(a)
Interpretation:
The
pH definition:
The concentration of hydrogen ion is measured using
The
On rearranging, the concentration of hydrogen ion
(a)

Answer to Problem 6D.6E
The
Explanation of Solution
Pyridine is a weak base when it is dissolved in water it ionized as positive and negative ions and it is given below.
The equilibrium expression for the above reaction is given below.
Initial concentration | 0.082 | 0 | 0 |
Change in concentration | -x | +x | +x |
Equilibrium concentration | 0.082-x | x | x |
The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.
Pyridine
Now, the
Therefore, the calculated
The general equilibrium expression to find out the pH of the solution is given below,
Therefore, the calculated
The percentage deprotonation is calculated using the concentration of hydronium ion divided by the initial concentration of lactic acid and the respective equation is given below.
Concentration of
Initial concentration of
Substitute the obtained values in above equation
Therefore, the percentage deprotonation of 0.082 M aqueous pyridine is 0.015%
(b)
Interpretation:
The
Concept introduction:
Refer to part (a).
(b)

Answer to Problem 6D.6E
The
Explanation of Solution
Hydroxylamine is a weak base when it is dissolved in water it ionized as positive and negative ions and it is given below.
The equilibrium expression for the above reaction is given below.
Initial concentration | 0.0103 | 0 | 0 |
Change in concentration | -x | +x | +x |
Equilibrium concentration | 0.0103-x | x | x |
The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.
Nicotine
Now, the
Therefore, the calculated
The general equilibrium expression to find out the pH of the solution is given below,
Therefore, the calculated
The percentage deprotonation is calculated using the concentration of hydronium ion divided by the initial concentration of lactic acid and the respective equation is given below.
Concentration of
Initial concentration of
Substitute the obtained values in above equation
Therefore, the percentage deprotonation of 0.0103 M aqueous nicotine is 1.03%
(c)
Interpretation:
The
Concept introduction:
Refer to part (a).
(c)

Answer to Problem 6D.6E
The
Explanation of Solution
Quinine is a weak base when it is dissolved in water it ionized as positive and negative ions and it is given below.
The equilibrium expression for the above reaction is given below.
Initial concentration | 0.060 | 0 | 0 |
Change in concentration | -x | +x | +x |
Equilibrium concentration | 0.060-x | x | x |
The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.
The
Therefore, the
The
Hence, the
The concentration of hydroxide ion is calculated using the given formula as,
The obtained quinine
Therefore, the concentration of
Now, the
Therefore, the calculated
The general equilibrium expression to find out the pH of the solution is given below,
Therefore, the calculated
The percentage deprotonation is calculated using the concentration of hydronium ion divided by the initial concentration of lactic acid and the respective equation is given below.
Concentration of
Initial concentration of
Substitute the obtained values in above equation
Therefore, the percentage deprotonation of 0.060 M aqueous quinine is 0.743%
(d)
Interpretation:
The
Concept introduction:
Refer to part (a).
(d)

Answer to Problem 6D.6E
The
Explanation of Solution
If strychnine is dissolved in water it ionized as positive and negative ions and it is given below.
The equilibrium expression for the above reaction is given below.
Initial concentration | 0.045 | 0 | 0 |
Change in concentration | -x | +x | +x |
Equilibrium concentration | 0.045-x | x | x |
The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.
Strychnine
Hence, the
The concentration of hydroxide ion is calculated using the given formula as,
The obtained quinine
Therefore, the concentration of
Now, the
Therefore, the calculated
The general equilibrium expression to find out the pH of the solution is given below,
Therefore, the calculated
The percentage deprotonation is calculated using the concentration of hydronium ion divided by the initial concentration of lactic acid and the respective equation is given below.
Concentration of
Initial concentration of
Substitute the obtained values in above equation
Therefore, the percentage deprotonation of 0.045 M aqueous codeine is 0.636%
Want to see more full solutions like this?
Chapter 6 Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
- Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electrons-pushing arrows for the following reaction or mechanistic step(s).arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. I I I H Select to Add Arrows HCI, CH3CH2OHarrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and the follow the arrows to draw the intermediate and product in this reaction or mechanistic step(s).arrow_forward
- Curved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the curved arrows to draw the intermediates and product of the following reaction or mechanistic step(s).arrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the arrows to draw the intermediate and the product in this reaction or mechanistic step(s).arrow_forwardLook at the following pairs of structures carefully to identify them as representing a) completely different compounds, b) compounds that are structural isomers of each other, c) compounds that are geometric isomers of each other, d) conformers of the same compound (part of structure rotated around a single bond) or e) the same structure.arrow_forward
- Given 10.0 g of NaOH, what volume of a 0.100 M solution of H2SO4 would be required to exactly react all the NaOH?arrow_forward3.50 g of Li are combined with 3.50 g of N2. What is the maximum mass of Li3N that can be produced? 6 Li + N2 ---> 2 Li3Narrow_forward3.50 g of Li are combined with 3.50 g of N2. What is the maximum mass of Li3N that can be produced? 6 Li + N2 ---> 2 Li3Narrow_forward
- Concentration Trial1 Concentration of iodide solution (mA) 255.8 Concentration of thiosulfate solution (mM) 47.0 Concentration of hydrogen peroxide solution (mM) 110.1 Temperature of iodide solution ('C) 25.0 Volume of iodide solution (1) used (mL) 10.0 Volume of thiosulfate solution (5:03) used (mL) Volume of DI water used (mL) Volume of hydrogen peroxide solution (H₂O₂) used (mL) 1.0 2.5 7.5 Time (s) 16.9 Dark blue Observations Initial concentration of iodide in reaction (mA) Initial concentration of thiosulfate in reaction (mA) Initial concentration of hydrogen peroxide in reaction (mA) Initial Rate (mA's)arrow_forwardDraw the condensed or line-angle structure for an alkene with the formula C5H10. Note: Avoid selecting cis-/trans- isomers in this exercise. Draw two additional condensed or line-angle structures for alkenes with the formula C5H10. Record the name of the isomers in Data Table 1. Repeat steps for 2 cyclic isomers of C5H10arrow_forwardExplain why the following names of the structures are incorrect. CH2CH3 CH3-C=CH-CH2-CH3 a. 2-ethyl-2-pentene CH3 | CH3-CH-CH2-CH=CH2 b. 2-methyl-4-pentenearrow_forward
- Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning


