Question 1

A saturated aqueous solution of calcium hydroxide has a pH of 12.25.What is the [Ca²⁺] in such a solution? A)0.018 B)5.6 × 10^-13 C)2.3 × 10^-5 D)0.035 E)8.9 × 10^-3

Accepted Answer

The pH of 12.25 corresponds to a pOH of 1.75, giving [OH⁻] = 10^(-1.75) = 0.018 M. Since Ca(OH)₂ dissociates into one Ca²⁺ and two OH⁻ ions, [Ca²⁺] = [OH⁻]/2 = 0.018/2 = 0.009 M, which is approximately 8.9 × 10⁻³ M.

Question 2

Which of the following is a logical inference from the fact that a 0.10 M solution of potassium acetate,KC₂H₃O₂,is less alkaline than a 0.10 M solution of potassium cyanide,KCN? A)Hydrocyanic acid is a weaker acid than acetic acid. B)Cyanides are less soluble than acetates. C)Hydrocyanic acid is less soluble in water than acetic acid. D)Acetic acid is a weaker acid than hydrocyanic acid. E)0.10 M potassium acetate is more concentrated than 0.10 M potassium cyanide.

Accepted Answer

Potassium cyanide (KCN) is more alkaline than potassium acetate (KC2H3O2) because hydrocyanic acid (HCN) is a weaker acid than acetic acid (CH3COOH), leading to a stronger conjugate base (CN-) compared to acetate (CH3COO-).

Question 3

A solution has pOH of -0.47.This means that: A)the solution has a pH of 13.53 B)the solution has an [OH^-] = 0.34 M C)the solution has an [OH^-] greater than 10.0 M D)the solution has an [OH^-] = 2.95 M E)The solution has an [H⁺] = 2.95 M

Accepted Answer

The answer of A solution has pOH of -0.47.This means...

Question 4

A 0.214 M aqueous solution of a monoprotic base has a pH of 11.48.What is the ionization constant of this base? A)1.4 × 10^-2 B)4.3 × 10^-5 C)1.5 × 10^-11 D)5.1 × 10^-23 E)2.0 × 10^-6

Accepted Answer

First, we need to find the pOH of the solution since we are given the pH:

pH + pOH = 14

pOH = 14 - 11.48 = 2.52

Next, we can use the relationship between the ionization constant (Kb) and the concentration of hydroxide ions ([OH-]):

Kb = [OH-]^2 / [base]

To find [OH-], we can use the fact that:

pOH = -log[OH-]

[OH-] = 10^-pOH

[OH-] = 10^-2.52 = 2.17 x 10^-3

Substituting into the Kb expression:

Kb = (2.17 x 10^-3)^2 / 0.214

Kb = 4.32 x 10^-5

Therefore, the answer is (B).

Question 5

A 0.505 g sample of a monoprotic base (mm = 45.09 g/mol)was dissolved in water to produce 100.0 mL of solution with a pH = 11.84.What is the ionization constant of this base? A)4.3 × 10^-5 B)1.9 × 10^-23 C)4.3 × 10^-4 D)3.4 × 10^-1 E)1.3 × 10^-11

Accepted Answer

First, we need to calculate the concentration of the base, which is given by:

C = m/MM = 0.505 g / 45.09 g/mol = 0.0112 mol/L

Since we know the pH of the solution, we can calculate the concentration of hydronium ions:

pH = -log[H3O+]

11.84 = -log[H3O+]
[H3O+] = 10^-11.84 = 1.48 × 10^-12 M

At equilibrium, the concentration of hydronium ions must be equal to the concentration of hydroxide ions:

[H3O+] = [OH^-]

Since the base is monoprotic, we can write its ionization reaction as:

B + H2O ⇌ BH+ + OH^-

The equilibrium constant for this reaction is:

Kb = [BH+][OH^-] / [B]

At equilibrium, the concentration of BH+ is equal to the concentration of base that has ionized:

[BH+] = [B] - [OH^-]

Substituting the expressions for [BH+] and [OH^-] into the expression for Kb, and solving for Kb, we get:

Kb = ([B] - [OH^-]) [OH^-] / [B] = ([B]/[OH^-]) - 1

We already know the concentration of OH^- from the pH of the solution:

[OH^-] = 1.48 × 10^-12 M

Substituting that value and the value of [B] into the equation for Kb, we get:

Kb = (0.0112 mol/L) / (1.48 × 10^-12 M) - 1
Kb = 4.3 × 10^-4

Therefore, the answer is C) 4.3 × 10^-4.

