Which of the following gives the best estimate for the pH of a 1 x 10⁻⁵ M HClO₄(aq) solution at 25 °C?

pH ≈ 5.0 because HClO₄ is a strong acid.

pH ≈ 3.0 because HClO₄ is a strong acid.

pH ≈ 9.0 because HClO₄ is a weak base.

pH ≈ 11.0 because HClO₄ is a strong base.

The equilibrium for the acid ionization of HCOOH is represented by the equation above, and the table gives the percent ionization for HCOOH at different initial concentrations of the weak acid at 25°C. Based on the information, which of the following is true for a 0.125 M aqueous solution of HCOOH?

It has a lower percent ionization, a higher [H₃O⁺]_eq, and a higher pH than a 0.100 M HCOOH solution does.

It has a higher percent ionization, a lower [H₃O⁺]_eq, and a lower pH than a 0.150 M HCOOH solution does.

It has a higher percent ionization, a lower [H₃O⁺]_eq, and a higher pOH than a 0.150 M HCOOH solution does.

It has a lower percent ionization, a higher [H₃O⁺]_eq, and a higher pOH than a 0.100 M HCOOH solution does.

The reaction between NH3 and water is represented above. A solution that is initially 0.200M in NH3 has a pH of 11.28. Which of the following correctly predicts the pH of a solution for which [NH3]initial=0.100�, and why?

The pH will be lower than 11.28 because the equilibrium concentration of OH- ions decreases when the initial concentration of the base decreases.

The pH will be lower than 11.28 because decreasing the initial concentration of NH3 increases the equilibrium concentration of the conjugate acid NH4+.

The pH will be higher than 11.28 because decreasing the initial concentration of NH3 decreases the equilibrium concentration of the conjugate acid NH4+.

The pH will be higher than 11.28 because the equilibrium concentration of OH- ions increases when the initial concentration of the base decreases.

The acid ionization equilibrium for HClO at 25°C is shown above. A 0.100 M aqueous solution of this acid has a pH of about 4.2. If a solution with an initial concentration of HClO of 0.200 M is allowed to reach equilibrium at the same temperature, which of the following correctly predicts its pH, and why?D

The pH will be lower than 4.2 because increasing the concentration of the weak acid produces more H3O+ to establish equilibrium.

The pH will be higher than 4.2 because increasing the concentration of the weak acid decreases the percent ionization.

The pH will be higher than 4.2 because, for weak acid solutions, the pH is directly proportional to the initial concentration of the weak acid.

The pH will be lower than 4.2 because the Ka of the weak acid is inversely proportional to the initial concentration of the weak acid.

The weak acid CH3COOH has a pKa of 4.76. A solution is prepared by mixing 500. mL of 0.150 M CH3COOH (aq) and 0.0200 mol of NaOH(s). Which of the following can be used to calculate the pH of the solution?

pH = 4.76 + log(0.0200 / 0.0550) = 4.32

pH = 4.76 + log(0.0200 / 0.150) = 3.88

pH = 4.76 + log(0.0200 / 0.0750) = 4.19

pH = 4.76 + log(0.0200 / 0.0750) = 4.86

Which of the following provides the correct mathematical expression to calculate the pH of a solution made by mixing 10.0 mL of 1.00 M HCl and 11.0 mL of 1.00 M NaOH at 25° C?

pH = 14.00 + log(0.0010 / 0.0210) = 12.68

pH = 14.00 + log(0.0010) = 11.00

pH = 14.00 + log(0.0010 / 0.0110) = 12.96

A buffer solution is formed by mixing equal volumes of 0.12 M NH3(aq) and 0.10 M HCl(aq), which reduces the concentration of both solutions by one half. Based on the pKa data given in the table, which of the following gives the pH of the buffer solution?

pH = 9.25 + log(0.010/0.050) = 8.55

pH = 9.25 + log(0.050/0.010) = 9.32

pH = 14.00 − (−log(0.010)) = 12.00

The table above provides information on two weak acids. Which of the following explains the difference in their acid strength?

Acid 2 is a stronger acid because it has a more stable conjugate base than acid 1 due to the greater number of electronegative Br atoms.

Acid 1 is a stronger acid because it has more acidic hydrogen atoms than acid 2.

Acid 1 is a stronger acid because it can make more hydrogen bonds than acid 2.

Acid 2 is a stronger acid because it is a larger, more polarizable molecule with stronger intermolecular forces than acid 1.

