First read the text sections 10.5 and 10.6 on pH.This is a Dry Lab since we cannot meet on Campus. After calibrating a pH meter youwould have measured the pH of four different Basic Aqueous Solutions and examinedthe data for trends. You would have calculated the expected pH of each of the strongbase solutions by assuming that for every NaOH formula unit that dissolves, one OH-ion is released. Thus, we can assume that the OH- Molarity equals the NaOH Molarity.Using the OH- Molarity and the water ionization equilibrium expression (Kw= 1.0 X 10to the -14 power H+ Molarity X OH- Molarity),wwwwe can calculate the H+ Molarity of the solution.Using the H+ Molarity, we can calculate the solution's pH.Thus a 1.0 X 10 to the 4 power NaOH solution has a pH of 10. The calculated pHvalue is arrived at as follows.By rearranging the water ionization equilibrium expression and plugging in the OH-Molarity, we can solve for H+ Molarity. H+ M = 1.0 X 10 to the -14 powerdivided by 1.0 X 10 to the -4 power 1.0 X 10 to the 10 power.pH= -log (1.0 X 10 to the -10 power)= 10ICalculate the expected pH of these three NaOH solutions, complete the table, answerthe questions and email it to me at jpanek@mc3.edu.NaOH Aqueous Solution0.10 M0.010 M0.0010 MCalculated expected pHExperimental pH (typical)12.85Answer these Questions. #1 What happens to the pH as OH- Molarity increases ?11.9511.05

First we need to calculate the expected pH using the given molarity of NaOH solution. Then we can…

Calculate the expected pH of these three NaOH solutions, complete the table, answer the questions and email it to me at jpanek@mc3.edu. NaOH Aqueous Solution 0.10 M 0.010 M 0.0010 M Calculated expected pH Experimental pH (typical) 12.85 11.95 Answer these Questions. #1 What happens to the pH as OH- Molarity increases ? 11.05

Calculate the expected pH of these three NaOH solutions, complete the table, answer the questions and email it to me at jpanek@mc3.edu. NaOH Aqueous Solution 0.10 M 0.010 M 0.0010 M Calculated expected pH Experimental pH (typical) 12.85 11.95 Answer these Questions. #1 What happens to the pH as OH- Molarity increases ? 11.05

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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In a solution with pH = 2.50, the H30* concentration is M, the OH concentration is M, and the pOH is Round the molarities to 2 sig figs and use scientific notation. Do not include units. Examples: • Input 5.51 x 10-6 as 5.5E-6 • Input 8.93 × 1010 as 8.9E10 Round the pOH to 2 decimal places and use decimal notation. Examples: • Input 12.545 as 12.55 • Input 1.112 as 1.11
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PassignmentProblemiD=182385374 pl = H-C-H Provide Feedback H.
I JUST GOT an answer, but i think it is wrong, in the answer you write: 0.233 mol/0.1 L = 0.0233 M But I get 2.33 !! that will change the pH. Please give me my question back if the previous answer was wrong. Step 1 Given that, Percentage of acetic acid solution = 14% It means, 14 g of acetic acid dissolved in 100 g or 100 mL of water. Moles of acetic acid in 14 g = 14 g/60.052 g/mol = 0.233 mol Molarity of 0.233 mol of 100 mL acetic acid solution = 0.233 mol/0.1 L = 0.0233 M Ka value of acetic acid = 1.7×10-5 Step 2 Now, acetic acid dissociates in water as follows; CH3COOH <----> CH3COO- + H+ Ka = [CH3COO- ][H+]/[CH3COOH] Let, [CH3COO- ] = [H+] = x Then we have, 1.7×10-5 = x.x/0.0233M x = 0.629×10-3 M Now, pH of the acetic acid solution will be pH = -log[H+] = -log0.629×10-3 M = 2.8
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Cancel RESET Done The pH of orange juice is typically 3.5, whereas the pH of tomato juice is usually 4.5. This statement implies that the molar concentration of hydronium ions in tomato juice is times that of hydronium ions in orange juice. [Provide a numerical answer such as "10" or "0.005". Do not use scientific notation. Note that for numbers less than 1, use the cosmetic zero before the decimal place "0.005"-- do not write ".005"] · $ Adjust Filters Crop | | | |
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Which is the correct pH and pOH of each of the solutions at 25 °C for which the substances ionize completely? (Choose all that apply) Question 29 options: a) 0.0031 M Ca(OH)2: pH = 11.79 and pOH= 2.21 b) 0.0163 M NaOH: pH = 12.16 and pOH= 1.845 c) 0.400 M HCl: pH = 0.699 and pOH= 13.30 d) 3.0 M HClO4: pH =-0.48 and pOH= 14.48
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