Acids Bases and Salts Multiple Choice Questions Organic Chemistry with Answers For the following reaction of hypochlorous acid (HOCl) ...
Acids Bases and Salts Multiple Choice Questions Organic Chemistry with Answers
- For the following reaction of hypochlorous acid (HOCl)
with water,
HOCl
+ H2O <==> H3O+ + OCl-
what would be the effect of adding
sodium hypochlorite (NaOCl) to the reaction at equilibrium?
- The concentrations of both HOCl and H3O+
would increase.
- The concentrations of both HOCl and H3O+
would decrease.
- The concentration of HOCl would increase and the
concentration of H3O+ would decrease.
- The concentration of HOCl would decrease and the
concentration of H3O+ would increase.
- There would be no change because sodium hypochlorite
is a salt without any acidic or basic properties.
- Calculate the molar hydronium ion concentration in a
solution containing 0.23 M hypochlorous acid (HOCl), a monoprotic weak
acid used in bleach solutions. For HOCl, Ka = 2.9 x 10-8.
- Given three separate solutions containing equal
concentrations of formic acid (Ka = 1.7 x 10-4),
phenol (Ka = 1.3 x 10-10), and acetic acid (Ka
= 1.8 x 10-5), select the response below that has the acids
arranged in order of increasing percent dissociation at equilibrium.
- formic < phenol < acetic
- formic < acetic < phenol
- acetic < formic < phenol
- phenol < acetic < formic
- No response is correct.
- From the following choices, select the one that would
be the most basic (least acidic).
- 0.1 M hydrochloric acid (a strong acid)
- 0.1 M acetic acid (a weak acid)
- 0.1 M sodium acetate (the salt of a weak acid)
- 0.1 M ammonium chloride (the salt of a weak base)
- pure water
- If 10 mL a 1.0 x 10-4 M solution of a strong
acid were added to 100 mL each of one solution containing 1.8 x 10-5
M hydrochloric acid and a second solution containing 1.0 M acetic acid (Ka
= 1.8 x 10-5) plus 1.0 M sodium acetate, it is expected that
the:
- pH of both solutions would remain unchanged.
- change in pH would be very large in both solutions.
- change in pH would be the same in both solutions.
- change in pH would be larger in the solution
containing acetic acid and sodium acetate.
- change in pH would be larger in the HCl solution.
- Addition of a strong acid to a solution of acetic acid
at equilibrium (HOAc + H2O <=> H3O+
+ OAc-) would cause the:
- acetate ion concentration to decrease.
- acetate ion concentration to increase.
- pH to increase.
- hydroxide ion concentration to increase.
- None of the above is correct.
- Calculate to a first approximation the molar
concentration of hydronium ion in a 0.171 M solution of benzoic acid
(HOBz, a monoprotic weak acid with Ka = 6.5 x 10-5).
- Given that the acid dissociation constant for benzoic
acid (HOBz) is Ka = 6.5 x 10-5, calculate the basic
dissociation constant, Kb , of the benzoate ion (OBz-).
- Benzoic acid, C6H5CO2H,
is a weak acid (Ka = 6.3 x 10-5). Calculate the
initial concentration (in M) of benzoic acid that is required to produce
an aqueous solution of benzoic acid that has a pH of 2.54.
- Which of the following weak acid dissociation constants
would result in the smallest degree of dissociation?
- Ka = 1.0 x 10-2
- Ka = 1.0 x 10-3
- Ka = 1.0 x 10-4
- Ka = 1.0 x 10-5
- Addition of sodium acetate to an acetic acid solution
at equilibrium will cause:
- no change in H3O+ concentration.
- H3O+ concentration to decrease.
- H3O+ concentration to increase.
- concentrations of all species to increase.
- a decrease in hydroxide concentrations.
- What is the H3O+ concentration in
a 0.17 M solution of a weak acid, HA, with a dissociation constant of 3.21
x 10-6.
- Calculate the hydronium ion concentration in a solution
that contains 0.21 M acetic acid and 0.17 M sodium acetate. For acetic
acid, Ka = 1.8 x 10-5.
- What is the concentration of the HPO42-
ion in a 0.078 M solution of phosphoric acid (H3PO4)?
[Ka1 = 7.5 x 10-3; Ka2 = 6.3 x 10-8;
Ka3 = 4.8 x 10-13]
- Why is it necessary to take the acid-base properties of
water into account when computing the hydronium ion concentration of very
dilute solutions of strong acids?