The equilibrium reactions for diprotic oxoacids with a general formula H2XO4 are represented by the equations above. The acid ionization constants for H2SeO4 and H2TeO4 are provided in the table. Which of the following best explains the difference in strength for these two acids?

H2TeO4 is weaker because Te is less electronegative than Se, resulting in less stable conjugate bases (HTeO4- and TeO42-) than those for H2SeO4.

H2SeO4 is weaker because Se has a smaller positive formal charge than Te, resulting in a decrease in its ability to transfer an H+ to H2O.

H2TeO4 is weaker because Te has a smaller positive formal charge than Te, resulting in a decrease in its ability to transfer an H+ to H2O.

H2SeO4 is weaker because Se is more electronegative than Te, resulting in more stable conjugate bases (HSeO4- and SeO42-) than those for H2TeO4.

Lewis diagrams of the weak bases NH3 and NF3 are shown above. Based on these diagrams, which of the following predictions of their relative base strength is correct, and why?

NF3 is a weaker base than NH3 because of the greater electronegativity of F compared with H.

NF3 is a stronger base than NH3 because more than one resonance structure exists for its conjugate acid NF3H+, making it more stable than NH4+.

NF3 is a stronger base than NH3 because of the greater electronegativity of F compared with H.

NF3 is a weaker base than NH3 because more than one resonance structure exists for its conjugate acid NF3H+, making it more stable than NH4+.

The weak acid ionization equilibrium for C2H3COOH is represented by the equation above. A student measures the pH of C2H3COOH(aq) using a probe and a pH meter in the experimental setup shown. Based on the information given, which of the following is true?

[C2H3COOH] > [C2H3COO⁻] since the pKa of the weak acid is greater than the pH of the solution.

[C2H3COOH] > [C2H3COO⁻] since the pKa of the weak acid is less than pKw.

The acid equilibrium for CH3CH2COOH and the base equilibrium for CH3CH2COO− are represented above. One liter of a buffer solution with pH = 4.85 is made by mixing 0.100 M CH3CH2COOH and 0.100 M NaCH3CH2COO. If 10.0 mL of 0.500 M NaOH is added to the buffer, which of the following is most likely the resulting pH, and why?

The pH will be slightly greater than 4.85, because some CH3CH2COOH will react with the added base, resulting in a slight decrease in [H3O+].

The pH will be much less than 4.85, because the buffer will respond to the addition of NaOH by producing more CH3CH2COOH.

The pH will be much greater than 4.85, because there is a large increase in the concentration of the weak base, CH3CH2COO−.

The pH will be slightly less than 4.85, because the addition of NaOH reduces the autoionization of H2O.

A buffer solution that is 0.100 M in both HCOOH and HCOOK has a pH = 3.75. A student says that if a very small amount of 0.100 M HCl is added to the buffer, the pH will decrease by a very small amount. Which of the following best supports the student’s claim?

HCOO− will accept a proton from HCl to produce more HCOOH and H2O.

HCOOH will accept a proton from HCl to produce more HCOO− and H2O.

HCOO− will donate a proton to HCl to produce more HCOOH and H2O.

HCOOH will donate a proton to HCl to produce more HCOO− and H2O.

To prepare a buffer solution for an experiment, a student measured out 53.49 g of NH4Cl(s) (molar mass 53.49 g/mol) and added it to 1.0 L of 1.0 M NH3(aq). However, in the process of adding the NH4Cl(s) to the NH3(aq), the student spilled some of the NH4Cl(s) onto the bench top. As a result, only about 50.0 g of NH4Cl(s) was actually added to the 1.0 M NH3(aq). Which of the following best describes how the buffer capacity of the solution is affected as a result of the spill?

The solution has a greater buffer capacity for the addition of base than for acid, because [NH3] > [NH4+].

The solution has a greater buffer capacity for the addition of base than for acid, because [NH3]

The solution has a greater buffer capacity for the addition of acid than for base, because [NH3]

The solution has a greater buffer capacity for the addition of acid than for base, because [NH3] > [NH4+].

The acid ionization equilibrium for HC3H5O2 is represented by the equation above. A mixture of 1.00 L of 0.100 M HC3H5O2 and 0.500 L of 0.100 M NaOH will produce a buffer solution with a pH = 4.87. If the NaOH solution was mislabeled and was 1.00 M instead of 0.100 M, which of the following would be true?

The pH of the resulting solution would be much higher than 4.87 because the weak acid would be completely neutralized by the larger amount of NaOH added.

