Question 1

A buffer solution with a pH of 4.63 is prepared with 0.14 M formic acid and ________ M sodium formate. The K_a of formic acid is 1.8 × 10^-4. A)1.1 B)2.1 C)5.4 × 10^-6 D)3.0 × 10^-8 E)0.54

Accepted Answer

The correct answer is A because the pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid. In this case, the pKa of formic acid is 1.8 × 10-4, and the pH of the buffer solution is 4.63. Therefore, log([A-]/[HA]) = 4.63 - 1.8 × 10-4 = 2.83. This means that the concentration of the conjugate base is 2.83 times greater than the concentration of the acid. Since the total concentration of the buffer solution is 0.14 M, the concentration of the conjugate base is 0.14 M / 2.83 = 0.05 M. Therefore, the concentration of sodium formate is 0.05 M.

Question 2

A buffer solution with a pH of 4.78 is prepared with ________ M formic acid and 0.90 M sodium formate. The K_a of formic acid is 1.8 × 10^-4. A)0.083 B)0.17 C)3.3 × 10³ D)9.8 E)0.041

Accepted Answer

Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa = -log(Ka). For formic acid, pKa = -log(1.8 × 10^-4) ≈ 3.74. Plugging in the values: 4.78 = 3.74 + log(0.90/[HA]). Solving for [HA], we get approximately 0.083 M.

Question 3

Calculate the pH of a solution prepared by dissolving 0.850 mol of NH₃ and 0.300 mol of NH₄Cl in water sufficient to yield 1.00 L of solution. The K_b of ammonia is 1.77 × 10^-5. A)5.204 B)4.300 C)9.700 D)8.781 E)8.796

Accepted Answer

The answer of Calculate the pH of a solution prepared...

Question 4

The pH of a solution that contains 0.800 M acetic acid (K_a = 1.76 × 10^-⁵)and 0.172 M sodium acetate is ________. A)4.087 B)5.422 C)8.578 D)8.370 E)9.913

Accepted Answer

The solution is a buffer system consisting of acetic acid and its conjugate base, sodium acetate. The pH can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Here, pKa = -log(1.76 × 10^-5) = 4.755, [A-] = 0.172 M, and [HA] = 0.800 M. Substituting these values gives pH = 4.755 + log(0.172/0.800) = 4.087.

Question 5

A solution is prepared by dissolving 0.23 mol of hypochlorous acid and 0.27 mol of sodium hypochlorite in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly. The pH does not decrease drastically because the HCl reacts with the ________ present in the buffer solution. The K_a of hypochlorous acid is 1.36 × 10^-3. A)H₂O B)H₃O⁺ C)hypochlorite ion D)hypochlorous acid E)This is a buffer solution: the pH does not change upon addition of acid or base.

Accepted Answer

The HCl reacts with the hypochlorite ion (ClO⁻) present in the buffer solution, which helps to resist drastic changes in pH.

Question 6

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.
A)0.0735
B)0.0762
C)0.0980
D)0.0709
E)0.00253

Accepted Answer

Before the addition of HCl, we can assume that the dissociation of hydrogen fluoride is negligible so the concentration of fluoride ions is 0.100 M.

After the addition of 9.00 mL of 0.0100 M HCl, the total volume of the solution becomes 34.0 mL. The moles of HCl added can be calculated as:

(9.00 mL) x (0.0100 mol/L HCl) = 0.0900 mmol HCl

This HCl will react with the fluoride ions to form more hydrogen fluoride:

HCl + F- → HF + Cl-

The reaction will consume the same amount of moles of fluoride ions as HCl added, assuming complete reaction. So, the concentration of fluoride ions will decrease by:

0.0900 mmol / (0.0340 L) = 2.65 x 10^-3 M

The final concentration of fluoride ions will be:

0.100 M - 2.65 x 10^-3 M = 0.0709 M

Thus, the answer is D) 0.0709.

Question 7

A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The K_a of benzoic acid is 4.50 × 10^-4. A)11.20 B)9.80 C)4.20 D)7.00 E)8.54

Accepted Answer

At the equivalence point, benzoic acid is completely converted to its conjugate base, benzoate ion. The pH is determined by the hydrolysis of the benzoate ion. Using the Kb calculated from Kw/Ka, the pH is approximately 8.54.

Question 8

Calculate the pH of a solution prepared by dissolving 0.250 mol of benzoic acid (C₇H₅O₂H)and 0.150 mol of sodium benzoate (NaC₇H₅O₂)in water sufficient to yield 1.00 L of solution. The K_a of benzoic acid is 6.50 × 10^-5. A)4.409 B)3.965 C)10.035 D)9.591 E)5.190

Accepted Answer

We can use the Henderson-Hasselbalch equation to solve this problem:

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

where pKa is the dissociation constant of benzoic acid, A- is the concentration of benzoate ions, and HA is the concentration of benzoic acid.

First, we need to calculate the concentrations of A- and HA in the solution.

[A-] = 0.150 mol / 1.00 L = 0.150 M

[HA] = 0.250 mol / 1.00 L = 0.250 M

Next, we need to calculate the pKa of benzoic acid.

pKa = -log(Ka) = -log(6.50 × 10-5) = 4.19

Now we can plug in the values into the Henderson-Hasselbalch equation:

pH = 4.19 + log(0.150 / 0.250) = 3.965

Therefore, the pH of the solution is 3.965. The best choice is B.

Question 9

What is the pH of a buffer solution that is 0.172 M in hypochlorous acid (HClO)and 0.131 M in sodium hypochlorite? The K_a of hypochlorous acid is 3.8 × 10^-⁸. A)14.12 B)6.70 C)9.07 D)7.54 E)7.30

Accepted Answer

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Here, pKa = -log(3.8 × 10^-8) ≈ 7.42. Plugging in the concentrations, pH = 7.42 + log(0.131/0.172) ≈ 7.30.

Question 10

How many milliliters of 0.0839 M NaOH are required to titrate 25.0 mL of 0.0990 M HBr to the equivalence point?
A)29.5
B)0.332
C)4.57
D)0.208
E)21.2

Accepted Answer

The correct answer is A because the number of moles of NaOH required to titrate the HBr is equal to the number of moles of HBr present in the solution. The number of moles of HBr is equal to the molarity of the HBr solution multiplied by the volume of the solution in liters. The number of moles of NaOH required to titrate the HBr is equal to the molarity of the NaOH solution multiplied by the volume of the solution in liters. The volume of the NaOH solution required to titrate the HBr is equal to the number of moles of NaOH required to titrate the HBr divided by the molarity of the NaOH solution.

Question 11

Calculate the pH of a solution prepared by dissolving 0.270 mol of formic acid (HCO₂H)and 0.260 mol of sodium formate (NaCO₂H)in water sufficient to yield 1.00 L of solution. The K_a of formic acid is 1.77 × 10^-4. A)2.099 B)10.264 C)3.736 D)2.307 E)3.952

Accepted Answer

First, we need to find the concentration of formic acid and formate ion in the solution:
[HCOS1U1B12S1U1B0H] = 0.270 mol/1.00 L = 0.270 M
[NaCOS1U1B12S1U1B0H] = 0.260 mol/1.00 L = 0.260 M

Since sodium formate is a strong electrolyte, it will dissociate completely in water to form Na+ and formate ion. Therefore, the concentration of formate ion is equal to the concentration of sodium formate:
[COS1U1B12S1U1B0^-] = 0.260 M

Formic acid is a weak acid, so we can use the formula for the dissociation constant of a weak acid (Ka) to calculate the concentration of H+ ions in the solution:

Ka = [H+][COS1U1B12S1U1B0^-]/[HCOS1U1B12S1U1B0H]

We know the value of Ka (1.77 × 10S1U1P1-4S1S1P0) and the concentrations of formic acid and formate ion, so we can rearrange the equation to solve for [H+]:

[H+] = sqrt(Ka*[HCOS1U1B12S1U1B0H]/[COS1U1B12S1U1B0^-])

[H+] = sqrt((1.77 × 10S1U1P1-4S1S1P0)*(0.270)/(0.260))

[H+] = 1.88 × 10S1U1P-4S1S1P0

Finally, we can use the equation for pH to find the pH of the solution:

pH = -log[H+]

pH = -log(1.88 × 10S1U1P-4S1S1P0)

pH = 3.736

Therefore, the pH of the solution is 3.736, which corresponds to choice C.

Question 12

The K_a of acetic acid is 1.76 × 10^-⁵. The pH of a buffer prepared by combining 15.0 mL of 1.00 M potassium acetate and 50.0 mL of 1.00 M acetic acid is ________. A)1.705 B)0.851 C)3.406 D)4.232 E)2.383

Accepted Answer

The answer of The K_a of acetic acid is 1.76...

