**Question #7:**Calculate the pH of a titration at the point when 15.0 mL of 0.15 M NaOH is added to 30.0 mL of 0.20 M HNO₃.**Explanation:**In this problem, we are calculating the pH during the titration of a strong base (NaOH) with a strong acid (HNO₃). At this point in the titration, the base has been added to the acid, and we need to determine the resulting pH of the solution.1. **Calculate moles of NaOH and HNO₃:** - Moles of NaOH = Volume (L) × Molarity = 0.015 L × 0.15 mol/L = 0.00225 mol - Moles of HNO₃ = Volume (L) × Molarity = 0.030 L × 0.20 mol/L = 0.006 mol2. **Determine the limiting reactant:** - The moles of NaOH (0.00225) are less than the moles of HNO₃ (0.006), so NaOH is the limiting reactant.3. **Calculate the moles of HNO₃ remaining:** - Moles of HNO₃ remaining = 0.006 mol - 0.00225 mol = 0.00375 mol4. **Calculate the concentration of HNO₃ in the solution:** - Total volume = 15.0 mL + 30.0 mL = 45.0 mL = 0.045 L - Concentration of HNO₃ = 0.00375 mol / 0.045 L = 0.0833 M5. **Calculate the pH of the solution:** - HNO₃ is a strong acid, so it completely dissociates. - Therefore, [H⁺] = 0.0833 M - pH = -log[H⁺] = -log(0.0833) ≈ 1.08The pH at this point of the titration is approximately 1.08.

Given -> Volume of NaOH = 15.0 ml Molarity of NaOH= 0.15 M Volume of HNO3 = 30.0 ml Molarity of…

Answered: Calculate the pH of a titration at the…

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Calculate the pH of a titration at the point when 15.0 mL of 0.15 M NaOH is added to 30.0 mL of 0.20 M HNO3 10.

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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