The pH of the resulting solution would still be 4.87 because buffer solutions regulate changes in pH regardless of how much NaOH is added.

The pH of the resulting solution would be somewhat higher than 4.87 because adding NaOH that is 10 times more concentrated makes the concentration of the conjugate base 10 times larger.

The pH of the resulting solution would still be 4.87 because the solution contains a large volume of the weak acid to react with the NaOH and would maintain the same pH.

The acid ionization equilibrium for HNO2 is represented by the equation above. A 250.0 mL buffer solution is prepared by mixing 125.0 mL of 0.20 M HNO2 and 125.0 mL of 0.10 M NaOH. To test the buffer capacity, the pH is measured and recorded in the table for four samples of the buffer and one sample of a mixture of the buffer and HCl. Which of the following best helps explain why the pH of sample 4 is lower than the pH of the other samples containing only buffer solution?

After measuring the pH of the more acidic sample 3, the pH probe was not rinsed and wiped, resulting in the neutralization of a very small amount of the conjugate base in sample 4.

Prior to measuring the pH of sample 4, some water evaporated, resulting in an increase in the concentration of HNO2 in the sample.

Prior to measuring the pH of sample 4, the room temperature dropped suddenly, resulting in an increase in the pKa for HNO2.

The volume used to measure the pH of sample 4 was less than 25.0 mL, resulting in a decrease in the concentration of NO2-.

Which of the following pairs of mathematical expressions can be used to correctly calculate the pH and pOH of a 0.0015 M KOH(aq) solution at 25°C?

pH = 14.00 − ( − log(0.0015)) and pOH = −log(0.0015)

pH = −log(14.00 − 0.0015) and pOH = −log(0.0015)

pH = 14.00 − ( − log(0.0015)) and pOH = 14.00 − ( − log(0.0015))

pH = − ( − log(0.0015)) and pOH = 14.00 − ( − log(0.0015))

Which of the following statements about the pH of 0.010 M HClO4 is correct?

pH = 2.00, because [H⁺] = 1.0 x 10⁻² M.

pH = 2.00, because [H⁺] = 2.0 x 10⁻² M.

pH > 2.00, because HClO4 is a strong acid.

A student performs an acid-base titration and plots the experimental results in the graph above. Which of the following statements best explains the experimental findings?

A weak acid was titrated with a strong base, as evidenced by the equivalence point at pH > 7.

A strong acid was titrated with a strong base, as evidenced by the equivalence point at pH = 7.

A strong acid was titrated with a strong base, as evidenced by the equivalence point at pH > 7.

A weak acid was titrated with a weak base, as evidenced by the equivalence point at pH approximately 7.

$$HF\left(aq\right)+H_2O\left(l\right)\ \leftrightarrow\ H_3O^+\left(aq\right)+F^-\left(aq\right)\ \ \ \ Ka\ =6.3\times10^{-4}\ at\ 25\degree C$$ The acid ionization equilibrium for HF is represented by the chemical equation above. A student claims that the pH of a solution that contains 0.100 M HF (aq) and 0.100 M NaF (aq) will change only slightly when small amounts of acids or bases are added. Which of the following pairs of equations can the student use to justify the claim?

HF(aq) + OH⁻(aq) → H₂O(l) + F⁻(aq) and F⁻(aq) + H₃O⁺(aq) → HF(aq) + H₂O(l)

HF(aq) + OH⁻(aq) → H₂O(l) + F⁻(aq) and OH⁻(aq) + H₃O⁺(aq) → 2 H₂O(l)

H₃O⁺(aq) + OH⁻(aq) → 2 H₂O(l) and F⁻(aq) + H₃O⁺(aq) → HF(aq) + H₂O(l)

HF(aq) + OH⁻(aq) → H₂O(l) + F⁻(aq) and F⁻(aq) + H₂O(l) → HF(aq) + OH⁻(aq)

Which of the following equations can be used to correctly calculate the pH of a solution at 25 °C that is 0.100 M (CH₃)₂NH(aq) and 0.100 M (CH₃)₂NH₂Cl(aq)?

pH = −log(1.0×10⁻¹⁴ / 5.4×10⁻⁴)

pH = 14.00 + ( − log 5.4 × 10⁻⁴)

pH = −log(5.4×10⁻⁴ / 1.0×10⁻¹⁴)

pH = ( − log 5.4 × 10⁻⁴) + log(0.100)

A buffer solution is made up of acetic acid (CH₃COOH) and sodium acetate (NaCH₃COO). Which of the following equilibria could be used to support the claim that the addition of a small amount of NaOH to the buffer will result in only a very small change in pH?

CH₃COOH (aq) + OH⁻ (aq) ⇌ CH₃COO⁻ (aq) + H₂O (l)

CH₃COOH (aq) + CH₃COO⁻ (aq) ⇌ CH₃COO⁻ (aq) + CH₃COOH (aq)

CH₃COOH (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + CH₃COO⁻ (aq)

AP Chemistry Unit 8 Test Review

Authored by Sarah Callo

Science

12th Grade

NGSS covered

Used 13+ times

AI Actions

Add similar questions

Adjust reading levels

Convert to real-world scenario

Translate activity

More...

Content View
Student View

45 questions

Show all answers

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Buffer solutions containing Na2CO3 and NaHCO3 range in pH from 10.0 to 11.0. The chemical equation below represents the equilibrium between CO32−and H2O, and the table lists the composition of four different buffer solutions at 25°C.
Which of the following chemical equilibrium equations best shows what happens in the buffer solutions to minimize the change in pH when a small amount of a strong base is added?

H₃O⁺(aq) + OH⁻(aq) ⇌ 2H₂O(l)

HCO₃⁻(aq) + OH⁻(aq) ⇌ CO₃²⁻(aq) + H₂O(l)

CO₃²⁻(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + OH⁻(aq)

CO₃²⁻(aq) + H₃O⁺(aq) ⇌ HCO₃⁻(aq) + H₂O(l)

Answer explanation

The species HCO3− partially consumes the added OH−. Therefore, HCO3−(aq)+OH−(aq)⇄CO32−(aq)+H2O(l) is the chemical equilibrium equation that best shows how the buffer minimizes change in pH when base is added.

Tags

NGSS.HS-PS1-6

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Buffer solutions containing Na2CO3 and NaHCO3 range in pH from 10.0 to 11.0. The chemical equation below represents the equilibrium between CO32−and H2O, and the table lists the composition of four different buffer solutions at 25°C.
Which of the following mathematical expressions can be used to determine the approximate pH of buffer 1?

pH = −log(2.1 x 10⁻⁴) + log(0.100 / 0.150) = 3.50

pH = −log(2.1 x 10⁻⁴) + log(0.150 / 0.100) = 3.85

pH = [14.00 + log(2.1 x 10⁻⁴)] + log(0.100 / 0.150) = 10.15

pH = [14.00 + log(2.1 x 10⁻⁴)] + log(0.150 / 0.100) = 10.50

Answer explanation

The species HCO3− partially consumes the added OH−. Therefore, HCO3−(aq)+OH−(aq)⇄CO32−(aq)+H2O(l) is the chemical equilibrium equation that best shows how the buffer minimizes change in pH when base is added.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Buffer solutions containing Na2CO3 and NaHCO3 range in pH from 10.0 to 11.0. The chemical equation below represents the equilibrium between CO32−and H2O, and the table lists the composition of four different buffer solutions at 25°C.
Which mathematical expression can be used to explain why buffer 2 and buffer 3 have the same pH?

log(0.200 / 0.200) = log(0.100 / 0.100) = log(1)

−log(Kₐ) − [ − log(KB)] = pKa

0.200 M = 2 x (0.100 M)

0.200 M − 0.200 M = 0.100 M − 0.100 M

Answer explanation

Since the solutions are at 25°C, Kb×Ka=Kw, hence pKb+pKa=pKw=14.00. Using pH=pKa+log([A−][HA]) and substituting [14.00+log(2.1×10−4)] for pKw, pH=[14.00+log(2.1×10−4)]+log(0.1000.150)=10.15. This pH is reasonable, since the buffer has a higher concentration of the conjugate acid HCO3− compared with the weak base CO32−; thus, the pH should be slightly lower than 10.32 (compared with buffers 2 and 3).

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$H_2O\left(l\right)+H_2O\left(l\right)\leftrightarrow H_3O^+\left(aq\right)+OH^-\left(aq\right)\ \ \ \ K_w=7.0\times10^{-14}\ at\ 55\degree C$
Pure water autoionizes as shown in the equation above. Based on this information, which of the following is correct?

The autoionization equilibrium for pure water favors the formation of reactants more at 55 °C than at 25 °C.

The autoionization equilibrium for pure water produces the same amount of OH⁻ ions at 55 °C and 25 °C.

At 55 °C, pH = – log (√Kw) for pure water.

At 55 °C, pH = – log (Kw) for pure water.

Answer explanation

Regardless of the temperature, for pure water Kw=[H3O+][OH−]. At 55°C, [H3O+]=[OH−]=7.0×10−14, and pH =⁢−log(⁢Kw).

Tags

NGSS.HS-PS1-5

NGSS.HS-PS1-6

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Based on the information above, which of the following is true for a sample of pure water at 25 °C?

[H₃O⁺] = 7.0 M

[OH⁻] = 1.0 x 10⁻¹⁴ M

pH = 10⁻⁷

pOH = 7.00

Answer explanation

pOH=−log[OH−], which is equal to −log (1.0×10−7)=7.00 at 25°C.

Tags

NGSS.HS-ESS2-5

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The endothermic autoionization of pure water is represented by the chemical equation shown above. The pH of pure water is measured to be 7.00 at 25.0 °C and 6.02 at 100.0 °C. Which of the following statements best explains these observations?

At the higher temperature water dissociates less, [H₃O⁺] < [OH⁻], and the water becomes basic.

At the higher temperature water dissociates less, [H₃O⁺] > [OH⁻], and the water remains neutral.

At the higher temperature water dissociates more, [H₃O⁺] > [OH⁻], and the water becomes acidic.

At the higher temperature water dissociates more, [H₃O⁺] = [OH⁻], and the water remains neutral.

Answer explanation

Because the reaction is endothermic, increasing temperature will increase Kw. So the water dissociates more, and [H3O+] and [OH−] will always be equal in pure water.

Tags

NGSS.HS-PS1-5

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which of the following gives the best estimate for the pH of a 5 x 10⁻⁴ M Sr(OH)₂ solution at 25 °C?

pH ≈ 3.0 because Sr(OH)₂ is a strong acid.

pH ≈ 5.0 because Sr(OH)₂ is a weak acid.

pH ≈ 9.0 because Sr(OH)₂ is a weak base.

pH ≈ 11.0 because Sr(OH)₂ is a strong base.

Answer explanation

Sr(OH)2(aq) is a strong base, and its complete dissociation produces 2 moles of OH− ions per mole of Sr(OH)2 that dissolves. As a result, [OH−]=2×(5×10⁻⁴M)=1×10⁻³M and pOH≈3.0, so pH≈11.0.

Access all questions and much more by creating a free account

Create resources

Host any resource

Get auto-graded reports

Continue with Google

Continue with Email

Continue with Microsoft

or continue with

Facebook

Apple

Others

By signing up, you agree to our Terms of Service & Privacy Policy

Already have an account?

Similar Resources on Wayground

50 questions

Earth’s Energy Budget & Climate Systems

Quiz

•

9th Grade - University

40 questions

Integumentary System Worksheet

Quiz

•

10th Grade - University

40 questions

Ulangan Harian Kemagnetan

Quiz

•

9th - 12th Grade

45 questions

Health Science Foundations Ch1 Test Review

Quiz

•

9th - 12th Grade

40 questions

Examen Parcial Ecología y Medio Ambiente

Quiz

•

12th Grade

40 questions

Quis 1. Bab 1 dan Bab 2_Makluk hidup dan Lingkungan ( STS TW 3 )

Quiz

•

10th Grade - University

50 questions

Hair and Fiber

Quiz

•

9th - 12th Grade

40 questions

ôn tập chương este - lipit - 2023-2024

Quiz

•

12th Grade

Popular Resources on Wayground

10 questions

How much do you know about our Portrait of an Eagle?

Quiz

•

10th Grade

10 questions

Fast Food Slogans

Quiz

•

6th - 8th Grade

21 questions

Continents and Oceans

Quiz

•

6th Grade

20 questions

Parts of Speech

Quiz

•

5th Grade

16 questions

Subject & Predicate

Quiz

•

5th Grade

25 questions

Multiplication Facts

Quiz

•

5th Grade

12 questions

Map Skills

Quiz

•

3rd Grade

22 questions

Continents and Oceans

Quiz

•

5th Grade

Discover more resources for Science

22 questions

Periodic Table Review

Quiz

•

9th - 12th Grade

7 questions

Newton's Laws Video Quiz

Interactive video

•

9th - 12th Grade

17 questions

Unit 3A: Cells Vocab

Quiz

•

9th - 12th Grade

89 questions

Unit 1 (Ch 2 & 3) Test Review - Water/Ocean Currents

Quiz

•

9th - 12th Grade

58 questions

Biomolecule Test Corrections

Quiz

•

9th - 12th Grade

20 questions

Unit 2: Cell Energy Review

Quiz

•

9th - 12th Grade

17 questions

Explore Newton's Laws of Motion

Quiz

•

8th - 12th Grade

35 questions

Directional Terms, Body Cavities, Body Planes

Quiz

•

11th - 12th Grade

AP Chemistry Unit 8 Test Review

The species HCO3− partially consumes the added OH−. Therefore, HCO3−(aq)+OH−(aq)⇄CO32−(aq)+H2O(l) is the chemical equilibrium equation that best shows how the buffer minimizes change in pH when base is added.