- The hydroxide ion produced from the dissociation of
water reacts with most of the hydronium ion produced from the acid.
- The dissociation constant for water is larger in
dilute rather than in concentrated solutions of acids.
- The acids do not dissociate completely in dilute
solutions.
- The amount of hydronium ion produced by the
dissociation of water is significant compared to that produced by the
acid.
- The conjugate base of the strong acid reacts with the
hydroxide ion produced from the dissociation of water.
- Hypochlorite ion (OCl-) is the conjugate
base of hypochlorous acid (HOCl, Ka = 3.5 x 10-8).
What is the value of the base ionization equilibrium constant, Kb,
for hypochlorite ion?
- 3.5 x 10-22
- 3.5 x 10-8
- 2.9 x 10-7
- 2.9 x 107
- 4.7 x 109
- Calculate the pH of an aqueous solution prepared to
contain 1.3 x 10-3 M sodium nitrite (NaNO2) if the
acid dissociation equilibrium constant, Ka, for nitrous acid
(HNO2) is 5.1 x 10-4.
- 3.1
- 5.1
- 7.0
- 7.3
- 10.9
- Calculate the carbonate ion concentration in a 0.10 M
solution of the weak acid, carbonic acid (H2CO3).
The stepwise dissociation constants of carbonic acid are Ka1 =
4.5 x 10-7 and Ka2 = 4.7 x 10-11.
- 4.7 x 10-11 M
- 1.0 x 10-7 M
- 4.5 x 10-7 M
- 2.1 x 10-4 M
- 3.5 x 10-3 M
- The very first disinfectant used by Joseph Lister was
called "carbolic acid". This substance is now known as phenol
(PhOH). What is the H3O+ ion concentration in a 0.10
M solution of phenol? [PhOH: Ka = 1.0 x 10-10]
- 1.0 x 10-11
- 3.2 x 10-5
- 5.0 x 10-12
- 3.2 x 10-6
- The sweetener, saccharin, is a weak monoprotic acid
with Ka = 2.1 x 10-12. Calculate the H3O+
concentration in a solution that contains 1.0 x 10-2 mole of
saccharin in 1.00 L of otherwise pure water.
- 1.4 x 10-7
- 1.8 x 10-7
- 2.1 x 10-12
- 2.1 x 10-14
- When would the pH of a solution prepared by adding
sodium formate to formic acid be equal to the pKa of formic
acid, HCO2H?
- when [HCO2H] < [HCO2-]
- when [HCO2H] = [HCO2-]
- when [HCO2H] > [HCO2-]
- the pH of this buffer will never equal the pKa
of formic acid.
- Calculate the pH of a buffer prepared by mixing 0.10
mol of sodium formate and 0.05 mol of formic acid in 1.0 L of solution.
[HCO2H: Ka = 1.8 x 10-4]
- 1.8 x 10-4
- 3.44
- 4.05
- 5.31
- none of these
- For a weak diprotic acid, H2A, for which Ka1
= 2.1 x 10-7 and Ka2 = 4.3 x 10-13, the A2-
ion concentration at equilibrium will be:
- approximately equal to the initial concentration of H2A.
- roughly equal to Ka2.
- roughly equal to the HA- concentration.
- much larger than the HA- concentration.
- approximately equal to the H3O+
concentration.
- Many insects discharge sprays containing weak acids as
a means of defense. For example, some ants discharge a spray that contains
the weak acid, formic acid (HCO2H). Calculate the pH of a 0.14
M solution of formic acid. Ka (HCO2H) = 1.8 x 10-4.
- Calculate the pH of a solution prepared by dissolving
0.20 moles of benzoic acid (abbreviated HOBz) and 0.15 moles of sodium
benzoate (abbreviated NaOBz) in enough water to make 1.0 L of solution.
The acid-dissociation equilibrium constant for benzoic acid is Ka
= 6.3 x 10-5.
- Consider an aqueous solution of a weak acid. Explain
why the contribution of hydronium ion from the dissociation of water
(i.e., [H3O+]water) to the total
hydronium ion concentration is not equal to that for pure water (i.e., 1.0
x 10-7 M).
- Calculate the [OH-] (in M) for an acetic
acid solution (Ka = 1.8 x 10-5) having a pH of 6.32.
- Ascorbic acid is also known as Vitamin C. In a 0.10 M
solution of ascorbic acid 2.8% of the ascorbic acid will dissociate.
Consider the pH you would measure for a 0.25 M solution of ascorbic acid.
Which of the following statements is true?