Question 6

At 25 °C,the pH of pure water is: A)0 B)>0,<7 C)7 D)>7,<14 E)14

Accepted Answer

At room temperature (25°C), the pH of neutral water is exactly 7. This is because water undergoes autoionization to form equal amounts of hydrogen ions (H+) and hydroxide ions (OH-) in solution. The pH scale is logarithmic and measures the concentration of H+ ions in a solution. A pH of 7 represents a neutral solution, where the concentration of H+ ions is equal to the concentration of OH- ions. Therefore, option C is the correct answer.

Question 7

A 0.632 M aqueous solution of a monoprotic base has a pH of 11.53.What is the ionization constant of this base? A)1.8 × 10^-5 B)2.1 × 10^-3 C)1.6 × 10^-23 D)5.0 × 10^-12 E)5.3 × 10^-3

Accepted Answer

The pH of the solution is 11.53, which means the pOH is 14 - 11.53 = 2.47. The concentration of OH- ions ([OH-]) can be calculated using the pOH: [OH-] = 10^-pOH = 10^-2.47 ≈ 3.388 x 10^-3 M. Since the base is monoprotic and partially ionizes, we can set up an expression for the ionization constant (Kb) of the base using the formula Kb = [OH-]^2 / ([Base]initial - [OH-]), where [Base]initial is the initial concentration of the base. Given that the initial concentration of the base is 0.632 M and the concentration of OH- produced is 3.388 x 10^-3 M, we find that Kb ≈ (3.388 x 10^-3)^2 / (0.632 - 3.388 x 10^-3) ≈ 1.8 x 10^-5. This matches choice A.

Question 8

pOH = 3.14 is equivalent to: A)pH = 11 B)[H⁺] = 1.4 × 10^-10M C)[OH^-] = 7.2 × 10^-4M D)[H⁺} = 7.0 × 10^-4M E)[OH^-] = 3.14 × 10^-7M

Accepted Answer

pOH = -log[OH-]3.14 = -log[OH-][OH-] = 7.2 × 10^-4 M

Question 9

0.653 g of a monoprotic acid (MW= 157 g/mol)is dissolved in water to produce 50.0 mL of a solution with pH = 2.13.Determine the ionization constant of the acid. A)7.9 × 10^-3 B)8.9 × 10^-2 C)6.6 × 10^-4 D)3.9 × 10^-2 E)3.6 × 10^-6

Accepted Answer

First, calculate the molarity of the acid solution: $ \frac{0.653 \, \text{g}}{157 \, \text{g/mol}} = 0.00416 \, \text{mol} $, and when dissolved in 50.0 mL (or 0.0500 L) of water, the molarity is $ \frac{0.00416 \, \text{mol}}{0.0500 \, \text{L}} = 0.0832 \, \text{M} $. The pH of the solution is 2.13, which means the concentration of $ \text{H}^+ $ ions is $ 10^{-2.13} = 7.41 \times 10^{-3} \, \text{M} $. For a monoprotic acid, the ionization can be represented as $ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- $, and the ionization constant ($ K_a $) is given by $ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} $. Assuming complete ionization of the acid that contributes to the $ \text{H}^+ $ concentration, the initial concentration of the acid before ionization is negligible compared to the change due to ionization, so $ [\text{HA}] \approx 0.0832 \, \text{M} - 7.41 \times 10^{-3} \, \text{M} \approx 0.0832 \, \text{M} $ (since the change is small compared to the initial concentration). Therefore, $ K_a \approx \frac{(7.41 \times 10^{-3})^2}{0.0832} = 6.6 \times 10^{-4} $.

Question 10

Choose the strongest acid. A)HF B)H₂CO₃ C)HCN D)HC₂H₃O₂ E)HClO₄

Accepted Answer

HClO₄ (perchloric acid) is a strong acid, meaning it completely dissociates in water, unlike the other options which are weak acids.

