass of oxalic acid (H₂C₂O₄) = 2(1.01 g/mol) + 2(12.01 g/mol) + 4(16.00 g/mol) = 90.04 g/mol
   - Moles of oxalic acid (n) = mass (g) / molar mass (g/mol) = 0.231 g / 90.04 g/mol ≈ 0.00257 mol

2. **Reaction Stoichiometry**:
   - The balanced equation for the reaction between oxalic acid and sodium hydroxide (NaOH) is:
     \[ \text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \]
   - From the equation, 1 mole of oxalic acid reacts with 2 moles of NaOH.
   - Moles of NaOH required = 2 × moles of oxalic acid = 2 × 0.00257 mol = 0.00514 mol

3. **Volume of Sodium Hydroxide Solution Used**:
   - Volume of sodium hydroxide solution used = 130.9 mL = 0.1309 L

4. **Calculate the Molarity (M) of Sodium Hydroxide Solution**
Transcribed Image Text:**Titration Problem: Standardizing a Sodium Hydroxide Solution** A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 231. mg of oxalic acid (H₂C₂O₄), 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 his sodium hydroxide solution. When the titration reaches the equivalence point, the student finds he has used 130.9 mL of sodium hydroxide solution. **Objective:** Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 3 significant digits. **Calculation Steps:** 1. **Determine the Moles of Oxalic Acid**: - Given mass of oxalic acid = 231 mg = 0.231 g - Molar mass of oxalic acid (H₂C₂O₄) = 2(1.01 g/mol) + 2(12.01 g/mol) + 4(16.00 g/mol) = 90.04 g/mol - Moles of oxalic acid (n) = mass (g) / molar mass (g/mol) = 0.231 g / 90.04 g/mol ≈ 0.00257 mol 2. **Reaction Stoichiometry**: - The balanced equation for the reaction between oxalic acid and sodium hydroxide (NaOH) is: \[ \text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \] - From the equation, 1 mole of oxalic acid reacts with 2 moles of NaOH. - Moles of NaOH required = 2 × moles of oxalic acid = 2 × 0.00257 mol = 0.00514 mol 3. **Volume of Sodium Hydroxide Solution Used**: - Volume of sodium hydroxide solution used = 130.9 mL = 0.1309 L 4. **Calculate the Molarity (M) of Sodium Hydroxide Solution**
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 1 images

Blurred answer
Knowledge Booster
Quality Assurance and Calibration Methods
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY