Concept explainers
Sodium benzoate is a salt of benzoic acid, C6H5COOH. A 0.15 M solution of this salt has a pOH of 5.31 at room temperature.
- a Calculate the value for the equilibrium constant for the reaction
- b What is the Ka value for benzoic acid?
- c Benzoic acid has a low solubility in water. What is its molar solubility if a saturated solution has a pH of 2.83 at room temperature?
(a)
Interpretation:
A 0.15 M solution of sodium benzoate has a pOH of 5.31
The value for the equilibrium constant for the given reaction
Concept Introduction:
Acid ionization constant
The ionization of a weak acid
The equilibrium expression for the above reaction is given below.
Where,
Relationship between
pH definition:
The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion
On rearranging,
Answer to Problem 16.114QP
The value for the equilibrium constant for the given reaction
Explanation of Solution
To Calculate: The value for the equilibrium constant for the given reaction
Given data:
Sodium benzoate is a salt of benzoic acid
A 0.15 M solution of this salt has a pOH of 5.31
The given reaction is
Equilibrium constant for the given reaction:
The hydroxide ion concentration is found from the given pOH as follows,
Construct an equilibrium table for the given reaction:
The initial concentration of
|
|||
Initial |
0.15
0.15-x |
0.00 | 0.00 |
Change |
|
|
|
Equilibrium |
x | x |
Here, x gives the concentration of hydroxide ion that reacted,
Substitute the equilibrium concentrations into the equilibrium-constant expression:
Therefore, the equilibrium-constant for the given reaction is
The value for the equilibrium constant for the given reaction
(b)
Interpretation:
A 0.15 M solution of sodium benzoate has a pOH of 5.31
The
Concept Introduction:
Acid ionization constant
The ionization of a weak acid
The equilibrium expression for the above reaction is given below.
Where,
Relationship between
pH definition:
The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion
On rearranging,
Answer to Problem 16.114QP
The
Explanation of Solution
To Calculate: The
The
The
(c)
Interpretation:
A 0.15 M solution of sodium benzoate has a pOH of 5.31
Benzoic acid has a low solubility in water. If a saturated solution has a pH of 2.83, the molar solubility of the solution has to be calculated.
Concept Introduction:
Acid ionization constant
The ionization of a weak acid
The equilibrium expression for the above reaction is given below.
Where,
Relationship between
pH definition:
The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion
On rearranging,
Answer to Problem 16.114QP
Benzoic acid has a low solubility in water. If a saturated solution has a pH of 2.83, the molar solubility of the solution is 0.036 M
Explanation of Solution
To Calculate: Benzoic acid has a low solubility in water. If a saturated solution has a pH of 2.83, the molar solubility of the solution
Use equilibrium-constant expression and
The reaction is
From the pH, calculate the hydronium ion and benzoate ion concentrations.
The concentration of benzoate ion is same as the concentration of hydronium ion
Substitute the above values in equilibrium-constant expression.
The molar solubility will be the total dissolved benzoic acid, molecular and dissociated:
Benzoic acid has a low solubility in water. If a saturated solution has a pH of 2.83, the molar solubility of the solution was calculated as 0.036 M
Want to see more full solutions like this?
Chapter 16 Solutions
Lab Manual Experiments in General Chemistry
- Methylammonium chloride is a salt of methylamine, CH3NH2. A 0.10 M solution of this salt has a pH of 5.82. a Calculate the value for the equilibrium constant for the reaction CH3NH3++H2OCH3NH2+H3O+ b What is the Kb value for methylamine? c What is the pH of a solution in which 0.450 mol of solid methylammonium chloride is added to 1.00 L of a 0.250 M solution of methylamine? Assume no volume change.arrow_forwardEstimate the pH that results when the following two solutions are mixed. a) 50 mL of 0.3 M CH3COOH and 50 mL of 0.4 M KOH b) 100 mL of 0.3 M CH3COOH and 50 mL of 0.4 M NaOH c) 150 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2 d) 200 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2arrow_forwardFor conjugate acidbase pairs, how are Ka and Kb related? Consider the reaction of acetic acid in water CH3CO2H(aq)+H2O(l)CH3CO2(aq)+H3O+(aq) where Ka = 1.8 105 a. Which two bases are competing for the proton? b. Which is the stronger base? c. In light of your answer to part b. why do we classify the acetate ion (CH3CO2) as a weak base? Use an appropriate reaction to justify your answer. In general, as base strength increases, conjugate acid strength decreases. Explain why the conjugate acid of the weak base NH3 is a weak acid. To summarize, the conjugate base of a weak acid is a weak base and the conjugate acid of a weak base is a weak acid (weak gives you weak). Assuming Ka for a monoprotic strong acid is 1 106, calculate Kb for the conjugate base of this strong acid. Why do conjugate bases of strong acids have no basic properties in water? List the conjugate bases of the six common strong acids. To tie it all together, some instructors have students think of Li+, K+, Rb+, Cs+, Ca2+, Sr2+, and Ba2+ as the conjugate acids of the strong bases LiOH, KOH. RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. Although not technically correct, the conjugate acid strength of these cations is similar to the conjugate base strength of the strong acids. That is, these cations have no acidic properties in water; similarly, the conjugate bases of strong acids have no basic properties (strong gives you worthless). Fill in the blanks with the correct response. The conjugate base of a weak acid is a_____base. The conjugate acid of a weak base is a_____acid. The conjugate base of a strong acid is a_____base. The conjugate acid of a strong base is a_____ acid. (Hint: Weak gives you weak and strong gives you worthless.)arrow_forward
- The base ethylamine (CH3CH2NH2) has a Kb of. A closely related base, ethanolamine(HOCH2CH2NH2), has a Kb of 3.2105. (a) Which of the two bases is stronger? (b) Calculate the pH of a 0.10M solution of the strong base?arrow_forwardCalculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH3NH2 and 0.10 M in C5H5N (Kb=1.7109).arrow_forwardWrite an equation for each of the following buffering actions. a. the response of a HPO42/PO43 buffer to the addition of OH ions b. the response of a HF/F buffer to the addition of OH ions c. the response of a HCN/CN buffer to the addition of H3O+ ions d. the response of a H3PO4/H2PO4 buffer to the addition of H3O+ ionsarrow_forward
- . The concepts of acid-base equilibria were developed in this chapter for aqueous solutions (in aqueous solutions, water is the solvent and is intimately involved in the equilibria). However, the Brønsted-Lowry acid-base theory can be extended easily to other solvents. One such solvent that has been investigated in depth is liquid ammonia. NH3. a. Write a chemical equation indicating how HCl behaves as an acid in liquid ammonia. b. Write a chemical equation indicating how OH- behaves as a base in liquid ammonia.arrow_forwardIn each of the following acid-base reactions, identify the Brnsted acid and base on the left and their conjugate partners on the right. (a) C2H5N(aq) + CH3CO2H(aq) C5H5NH+(aq) + CH3CO2(aq) (b) N2H4(aq) + HSO4(aq) N2H5+(aq) + SO42(aq) (c) [Al(H2O)6]3+ (aq) + OH(aq) [Al(H2O)5OH]2+ (aq) + H2O+()arrow_forwardWhat is the pH of a solution obtained by adding 13.0 g of NaOH to 795 mL of a 0.200 M solution of Sr(OH)2? Assume no volume change after NaOH is added.arrow_forward
- The active ingredient formed by aspirin in the body is salicylic acid, C6H4OH(CO2H). The carboxyl group. (-CO2H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-M aqueous solution of C6H4OH(CO2H).arrow_forwardCalculate [CO32] in a 0.010-M solution of CO2 in water (usually written as H2CO3). If all the CO32 in this solution comes from the reaction HCO3(aq)H+(aq)+CO32(aq) what percentage of the H+ ions in the solution is a result of the dissociation of HCO3? When acid is added to a solution of sodium hydrogen carbonate (NaHCO3), vigorous bubbling occurs. How is this reaction related to the existence of carbonic acid (H2CO3) molecules in aqueous solution?arrow_forwardWeak base B has a pKb of 6.78 and weak acid HA has a pKa of 5.12. a Which is the stronger base, B or A? b Which is the stronger acid, HA or BH+? c Consider the following reaction: B(aq)+HA(aq)BH+(aq)+A(aq) Based on the information about the acid/base strengths for the species in this reaction, is this reaction favored to proceed more to the right or more to the left? Why? d An aqueous solution is made in which the concentration of weak base B is one half the concentration of its acidic salt, BHCl, where BH+ is the conjugate weak add of B. Calculate the pH of the solution. e An aqueous solution is made in which the concentration of weak acid HA twice the concentration of the sodium salt of the weak acid, NaA. Calculate the pH of the solution. f Assume the conjugate pairs B/BH+ and HA/A are capable of being used as color-based end point indicators in acidbase titrations, where B is the base form indicator and BH is the acid form indicator, and HA is the acid form indicator and A is the base form indicator. Select the indicator pair that would be best to use in each of the following titrations: (1) Titration of a strong acid with a strong base. (i) B/BH+ (ii) HA/A (2) Titration of a weak base with a strong acid. (i) B/BH+ (ii) HA/Aarrow_forward
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningIntroductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage LearningGeneral, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning