**Lab Report #4-2-1:****Task:** Using a 0.1 M NH₄Cl solution and the given pH data, calculate the concentration of \([H^+]\) and \([OH^-]\).**Options:**- ⃝ \([H^+]\) = \(3.2 \times 10^{-5}\) M, \([OH^-]\) = \(3.2 \times 10^{-10}\) M- ⃝ \([H^+]\) = \(2.2 \times 10^{-9}\) M, \([OH^-]\) = \(4.5 \times 10^{-6}\) M- ⃝ \([H^+]\) = \(3.2 \times 10^{-10}\) M, \([OH^-]\) = \(3.2 \times 10^{-5}\) M- ⃝ \([H^+]\) = \(1.0 \times 10^{-7}\) M, \([OH^-]\) = \(1.0 \times 10^{-7}\) M ### Table of pH Values for Various SolutionsThe following table presents the pH values of different solutions, both before and after specific treatments such as boiling.| Solutions | pH Values ||-------------------|-----------|| H₂O (unboiled) | 3.5 || H₂O (boiled) | 7.0 || NaCl | 7.0 || NaC₂H₃O₂ | 9.1 || NH₄Cl | 4.5 || NaHCO₃ | 9.5 || Na₃PO₄ | 11.9 || Na₂CO₃ | 11.0 |#### Explanation of the Table:- **H₂O (unboiled)**: This is water that has not undergone the boiling process, with a pH of 3.5, indicating it is slightly acidic.- **H₂O (boiled)**: Water that has been boiled shows a neutral pH of 7.0.- **NaCl (Sodium Chloride)**: This solution has a neutral pH of 7.0.- **NaC₂H₃O₂ (Sodium Acetate)**: This solution is slightly basic with a pH of 9.1.- **NH₄Cl (Ammonium Chloride)**: This solution has a pH of 4.5, showing it is mildly acidic.- **NaHCO₃ (Sodium Bicarbonate)**: This solution has a basic pH of 9.5.- **Na₃PO₄ (Trisodium Phosphate)**: This solution is strongly basic with a pH of 11.9.- **Na₂CO₃ (Sodium Carbonate)**: This solution also shows a strong basicity with a pH of 11.0.Use this table to understand the acidity or basicity of common solutions, which is critical in various applications, including industrial processes, cooking, and scientific research.

pH of a solution measures acidity of that solution. If pH >7 solution is basic. If pH <7…

Lab Report #4-2-1: 0.1 M NH4CI solution, using the given pH data, calculate: [H*], [OH"]: O [H*] = 3.2 x 10-5 M, [OH"]= 3.2 x 10-10 M O [H*] = 2.2 x 10-9 M, [OH¯]= 4.5 x 10-6 M O [H*] = 3.2 x 10-10 M, [OH"]= 3.2 x 10-5 M %3D [H*] = 1.0 x 10-7 M, [OH"]= 1.0 x 10-7 M

Lab Report #4-2-1: 0.1 M NH4CI solution, using the given pH data, calculate: [H], [OH"]: O [H] = 3.2 x 10-5 M, [OH"]= 3.2 x 10-10 M O [H] = 2.2 x 10-9 M, [OH¯]= 4.5 x 10-6 M O [H] = 3.2 x 10-10 M, [OH"]= 3.2 x 10-5 M %3D [H*] = 1.0 x 10-7 M, [OH"]= 1.0 x 10-7 M

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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