Question 1

A solution is prepared by dissolving 0.23 mol of hydrazoic acid and 0.27 mol of sodium azide 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_aof hydrazoic acid is 1.9 × 10^-⁵. A) azide B) This is a buffer solution: The pH does not change upon addition of acid or base. C) H₃O⁺ D) H₂O E) hydrazoic acid

Accepted Answer

The answer of A solution is prepared by dissolving 0.23...

Question 2

A solution is prepared by dissolving 0.23 mol of chloroacetic acid and 0.27 mol of sodium chloroacetate 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_aof chloroacetic acid is 1.36 × 10^-³. A) H₃O⁺ B) chloroacetic acid C) H₂O D) This is a buffer solution: The pH does not change upon addition of acid or base. E) chloroacetate ion

Accepted Answer

The HCl reacts with the chloroacetate ion in the buffer solution, which helps to resist drastic changes in pH.

Question 3

The molar solubility of _ is not affected by the pH of the solution. A) Na₃PO₄ B) AlCl₃ C) NaF D) KNO₃ E) MnS

Accepted Answer

The molar solubility of KNO₃ is not affected by the pH of the solution because it is a salt of a strong acid (HNO₃) and a strong base (KOH), which do not react with H⁺ or OH⁻ ions in solution.

Question 4

A 25.0 mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution. What is the pH after 26.0 mL of base is added? The K_aof hypochlorous acid is 3.0 × 10^-⁸. A) 11.47 B) 2.54 C) 7.54 D) 7.51 E) 7.00

Accepted Answer

The correct answer is A because the pH after 26.0 mL of base is added is 11.47. This 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 hypochlorous acid is 3.0 × 10-8, the concentration of the conjugate base (NaOH) is 0.150 M, and the concentration of the acid (HOCl) is 0.150 M - 0.026 M = 0.124 M. Plugging these values into the equation gives a pH of 11.47.

Question 5

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

Accepted Answer

Hydrofluoric acid (HF) is a weak acid that ionizes partially in water to form its conjugate base, fluoride ions (F-). The addition of a strong acid or a strong base to this solution will not significantly change the pH since the buffer can resist small changes in pH. Therefore, the addition of hydrofluoric acid to water produces a buffer solution containing fluoride ions and their conjugate acid, which is sodium fluoride (NaF).

Question 6

Which of the following could be added to a solution of potassium fluoride to prepare a buffer?
A) ammonia
B) hydrochloric acid
C) sodium hydroxide
D) sodium fluoride
E) potassium acetate

Accepted Answer

Potassium fluoride (KF) contains the fluoride ion (F-), which can act as a weak base. Adding hydrochloric acid (HCl) will provide H+ ions that can react with F- to form HF, a weak acid, thus creating a buffer system of HF/F-.

Question 7

What is the pH of a buffer solution that is 0.211 M in lactic acid and 0.111 M in sodium lactate? The K_aof lactic acid is 1.37 × 10^-⁴. A) 4.13 B) 3.57 C) 5.48 D) 14.28 E) 10.43

Accepted Answer

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Here, pKa = -log(1.37 × 10^-4) ≈ 3.86. Plugging in the concentrations, pH = 3.86 + log(0.111/0.211) ≈ 3.57.

Question 8

How many milliliters of 0.120 M NaOH are required to titrate 50.0 mL of 0.0998 M butanoic acid to the equivalence point? The K_aof butanoic acid is 1.5 × 10^-⁵. A) 4.90 B) 41.6 C) 50.0 D) 4.65 E) 60.1

Accepted Answer

To find the volume of NaOH required, use the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of butanoic acid, and M2 and V2 are the molarity and volume of NaOH. Solving for V2 gives (0.0998 M)(50.0 mL) = (0.120 M)(V2), resulting in V2 = 41.6 mL.

Question 9

A 25.0 mL sample of 0.150 M nitrous acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The K_aof nitrous acid is 4.50 × 10^-⁴. A) 3.35 B) 7.00 C) 8.11 D) 10.35 E) 10.65

Accepted Answer

At the equivalence point, the number of moles of NaOH added will be equal to the number of moles of HNO2 in the original solution. 
moles of NaOH = 0.150 M x 25.0 mL = 0.00375 mol
moles of HNO2 = 0.150 M x 25.0 mL = 0.00375 mol
The reaction between NaOH and HNO2 is:
HNO2 + NaOH → NaNO2 + H2O
At the equivalence point, all HNO2 is converted to NaNO2. NaNO2 is a salt of a weak base and a strong acid, and it undergoes hydrolysis in water. The reaction is:
NaNO2 + H2O → HNO2 + NaOH
The hydrolysis of NaNO2 gives OH- ions, and the pH increases. 
The pH of the solution at the equivalence point can be calculated using the equation:
pH = pKb + log([salt]/[acid])
where pKb is the negative logarithm of the base dissociation constant of the conjugate base (NO2-), and [salt]/[acid] is the ratio of the concentrations of NaNO2 and HNO2, respectively. 
The pKb of NO2- can be calculated from the pKa of HNO2 as follows:
pKw = pKa + pKb
14.00 = 3.30 + pKb
pKb = 10.70
At the equivalence point, [salt]/[acid] = 1, so 
pH = pKb + log([salt]/[acid]) = pKb = 10.70
However, this is the pH of NaNO2 in water, which is not the same as the pH of the original solution because of dilution. 
Before the titration, the HNO2 solution has a pH of:
pH = pKa + log([NO2-]/[HNO2])
pKa = 3.30
[NO2-]/[HNO2] = 0, because there is no NO2- before the titration
pH = pKa = 3.30
At the equivalence point, the volume of the solution doubles to 50.0 mL, and the number of moles of HNO2 is still 0.00375 mol. Therefore, the concentration of HNO2 is now:
[HNO2] = 0.00375 mol / 0.050 L = 0.075 M
The concentration of OH- can be calculated from the expression for the base dissociation constant of water:
Kw = [H+][OH-]
1.0 x 10^-14 = (x)(x)
x = 1.0 x 10^-7 M
The concentration of OH- from the hydrolysis of NaNO2 is the same as the concentration of NaNO2, which is also 0.075 M.
The total concentration of OH- is 0.075 M + 1.0 x 10^-7 M = 0.075 M (to 3 significant figures). The pH can be calculated from the expression:
pH = 14.00 - log([H+])
[H+] = Kw / [OH-] = 1.33 x 10^-13 M
pH = 12.88 (to 3 significant figures)
However, this is the pH of the solution after excess NaOH has been added. The pH at the equivalence point is between the initial pH of 3.30 and the final pH of 12.88. Since NaNO2 is a basic salt, the pH at the equivalence point is greater than 7.00. The only choice greater than 7.00 is C, which is the correct answer.

Question 10

How many milliliters of 0.0850 M NaOH are required to titrate 25.0 mL of 0.0720 M HBr to the equivalence point?
A) 3.92
B) 21.2
C) 0.245
D) 0.153
E) 29.5

Accepted Answer

To find the volume of NaOH needed, use the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of HBr, and M2 and V2 are the molarity and volume of NaOH. (0.0720 M)(25.0 mL) = (0.0850 M)(V2). Solving for V2 gives 21.2 mL.

Question 11

In which one of the following solutions is silver chloride the most soluble? A) 0.744 M LiNO₃ B) 0.181 M HCl C) 0.181 M NaCl D) 0.0176 M NH₃ E) pure water

Accepted Answer

The solubility of silver chloride is highest in NH₃ solution due to the formation of the complex ion [Ag(NH₃)₂]⁺.

Question 12

In which of the following aqueous solutions would you expect AgBr to have the highest solubility? A) 0.10 M AgNO₃ B) 0.20 M NaBr C) 0.10 M LiBr D) 0.15 M KBr E) pure water

Accepted Answer

AgBr is least soluble in solutions containing common ions (Ag⁺ or Br⁻). Pure water has no common ions, so AgBr will have the highest solubility in pure water.

Question 13

In which of the following aqueous solutions would you expect PbCl₂to have the lowest solubility? A) pure water B) 0.015 M NaCl C) 0.020 M KCl D) 0.015 M PbNO₃ E) 0.020 M BaCl₂

Accepted Answer

The answer of In which of the following aqueous solutions...

Question 14

A 25.0 mL sample of a solution of a monoprotic acid is titrated with a 0.115 M NaOH solution. The titration curve above was obtained. Which of the following indicators would be best for this titration?
A) thymol blue
B) bromthymol blue
C) phenolpthalein
D) bromocresol purple
E) methyl red

Accepted Answer

The answer of A 25.0 mL sample of a solution...

Question 15

Which of the following could be added to a solution of acetic acid to prepare a buffer?
A) nitric acid
B) sodium hydroxide
C) more acetic acid
D) hydrochloric acid
E) None of the above can be added to an acetic acid solution to prepare a buffer.

Accepted Answer

A buffer solution can be prepared by adding a strong base like sodium hydroxide to acetic acid, which will partially neutralize the acetic acid and form its conjugate base, acetate, creating a buffer system.

Question 16

In which aqueous system is PbI₂least soluble? A) 0.5 M HI B) 1.0 M HNO₃ C) 0.2 M HI D) 0.8 M KI E) H₂O

Accepted Answer

PbI2 is least soluble in 0.8 M KI due to the common ion effect, where the presence of a common ion (I-) from KI reduces the solubility of PbI2.

