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1. Calculate the hydronium ion concentration of a solution with a pH of-1.07. 2. What are the pOH and the pH of a 0.0125M solution of potassium hydroxide, KOH? 3. The hydronium ion concentration of vinegar is approximately 4 x 10 M.

1. Calculate the hydronium ion concentration of a solution with a pH of-1.07. 2. What are the pOH and the pH of a 0.0125M solution of potassium hydroxide, KOH? 3. The hydronium ion concentration of vinegar is approximately 4 x 10 M.

Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter17: Equilibrium
Section: Chapter Questions
Problem 1ALQ: Consider an equilibrium mixture of four chemicals (A. B. C. and D. all gases) reacting in a closed...
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SOLVING TIME ONLY NUMBERS 1 -3

Like all equilibrium reactions, this reaction has an equilibrium constant. Because this is a special equilibrium constant, specific to the self-ionization of water, it is denoted Kw, it hasa value of 1.0 x 10-14. If we write out the actual equilibrium expression for Kw, we get the following: Kw = [H+][OH-] = 1.0 x 10-14 %3D However, because H* and OH are formed in a 1:1 molar ratio, we have: [H+] = [OH-] = V1.0 x 10-14 1.0 x 10-7M %3D %3D Now, note the definition of pH and pOH: [+H]601- = Hd Hd- %3D %3D Hod- =[-HO] [-HO]B01- = HOd [H30+] The term pH refers to the "potential of hydrogen ion." It was proposed by Danish biochemist Soren Sorensen in 1909 so that there could be a more convenient way to describehydronium and hydroxide ion concentrations in aqueous solutions since both concentrations tend to be extremely small. If we plug in the above value into our equation for pH, we find that: pH = - log(1.0 x 107) = 7.0 pOH =-log(1.0 x 107) = 7.0 Here we have the reason why neutral water has a pH of 7.0; it represents the condition at %3D %3D which the concentrations of H and OH are exactly equal in solution. Since how the equilibrium constant can be expressed was already established, if we take the negative logarithm of both sides of this equation, we get the following: = (*y) = - log[H+] +-log[OH-] -log (Kw) ([-HO][+H]) 801- -log (Kw) pKw = pH + p0H %3D However, because we know that pKw= 14, we can establish the following relationship: |3D This relationship always holds true for any aqueous solution, regardless of its level ofacidity or alkalinity. Utilizing this equation is a convenient way to quickly determine pOH frompH and vice versa, as well as to determine hydroxide concentration given hydrogen concentration, or vice versa. Activity 1: Solving Time! Directions: Solve the following problem. Show your complete solutions. 1. Calculate the hydronium ion concentration of a solution with a pH of-1.07. 2. What are the pOH and the pH of a 0.0125M solution of potassium hydroxide, KOH? 3. The hydronium ion concentration of vinegar is approximately 4 x 10 M. Activity 2: Is it Acidic or Basic? Directions: Complete the table by identifying whether the sample solution mentioned is either acidic or basic and rank it all from the most acidic to most basic (1-5), with 5 as the most basic. SOLUTIONS Acidic or Basic Rank 1.1 М HCI 1. 2. Ocean Water 3.1 M NAOH 4. Rainwater 2. 5. Baking soda
Transcribed Image Text:Like all equilibrium reactions, this reaction has an equilibrium constant. Because this is a special equilibrium constant, specific to the self-ionization of water, it is denoted Kw, it hasa value of 1.0 x 10-14. If we write out the actual equilibrium expression for Kw, we get the following: Kw = [H+][OH-] = 1.0 x 10-14 %3D However, because H* and OH are formed in a 1:1 molar ratio, we have: [H+] = [OH-] = V1.0 x 10-14 1.0 x 10-7M %3D %3D Now, note the definition of pH and pOH: [+H]601- = Hd Hd- %3D %3D Hod- =[-HO] [-HO]B01- = HOd [H30+] The term pH refers to the "potential of hydrogen ion." It was proposed by Danish biochemist Soren Sorensen in 1909 so that there could be a more convenient way to describehydronium and hydroxide ion concentrations in aqueous solutions since both concentrations tend to be extremely small. If we plug in the above value into our equation for pH, we find that: pH = - log(1.0 x 107) = 7.0 pOH =-log(1.0 x 107) = 7.0 Here we have the reason why neutral water has a pH of 7.0; it represents the condition at %3D %3D which the concentrations of H and OH are exactly equal in solution. Since how the equilibrium constant can be expressed was already established, if we take the negative logarithm of both sides of this equation, we get the following: = (*y) = - log[H+] +-log[OH-] -log (Kw) ([-HO][+H]) 801- -log (Kw) pKw = pH + p0H %3D However, because we know that pKw= 14, we can establish the following relationship: |3D This relationship always holds true for any aqueous solution, regardless of its level ofacidity or alkalinity. Utilizing this equation is a convenient way to quickly determine pOH frompH and vice versa, as well as to determine hydroxide concentration given hydrogen concentration, or vice versa. Activity 1: Solving Time! Directions: Solve the following problem. Show your complete solutions. 1. Calculate the hydronium ion concentration of a solution with a pH of-1.07. 2. What are the pOH and the pH of a 0.0125M solution of potassium hydroxide, KOH? 3. The hydronium ion concentration of vinegar is approximately 4 x 10 M. Activity 2: Is it Acidic or Basic? Directions: Complete the table by identifying whether the sample solution mentioned is either acidic or basic and rank it all from the most acidic to most basic (1-5), with 5 as the most basic. SOLUTIONS Acidic or Basic Rank 1.1 М HCI 1. 2. Ocean Water 3.1 M NAOH 4. Rainwater 2. 5. Baking soda
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