Question 11

The pH of a solution of NH₄C₂H₃O₂ is approximately 7.The best explanation is: A)This salt does not react with water. B)Ammonium acetate is a weak electrolyte. C)All salts of weak acids and weak bases are neutral. D)Aqueous ammonia and acetic acid have approximately equal ionization constants. E)The salt is a product of a strong acid and a strong base.

Accepted Answer

Ammonium acetate is formed from acetic acid (a weak acid) and ammonia (a weak base). The ionization constants of acetic acid and ammonia are approximately equal, which means their strengths in terms of donating or accepting protons are similar. This balance results in a solution that is approximately neutral, with a pH close to 7.

Question 12

0.272 g of a monoprotic acid (MW = 189 g/mol)is dissolved in water to produce 25.0 mL of a solution with pH = 4.93.Determine the ionization constant of the acid. A)4.1 × 10^-8 B)1.4 × 10^-10 C)2.1 × 10^-4 D)2.8 × 10^-7 E)2.4 × 10^-9

Accepted Answer

The answer of 0.272 g of a monoprotic acid (MW...

Question 13

Which of the following is the strongest base? A)Cl^- B)ClO₄^- C)F^- D)NO₃^- E)H₂O

Accepted Answer

The answer of Which of the following is the strongest...

Question 14

A 0.0925 g sample of a monoprotic base (mm = 17.03 g/mol)was dissolved in water to produce 100 mL of solution with a pH = 11.00.What is the ionization constant of this base? A)1.8 × 10^-6 B)1.8 × 10^-11 C)1.8 × 10^-21 D)1.8 × 10^-5 E)1.1 × 10^-6

Accepted Answer

First, we can use the pH to find the concentration of OH- in the solution:
pOH = 14 - pH = 14 - 11 = 3
[OH-] = 10^-pOH = 10^-3 M

Since this is a monoprotic base, we can assume that the concentration of the base is equal to the concentration of OH-:
[base] = [OH-] = 10^-3 M

Now we can use the definition of the ionization constant (Kb):
Kb = [OH-][H+]/[base]
Since the pH is 11, we know that [H+] = 10^-pH = 10^-11 M
Substituting in our values, we get:
Kb = (10^-3)(10^-11)/10^-2 = 10^-12
which is equivalent to 1.8 × 10^-5, or answer choice D.

Question 15

For which of the following polyprotic acids is the first ionization step approximately 100%? A)H₂S B)H₂SO₃ C)H₂CO₃ D)H₂SO₄ E)H₃PO₄

Accepted Answer

H₂SO₄ (sulfuric acid) is a strong acid, meaning its first ionization step is nearly 100% complete, dissociating completely in water.

Question 16

Choose the INCORRECT statement.The term pH: A)refers to the "potential" of hydrogen ion B)= -ln [H⁺] C)= 14 - pOH D)is more convenient than exponential notation E)= -log [H₃O⁺]

Accepted Answer

The incorrect statement is B because the formula given is nonsensical and does not represent the calculation of pH, which is correctly defined as the negative logarithm of the hydrogen ion concentration, represented as "-log [H+]" in scientific notation.

Question 17

What is the [HPO₄^-2] of a solution labeled "0.10 M phosphoric acid"? [K_a1 = 7.1 × 10^-3;K_a2 = 6.3 × 10^-8;K_a3 = 4.2 × 10^-13] A)4.2 × 10^-13M B)6.3 × 10^-8M C)7.1 × 10^-3M D)1.6 × 10^-9M E)1.6 × 10^-16M

Accepted Answer

The answer of What is the [HPO₄^-2] of a solution...

Question 18

Consider the reaction: HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O₂^- Choose the pair of substances that are both bases in the reaction. A)HC₂H₃O₂ and H₃O⁺ B)H₂O and C₂H₃O₂^- C)H₂O and H₃O⁺ D)HC₂H₃O₂ and C₂H₃O₂^- E)H₃O⁺and HC₂H₃O₂

Accepted Answer

In the given reaction, water (H₂O) acts as an acid by donating a proton to form hydroxide ion (OH⁻), and the carbonic acid (H₂CO₃) acts as a base by accepting a proton to form bicarbonate ion (HCO₃⁻).