Question 17

A 25.0- mL sample of 0.150 M hydrazoic acid is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The K_aof hydrazoic acid is 1.9 × 10^-⁵. A) 4.66 B) 4.78 C) 3.03 D) 1.34 E) 4.45

Accepted Answer

1. Write the balanced chemical equation:
HN3 + NaOH -> NaN3 + H2O

2. Calculate the number of moles of HN3 initially present:
moles HN3 = (0.150 mol/L) x (25.0 mL/1000 mL) = 0.00375 mol

3. Calculate the number of moles of NaOH added:
moles NaOH = (0.150 mol/L) x (13.3 mL/1000 mL) = 0.001995 mol

4. Determine the limiting reagent:
HN3 : NaOH = 1 : 1
Since the number of moles of NaOH added is less than the number of moles of HN3 initially present, HN3 is the limiting reagent, and the remaining moles of HN3 can be calculated as follows:

moles HN3 remaining = 0.00375 mol - 0.001995 mol = 0.001755 mol

5. Calculate the concentration of HN3 after the addition of NaOH:
volume HN3 remaining = (25.0 mL + 13.3 mL) / 1000 mL = 0.0383 L
[Moles HN3] = 0.001755 mol / 0.0383 L = 0.0458 M

6. Calculate the pKa of HN3: 
pKa = -log(Ka) = -log(1.9 x 10^-5) = 4.72

7. Calculate the pH 
Since HN3 is a weak acid, we will use an ICE table to solve for [H+] and pH:

HN3 + H2O -> H3O+ + N3-

I 0.0458 M    0          0           
C -x          x          x           
E 0.0458-x    x          x

Ka = (H3O+)(N3-)/(HN3) = x^2/(0.0458-x)
Using the quadratic formula gives x = [H3O+] = 1.18 x 10^-5 M
pH = -log[H3O+] = 4.78

Thus, the pH after 13.3 mL of NaOH is added is B) 4.78

Question 18

A solution of HF is titrated with a 0.150 M NaOH solution. Based on the table above, the best indicator for this reaction is . The K_aof hydrofluoric acid is 6.8 × 10^-⁴. A) methyl orange B) phenolpthalein C) methyl red D) thymol blue E) bromocresol purple

Accepted Answer

The answer of A solution of HF is titrated with...

Question 19

Which one of the following is not amphoteric? A) Al(OH)₃ B) Zn(OH)₂ C) Cr(OH)₃ D) Ca(OH)₂ E) Sn(OH)₂

Accepted Answer

Ca(OH)₂ is a strong base and does not exhibit amphoteric behavior, unlike the other hydroxides listed.

Question 20

The pH of a solution prepared by mixing 45 mL of 0.183 M KOH and 65 mL of 0.145 M HCl is
)
A) 70.145
B) 2.92
C) 0.74
D) 1.31
E) 1.97

Accepted Answer

The answer of The pH of a solution prepared by...

Quiz 17: Additional Aspects of Equilibria

The molar solubility of _ is not affected by the pH of the solution.

A 25.0 mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution. What is the pH after 26.0 mL of base is added? The Ka of hypochlorous acid is 3.0 × 10- 8.

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

Which of the following could be added to a solution of potassium fluoride to prepare a buffer?

What is the pH of a buffer solution that is 0.211 M in lactic acid and 0.111 M in sodium lactate? The Ka of lactic acid is 1.37 × 10- 4.

How many milliliters of 0.120 M NaOH are required to titrate 50.0 mL of 0.0998 M butanoic acid to the equivalence point? The Ka of butanoic acid is 1.5 × 10- 5.

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

How many milliliters of 0.0850 M NaOH are required to titrate 25.0 mL of 0.0720 M HBr to the equivalence point?

In which one of the following solutions is silver chloride the most soluble?

In which of the following aqueous solutions would you expect AgBr to have the highest solubility?

In which of the following aqueous solutions would you expect PbCl2to have the lowest solubility?

A 25.0 mL sample of a solution of a monoprotic acid is titrated with a 0.115 M NaOH solution. The titration curve above was obtained. Which of the following indicators would be best for this titration?

Which of the following could be added to a solution of acetic acid to prepare a buffer?

In which aqueous system is PbI2 least soluble?

A 25.0- mL sample of 0.150 M hydrazoic acid is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of hydrazoic acid is 1.9 × 10- 5.

A solution of HF is titrated with a 0.150 M NaOH solution. Based on the table above, the best indicator for this reaction is . The Ka of hydrofluoric acid is 6.8 × 10- 4.

Which one of the following is not amphoteric?

The pH of a solution prepared by mixing 45 mL of 0.183 M KOH and 65 mL of 0.145 M HCl is )

A 25.0 mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution. What is the pH after 26.0 mL of base is added? The K_aof hypochlorous acid is 3.0 × 10^-⁸.

What is the pH of a buffer solution that is 0.211 M in lactic acid and 0.111 M in sodium lactate? The K_aof lactic acid is 1.37 × 10^-⁴.

How many milliliters of 0.120 M NaOH are required to titrate 50.0 mL of 0.0998 M butanoic acid to the equivalence point? The K_aof butanoic acid is 1.5 × 10^-⁵.

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

In which of the following aqueous solutions would you expect PbCl₂to have the lowest solubility?

In which aqueous system is PbI₂least soluble?

A 25.0- mL sample of 0.150 M hydrazoic acid is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The K_aof hydrazoic acid is 1.9 × 10^-⁵.

A solution of HF is titrated with a 0.150 M NaOH solution. Based on the table above, the best indicator for this reaction is . The K_aof hydrofluoric acid is 6.8 × 10^-⁴.