**Exercise:** Construct a titration curve by plotting \(pH\) vs. \(V_{\text{NaOH}}\) added.**Reaction:** \[\text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)} + \text{NaCl}_{(aq)}\]---**Observations:**- **Volume of NaOH added (mL):** \(0.00, 5.00, 10.00, 15.00, 20.00, 25.00, 30.00, 35.00, 40.00, 45.00, 50.00\)- **[H⁺] Concentration:** Varies with each addition, starts at \(1 \times 10^{-1}\) M- **pH Values:** \(1.00, \ldots, \text{calculated}\)---**Diagram Description:**There is an illustration of a titration setup with:- A **buret** containing the titrant, \(0.1 \, \text{M NaOH}\).- A **flask** which holds the analyte, \(0.1 \, \text{M HCl}\), \(50 \, \text{mL}\).---**Hints:**The pH of the solution is calculated using:**(a) Initial Conditions:**- \(V_{\text{NaOH}} = 0\), \([\text{H}^+]\) = [HCl]**(b) After Initial Addition of NaOH (Before Equivalence Point):**\[ [\text{H}^+] = \frac{M \cdot V_{\text{HCl}} - M \cdot V_{\text{NaOH}}}{V_{\text{total}}} \]**At Equivalence Point (\(V_{\text{NaOH}} = 50 \, \text{mL}\)):**- All HCl is consumed, acidity from auto-ionization of water: \([\text{H}^+] = 1 \times 10^{-7}\)**(d) After Equivalence Point:**\[ [\text{H}^+] = \frac{1 \times 10^{-14}}{M \cdot V_{\text

The initial concentration of HCl = 0.1 M The initial concentration of NaOH = 0.1 M The initial…

Exercise Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 52.00 55.00 60.00 65.00 70.00 75.00 80.00 95 00 1×10-² Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H'] = [HCI] (b) After initial addition of NaOH and before the end point: MVC-M NOVNO V LV (e) At the end point (Vsaou 50 mL), all the HC1 is being consumed and the acidi of the solution comes from auto ionization of water which is: [H]=1*10 (d) After the end point: [H"]= 1x10-¹4 M XV W V. +K

Exercise Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 52.00 55.00 60.00 65.00 70.00 75.00 80.00 95 00 1×10-² Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H'] = [HCI] (b) After initial addition of NaOH and before the end point: MVC-M NOVNO V LV (e) At the end point (Vsaou 50 mL), all the HC1 is being consumed and the acidi of the solution comes from auto ionization of water which is: [H]=1*10 (d) After the end point: [H"]= 1x10-¹4 M XV W V. +K

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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How would you prepare 1.40 L of a buffer with a pH of 9.07 from 0.170 M Na CO, and 0.240 M HCl7 For the carbonic acid. Ka (H₂CO3) = 4.45 x 10 and Ka (H₂CO₂) = 4.69 × 10 Mix mL of 0,170 M Nag COs with mL of 0.240 M HCI Submit Answer Try Another Version S tem attempts remaining
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Exercises: 4. Suppose we titrate 20.0 mL of 0.100 MHCHO2 with 0.100 M NaOH. Calculate the pH for the following points in the titration. Ka = 1.8 × 104. Total mL of NaOH added 0.00 mL 10.00 mL 20.00 mL 40.00 mL mol of NaOH added mol of excess HCHO2 [H*]remaining, M PH pH POH
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