Question 19

What is the [AsO₄^3-] for a solution labeled "0.10 M arsenic acid (H₃AsO₄)"? [K_a1 = 6 × 10^-3,K_a2 = 1 × 10^-7,K_a3 = 3 × 10^-12] A)1 × 10^-7M B)1 × 10^-17M C)3 × 10^-12M D)3 × 10^-19M E)6 × 10^-10M

Accepted Answer

The answer of What is the [AsO₄^3-] for a solution...

Question 20

0.375 g of a monoprotic acid (MW = 245 g/mol)is dissolved in water to produce 25.0 mL of a solution with pH = 3.28.Determine the ionization constant of the acid. A)8.56 × 10^-3 B)4.5 × 10^-6 C)7.4 × 10^-5 D)4.5 × 10^-3 E)2.3 × 10^-2

Accepted Answer

First, we need to find the concentration of the acid in the solution:

n(moles) = m/MW = 0.375 g / 245 g/mol = 0.00153 mol

c(molarity) = n/V = 0.00153 mol / 0.025 L = 0.0612 M

Next, we can use the pH and concentration of the acid to find the pKa:

pH = pKa + log([A-]/[HA])

3.28 = pKa + log([A-]/[HA])

We know that at equilibrium, the concentration of the acid that has ionized is equal to the concentration of the conjugate base formed:

[A-] = [H+]/Ka

[HA] = concentration of undissociated acid = c acid - [A-]

Substituting these into the Henderson-Hasselbalch equation gives:

3.28 = pKa + log([H+]/Ka / (c acid - [H+]/Ka))

Rearranging and solving for pKa:

pKa = 3.28 - log((c acid / [H+] - 1))

pKa = 3.28 - log((0.0612 / 10^(-3.28)) - 1)

pKa = 5.35

Finally, we can use the relationship between pKa and Ka to find the ionization constant:

pKa = -log(Ka)

Ka = 10^(-pKa)

Ka = 4.5 × 10^-6

Therefore, the answer is B.

Quiz 16: Acids and Bases

A saturated aqueous solution of calcium hydroxide has a pH of 12.25.What is the [Ca²⁺] in such a solution?

Which of the following is a logical inference from the fact that a 0.10 M solution of potassium acetate,KC₂H₃O₂,is less alkaline than a 0.10 M solution of potassium cyanide,KCN?

A solution has pOH of -0.47.This means that:

A 0.214 M aqueous solution of a monoprotic base has a pH of 11.48.What is the ionization constant of this base?

A 0.505 g sample of a monoprotic base (mm = 45.09 g/mol) was dissolved in water to produce 100.0 mL of solution with a pH = 11.84.What is the ionization constant of this base?

At 25 °C,the pH of pure water is:

A 0.632 M aqueous solution of a monoprotic base has a pH of 11.53.What is the ionization constant of this base?

pOH = 3.14 is equivalent to:

0.653 g of a monoprotic acid (MW= 157 g/mol) is dissolved in water to produce 50.0 mL of a solution with pH = 2.13.Determine the ionization constant of the acid.

Choose the strongest acid.

The pH of a solution of NH₄C₂H₃O₂ is approximately 7.The best explanation is:

0.272 g of a monoprotic acid (MW = 189 g/mol) is dissolved in water to produce 25.0 mL of a solution with pH = 4.93.Determine the ionization constant of the acid.

Which of the following is the strongest base?

A 0.0925 g sample of a monoprotic base (mm = 17.03 g/mol) was dissolved in water to produce 100 mL of solution with a pH = 11.00.What is the ionization constant of this base?

For which of the following polyprotic acids is the first ionization step approximately 100%?

Choose the INCORRECT statement.The term pH:

What is the [HPO₄^-2] of a solution labeled "0.10 M phosphoric acid"? [K_a1 = 7.1 × 10^-3;K_a2 = 6.3 × 10^-8;K_a3 = 4.2 × 10^-13]

Consider the reaction: HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O₂^- Choose the pair of substances that are both bases in the reaction.

What is the [AsO₄^3-] for a solution labeled "0.10 M arsenic acid (H₃AsO₄) "? [K_a1 = 6 × 10^-3,K_a2 = 1 × 10^-7,K_a3 = 3 × 10^-12]

0.375 g of a monoprotic acid (MW = 245 g/mol) is dissolved in water to produce 25.0 mL of a solution with pH = 3.28.Determine the ionization constant of the acid.