Question 13

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of hydrogen fluoride after addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.
A)0.0953
B)0.0900
C)0.130
D)0.122
E)0.00976

Accepted Answer

The addition of HCl will convert some fluoride ions to hydrogen fluoride. Calculate moles of HCl added (9.00 mL * 0.0100 M = 0.00009 mol). Initial moles of fluoride ions = 0.100 M * 0.025 L = 0.0025 mol. Moles of fluoride ions after reaction = 0.0025 mol - 0.00009 mol = 0.00241 mol. Initial moles of hydrogen fluoride = 0.126 M * 0.025 L = 0.00315 mol. Moles of hydrogen fluoride after reaction = 0.00315 mol + 0.00009 mol = 0.00324 mol. Total volume = 25.0 mL + 9.00 mL = 34.0 mL = 0.034 L. Concentration of hydrogen fluoride = 0.00324 mol / 0.034 L = 0.0953 M.

Question 14

Of the following solutions, which has the greatest buffering capacity?
A)1.15 M HF and 0.624 M NaF
B)0.574 M HF and 0.312 M NaF
C)0.287 M HF and 0.156 M NaF
D)0.189 M HF and 0.103 M NaF
E)They are all buffer solutions and would all have the same capacity.

Accepted Answer

The buffering capacity of a solution is determined by the concentration and ratio of the weak acid and its conjugate base. In this case, HF and NaF are a weak acid and its conjugate base. The solution with the greatest buffering capacity will have a higher concentration of both HF and NaF. Option A has the highest concentration of both species, making it the solution with the greatest buffering capacity.

Question 15

A buffer solution with a pH of 4.31 is prepared with 1.0 M HC₂H₃O₂ and ________ M NaC₂H₃O₂. The Ka of HC₂H₃O₂is 1.8 × 10^-⁵. A)0.37 B)0.74 C)4.2 × 10^-6 D)8.8 × 10^-10 E)0.18

Accepted Answer

The correct answer is A because the Henderson-Hasselbalch equation can be used to calculate the concentration of NaC₂H₃O₂ in the buffer solution. The equation is:pH = pKa + log([A-]/[HA])where pH is the pH of the buffer solution, pKa is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.In this case, the pH of the buffer solution is 4.31, the pKa of HC₂H₃O₂ is 1.8 × 10^-5, and the concentration of HC₂H₃O₂ is 1.0 M. Substituting these values into the Henderson-Hasselbalch equation, we get:4.31 = 1.8 × 10^-5 + log([A-]/[HA])Solving for [A-], we get:[A-]/[HA] = 10^(4.31 - 1.8 × 10^-5) = 0.37Therefore, the concentration of NaC₂H₃O₂ in the buffer solution is 0.37 M.

Question 16

The K_bof ammonia is 1.76 × 10^-⁵. The pH of a buffer prepared by combining 50.0 mL of 1.00 M ammonia and 45.0 mL of 1.00 M ammonium nitrate is ________. A)4.632 B)9.291 C)4.742 D)9.372 E)none of the above

Accepted Answer

The pH of the buffer can be calculated using the Henderson-Hasselbalch equation. The pKa is calculated from the Kb of ammonia, and the concentrations of ammonia and ammonium ion are used to find the pH.

Question 17

A buffer solution with a pH of 4.40 is prepared with 0.78 M Na C₂H₃O₂ and ________ M HC₂H₃O₂.The K_a of HC₂H₃O₂ is 1.8 × 10^-5. A)1.7 B)3.5 C)4.1 × 10⁴ D)0.35 E)0.86

Accepted Answer

The answer of A buffer solution with a pH of...

Question 18

The addition of hydrofluoric acid and ________ to water produces a buffer solution. A)NaF B)HF C)NaNO₃ D)NaBr E)KI

Accepted Answer

When hydrofluoric acid, HF, is added to water, it partially dissociates to form fluoride ions and hydronium ions. The fluoride ions react with the added sodium fluoride, NaF, to form an equilibrium mixture of HF, F-, and HF2- ions. This mixture acts as a buffer solution, meaning it can resist changes in pH when small amounts of an acid or base are added. Therefore, the best choice is A, NaF.

Question 19

A solution is prepared by dissolving 0.23 mol of benzoic acid and 0.27 mol of sodium benzoate in water sufficient to yield 1.00 L of solution.The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the ________ present in the buffer solution. The K_a of benzoic acid is 6.3 × 10^-5. A)H₂O B)H₃O⁺ C)benzoate D)benzoic acid E)This is a buffer solution: the pH does not change upon addition of acid or base.

Accepted Answer

The addition of NaOH causes the pH to increase slightly because it reacts with the weak acid in the buffer solution, which is benzoic acid (H₃O). NaOH is a strong base and will react with the weak acid to form its conjugate base (benzoate, ⁺) and water. This reaction consumes some of the NaOH, preventing a drastic increase in pH. Thus, the correct answer is choice D, benzoic acid.

Question 20

What is the pH of a buffer solution that is 0.266 M in lactic acid and 0.111 M in sodium lactate? The K_a of lactic acid is 1.4 × 10^-4. A)14.38 B)10.53 C)5.38 D)3.47 E)4.23

Accepted Answer

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Here, pKa = -log(1.4 × 10^-4) ≈ 3.85. Plugging in the concentrations: pH = 3.85 + log(0.111/0.266) ≈ 3.47.

Quiz 17: Additional Aspects of Aqueous Equilibria

A buffer solution with a pH of 4.63 is prepared with 0.14 M formic acid and ________ M sodium formate. The K_a of formic acid is 1.8 × 10^-4.

A buffer solution with a pH of 4.78 is prepared with ________ M formic acid and 0.90 M sodium formate. The K_a of formic acid is 1.8 × 10^-4.

Calculate the pH of a solution prepared by dissolving 0.850 mol of NH₃ and 0.300 mol of NH₄Cl in water sufficient to yield 1.00 L of solution. The K_b of ammonia is 1.77 × 10^-5.

The pH of a solution that contains 0.800 M acetic acid (K_a = 1.76 × 10^-⁵) and 0.172 M sodium acetate is ________.

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.

A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The K_a of benzoic acid is 4.50 × 10^-4.

Calculate the pH of a solution prepared by dissolving 0.250 mol of benzoic acid (C₇H₅O₂H) and 0.150 mol of sodium benzoate (NaC₇H₅O₂) in water sufficient to yield 1.00 L of solution. The K_a of benzoic acid is 6.50 × 10^-5.

What is the pH of a buffer solution that is 0.172 M in hypochlorous acid (HClO) and 0.131 M in sodium hypochlorite? The K_a of hypochlorous acid is 3.8 × 10^-⁸.

How many milliliters of 0.0839 M NaOH are required to titrate 25.0 mL of 0.0990 M HBr to the equivalence point?

Calculate the pH of a solution prepared by dissolving 0.270 mol of formic acid (HCO₂H) and 0.260 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The K_a of formic acid is 1.77 × 10^-4.

The K_a of acetic acid is 1.76 × 10^-⁵. The pH of a buffer prepared by combining 15.0 mL of 1.00 M potassium acetate and 50.0 mL of 1.00 M acetic acid is ________.

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of hydrogen fluoride after addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.

Of the following solutions, which has the greatest buffering capacity?

A buffer solution with a pH of 4.31 is prepared with 1.0 M HC₂H₃O₂ and ________ M NaC₂H₃O₂. The Ka of HC₂H₃O₂is 1.8 × 10^-⁵.

The K_bof ammonia is 1.76 × 10^-⁵. The pH of a buffer prepared by combining 50.0 mL of 1.00 M ammonia and 45.0 mL of 1.00 M ammonium nitrate is ________.

A buffer solution with a pH of 4.40 is prepared with 0.78 M Na C₂H₃O₂ and ________ M HC₂H₃O₂.The K_a of HC₂H₃O₂ is 1.8 × 10^-5.

The addition of hydrofluoric acid and ________ to water produces a buffer solution.

What is the pH of a buffer solution that is 0.266 M in lactic acid and 0.111 M in sodium lactate? The K_a of lactic acid is 1.4 × 10^-4.

Quiz 17: Additional Aspects of Aqueous Equilibria

A buffer solution with a pH of 4.63 is prepared with 0.14 M formic acid and ________ M sodium formate. The Ka of formic acid is 1.8 × 10-4.

A buffer solution with a pH of 4.78 is prepared with ________ M formic acid and 0.90 M sodium formate. The Ka of formic acid is 1.8 × 10-4.

Calculate the pH of a solution prepared by dissolving 0.850 mol of NH3 and 0.300 mol of NH4Cl in water sufficient to yield 1.00 L of solution. The Kb of ammonia is 1.77 × 10-5.

The pH of a solution that contains 0.800 M acetic acid (Ka = 1.76 × 10-5) and 0.172 M sodium acetate is ________.

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.

A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of benzoic acid is 4.50 × 10-4.

Calculate the pH of a solution prepared by dissolving 0.250 mol of benzoic acid (C7H5O2H) and 0.150 mol of sodium benzoate (NaC7H5O2) in water sufficient to yield 1.00 L of solution. The Ka of benzoic acid is 6.50 × 10-5.

What is the pH of a buffer solution that is 0.172 M in hypochlorous acid (HClO) and 0.131 M in sodium hypochlorite? The Ka of hypochlorous acid is 3.8 × 10-8.