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What is the concentration of a 54.70 mL solution of HBr that is completely titrated by 32.00 mL of a 0.200 M NaOH solution? [HBr]= M

What is the concentration of a 54.70 mL solution of HBr that is completely titrated by 32.00 mL of a 0.200 M NaOH solution? [HBr]= M

Chemistry
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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### Titration Problem Example **Problem Statement:** What is the concentration of a 54.70 mL solution of HBr that is completely titrated by 32.00 mL of a 0.200 M NaOH solution? **Solution:** To find the concentration of HBr, use the titration formula involving mole ratios: \[ \text{Moles of NaOH} = \text{Moles of HBr} \] 1. **Calculate the Moles of NaOH:** \[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} \] \[ \text{Moles of NaOH} = 0.03200 \, \text{L} \times 0.200 \, \text{M} = 0.00640 \, \text{moles} \] 2. **Since HBr and NaOH react in a 1:1 mole ratio:** \[ \text{Moles of HBr} = 0.00640 \, \text{moles} \] 3. **Calculate the Concentration of HBr:** \[ \text{Concentration of HBr (M)} = \frac{\text{Moles of HBr}}{\text{Volume of HBr (L)}} \] \[ \text{Concentration of HBr (M)} = \frac{0.00640 \, \text{moles}}{0.05470 \, \text{L}} \approx 0.117 \, \text{M} \] **Answer:** \[ \text{[HBr]} = 0.117 \, \text{M} \]
Transcribed Image Text:### Titration Problem Example **Problem Statement:** What is the concentration of a 54.70 mL solution of HBr that is completely titrated by 32.00 mL of a 0.200 M NaOH solution? **Solution:** To find the concentration of HBr, use the titration formula involving mole ratios: \[ \text{Moles of NaOH} = \text{Moles of HBr} \] 1. **Calculate the Moles of NaOH:** \[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} \] \[ \text{Moles of NaOH} = 0.03200 \, \text{L} \times 0.200 \, \text{M} = 0.00640 \, \text{moles} \] 2. **Since HBr and NaOH react in a 1:1 mole ratio:** \[ \text{Moles of HBr} = 0.00640 \, \text{moles} \] 3. **Calculate the Concentration of HBr:** \[ \text{Concentration of HBr (M)} = \frac{\text{Moles of HBr}}{\text{Volume of HBr (L)}} \] \[ \text{Concentration of HBr (M)} = \frac{0.00640 \, \text{moles}}{0.05470 \, \text{L}} \approx 0.117 \, \text{M} \] **Answer:** \[ \text{[HBr]} = 0.117 \, \text{M} \]
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