In general chemistry we taught you to calculate the pH of a solution if you know the H30* concentration. We neglected to talk about the concept of "activity"; when other ions in solution affect the behavior of the your analyte (analytical target) ion. This behavior is what we refer to as the ion activity. Please note that this only applies to ions - charged species - not uncharged molecules. Using the chart of activity versus ionic strength found in chapter 8 (7th edition) of the text, answer the following for a solution that has the following concentrations: [HCI] = 0.0118 and [KNO3] = 0.017 1) - the activity coefficient of H30+ ions = 2) - the activity of H30* ions = 3) - neglecting activities the pH of the solution is = 3) - accounting for activities the pH of the solution is =

In general chemistry we taught you to calculate the pH of a solution if you know the H30* concentration. We neglected to talk about the concept of "activity"; when other ions in solution affect the behavior of the your analyte (analytical target) ion. This behavior is what we refer to as the ion activity. Please note that this only applies to ions - charged species - not uncharged molecules. Using the chart of activity versus ionic strength found in chapter 8 (7th edition) of the text, answer the following for a solution that has the following concentrations: [HCI] = 0.0118 and [KNO3] = 0.017 1) - the activity coefficient of H30+ ions = 2) - the activity of H30* ions = 3) - neglecting activities the pH of the solution is = 3) - accounting for activities the pH of the solution is =

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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Each row of the table below describes an aqueous solution at 25 °C. The second column of the table shows the initial components of the solution. • Use the checkboxes in the third column to explain the type of the initial solution. The fourth column describes a change in the solution. • Use the fifth column to predict how the change in the solution will change its pH. alo initial type solution initial components (check all that apply) effect of change on pH change (check one) O acidic pH higher H,0 basic add H C10, O pH lower A O neutral O pH the same acidic O pH higher H,0 add Na ClO, B O basic O pH lower O neutral pH the same O acidic pH higher Н, 0, КОН V basic add H Br O pH lower O neutral O pH the same O acidic O pH higher D Н,о, КОН O basic add K Br O pH lower O neutral pH the same Explanation Check © 2021 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accessibility
Each row of the table below describes an aqueous solution at 25 °C. The second column of the table shows the initial components of the solution. • Use the checkboxes in the third column to explain the type of the initial solution. The fourth column describes a change in the solution. • Use the fifth column to predict how the change in the solution will change its pH. solution initial components initial type (check all that apply) change O acidic A H₂O basic add H Cl O neutral acidic B H₂O, HI basic add NaOH O neutral acidic C H₂O, HI basic add NaI neutral acidic effect of change on pH (check one) O pH higher O pH lower O pH the same O pH higher O pH lower O pH the same O pH higher O pH lower O pH the same O pH higher D H₂O basic add K Cl O pH lower O neutral O pH the same
Each row of the table below describes an aqueous solution at 25 °C. The second column of the table shows the initial components of the solution. • Use the checkboxes in the third column to explain the type of the initial solution. The fourth column describes a change in the solution. • Use the fifth column to predict how the change in the solution will change its pH. initial type solution initial components (check all that apply) O acidic basic neutral A B с D H₂O H₂O, HCI H₂O H₂O, HCI acidic basic neutral acidic basic neutral acidic basic neutral change add HNO3 add KOH add NaNO3 add KCI change on pH (check one) O pH higher O pH lower O pH the same effect pH higher O pH lower pH the same O pH higher O pH lower O pH the same pH higher O pH lower O pH the same
In the laboratory, a general chemistry student measured the pH of a 0.386 M aqueous solution of nitrous acid to be 1.865. Use the information she obtained to determine the Ka for this acid. Ka(experiment) = In the laboratory, a general chemistry student measured the pH of a 0.431 M aqueous solution of hydrofluoric acid to be 1.741. Use the information she obtained to determine the Ka for this acid. Ka(experiment) =
Initial pH: 10.18 Kb: 5.13x10^-5
Consider the “window cleaner” problem, question 2 on page 12 of your lecture notes. Suppose that prior to the titration, the 10 mL sample of window cleaner was diluted by adding about 10 mL of DI water so that the solution could have better contact with the pH meter. What effect would this dilution have on the volume of titrant needed? (a) The amount of titrant would be the same because the moles of analyte was not changed.(b) The amount of titrant would be doubled because the analyte solution volume was doubled.(c) The amount of titrant would be halved because the was diluted by half.
For BrO−, enter an equation that shows how the anion acts as a base. Express your answer as a chemical equation including phases.
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 you would have measured the pH of four different Basic Aqueous Solutions and examined the data for trends. You would have calculated the expected pH of each of the strong base 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 10 to the -14 power = H+ Molarity X OH- Molarity), we 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 pH value 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 power divided by 1.0 X 10 to…
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In a 0.60 mM aqueous solution of acetic acid (CH,CO,H), what is the percentage of acetic acid that is dissociated? You can find some data that is useful for solving this problem in the ALEKS Data resource. Round your answer to 2 significant digits. | % x10
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Part A Water ionizes by the equation =H+(aq) + OH (ag) The extent of the reaction is small in pure water and dilute aqueous solutions. This reaction creates the following relationship between [H*] and [OH ]: What is the H† concentration for an aqueous solution with pOH = 2.73 at 25 °C? Express your answer to two significant figures and include the appropriate units. Kw = [H*][OH¯] • View Available Hint(s) Keep in mind that, like all equilibrium constants, the value of Kw changes with temperature. ? [H*] = Value Units