- The pH would show that the %-dissociation would be the
same in both ascorbic acid solutions.
- The pH would show that the %-dissociation would be
twice as much in the more concentrated acid solutions.
- The pH of the more concentrated solution would be
lower.
- You must know the Ka value for ascorbic
acid before determining which of the above selections is true.
- A buffer can be prepared by mixing:
- a strong acid and its conjugate base.
- a strong base and its conjugate acid.
- a weak acid and its conjugate base.
- a weak acid and a strong acid.
- all responses above are correct.
- Calculate the pH of a solution containing 0.1 M formic
acid (a monoprotic weak acid with Ka = 1.8 x 10-4 )
and 0.1 M sodium formate.
- Calculate the molar hydronium ion concentration, [H3O+],
in a 2.0 x 10-3 M solution of hypoiodious acid (HOI, Ka
= 2.3 x 10-11).
- Calculate the hydroxide ion concentration, [OH-],
in a 0.10 M solution of sodium formate. (For the formate ion, OF-,
Kb = 5.6 x 10-11.)
- The dissociation constant for nitrous acid, HNO2,
is Ka = 5.1 x 10-4. What is the dissociation
constant, Kb, for nitrite ion, NO2-, the
conjugate base of nitrous acid?
- Which of the following solutions would be best to
buffer a solution near pH = 4 ([H3O+] = 1.0 x 10-4).
- 1.0 x 10-4 M HCl
- 1.0 x 10-4 M NaOH
- A solution containing approximately equal
concentrations of formic acid (Ka = 1.8 x 10-4) and
sodium formate.
- A solution containing approximately equal
concentrations of hypochlorous acid (HOCl, Ka = 2.9 x 10-8)
and sodium hypochlorite (NaOCl).
- A solution containing approximately equal
concentrations of ammonia (Kb = 1.8 x 10-5) and
ammonium chloride.
- A 25.00 mL aliquot of a vinegar sample is diluted to
250.00 mL with water. Then a 25.00 mL aliquot of the diluted sample is
titrated with strong base, requiring 22.13 mL of 0.1027 M sodium hydroxide
to reach the endpoint. What is the molar concentration of acid in the
original vinegar sample before dilution?
- Assume that a 25.00-mL aliquot of a solution containing
a monoprotic weak acid is titrated with a standard solution of sodium
hydroxide, requiring 22.42 mL of titrant to reach the end point. At which
milliliter volume listed below will the pH be equal to the pKa
of the weak acid?
- Which of the following solutions would be an acid/base
buffer?
- 0.1 M HCl, a strong acid
- 0.1 M acetic acid, a weak acid
- 0.1 M sodium acetate
- 0.1 M acetic acid plus 0.1 M sodium acetate
- pure water
- Which of the following solutions would be the most
basic?
- 0.1 M HCl, a strong acid
- 0.1 M acetic acid, a weak acid
- 0.1 M sodium acetate
- 0.1 M acetic acid plus 0.1 M sodium acetate
- pure water
- Calculate the hydronium ion concentration, [H3O+],
in a 0.15 M solution of sodium formate. For formic acid, Ka =
1.8 x 10-4.
- In which of the following situations would the weak
acid dissociate to the largest extent (i. e., have the largest percent
dissociation)?
- 0.01 M formic acid, Ka = 1.8 x 10-4
- 0.1 M formic acid, Ka = 1.8 x 10-4
- 0.1 M acetic acid, Ka = 1.8 x 10-5
- 0.01 M formic acid (Ka = 1.8 x 10-4)
plus 0.15 M sodium formate
- 1.0 x 10-4 M phenol, a monoprotic weak acid
with Ka = 1.0 x 10-10
- In which of the following solutions would the
dissociation of water make the largest contribution to the total hydronium
ion concentration?
- 0.01 M formic acid, Ka = 1.8 x 10-4
- 0.1 M formic acid, Ka = 1.8 x 10-4
- 0.1 M acetic acid, Ka = 1.8 x 10-5
- 0.01 M formic acid (Ka = 1.8 x 10-4)
plus 0.15 M sodium formate
- 1.0 x 10-4 M phenol, a monoprotic weak acid
with Ka = 1.0 x 10-10
- A 25.0-mL aliquot of a monoprotic acid solution is
diluted to 100 mL. Then a 25.0-mL aliquot of this solution is titrated
with a standard sodium hydroxide solution, requiring 17.1 mL of 0.107 M
sodium hydroxide to reach the end point. What is the molar concentration
of the acid in the original solution?
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