The species HCO3− partially consumes the added OH−. Therefore, HCO3−(aq)+OH−(aq)⇄CO32−(aq)+H2O(l) is the chemical equilibrium equation that best shows how the buffer minimizes change in pH when base is added.

$H_2O\left(l\right)+H_2O\left(l\right)\leftrightarrow H_3O^+\left(aq\right)+OH^-\left(aq\right)\ \ \ \ K_w=7.0\times10^{-14}\ at\ 55\degree C$
Pure water autoionizes as shown in the equation above. Based on this information, which of the following is correct?

Regardless of the temperature, for pure water Kw=[H3O+][OH−]. At 55°C, [H3O+]=[OH−]=7.0×10−14, and pH =⁢−log(⁢Kw).

Based on the information above, which of the following is true for a sample of pure water at 25 °C?

pOH=−log[OH−], which is equal to −log (1.0×10−7)=7.00 at 25°C.

The endothermic autoionization of pure water is represented by the chemical equation shown above. The pH of pure water is measured to be 7.00 at 25.0 °C and 6.02 at 100.0 °C. Which of the following statements best explains these observations?

Because the reaction is endothermic, increasing temperature will increase Kw. So the water dissociates more, and [H3O+] and [OH−] will always be equal in pure water.

Which of the following gives the best estimate for the pH of a 5 x 10⁻⁴ M Sr(OH)₂ solution at 25 °C?

Sr(OH)2(aq) is a strong base, and its complete dissociation produces 2 moles of OH− ions per mole of Sr(OH)2 that dissolves. As a result, [OH−]=2×(5×10⁻⁴M)=1×10⁻³M and pOH≈3.0, so pH≈11.0.

Which of the following gives the best estimate for the pH of a 1 x 10⁻⁵ M HClO₄(aq) solution at 25 °C?

Since HClO4 is a strong acid, it dissociates completely, producing 1 mol of H3O+ per 1 mol of HClO4(aq). Therefore, pH=−log[H3O+]=−log[HClO4]_initial=−log(1×10−5)≈5.0. (In general, the negative log of 10^−x is x. Therefore, the negative log of 1×10−5 is 5.)

Which of the following is the correct mathematical relationship to use to calculate the pH of a 0.10 M aqueous HBr solution?

By definition, pH=−log[H3O+]. In this case, pH=−log(1.0×10−1), since for a strong acid like HBr, it is assumed that the acid ionizes 100% and [HBr]_initial=[H3O+].

The equilibrium for the acid ionization of HCOOH is represented by the equation above, and the table gives the percent ionization for HCOOH at different initial concentrations of the weak acid at 25°C. Based on the information, which of the following is true for a 0.125 M aqueous solution of HCOOH?

Based on the information given, when the initial concentration of the weak acid increases, the percent ionization decreases and the equilibrium concentration of H3O+ increases. A higher [H3O+]eq results in a lower [OH−]eq and a higher pOH.

Access all questions and much more by creating a free account

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Science

AP Chemistry Unit 8 Test Review

The species HCO3− partially consumes the added OH−. Therefore, HCO3−(aq)+OH−(aq)⇄CO32−(aq)+H2O(l) is the chemical equilibrium equation that best shows how the buffer minimizes change in pH when base is added.

The species HCO3− partially consumes the added OH−. Therefore, HCO3−(aq)+OH−(aq)⇄CO32−(aq)+H2O(l) is the chemical equilibrium equation that best shows how the buffer minimizes change in pH when base is added.

Regardless of the temperature, for pure water Kw=[H3O+][OH−]. At 55°C, [H3O+]=[OH−]=7.0×10−14, and pH =⁢−log(⁢Kw).

Based on the information above, which of the following is true for a sample of pure water at 25 °C?

pOH=−log[OH−], which is equal to −log (1.0×10−7)=7.00 at 25°C.

The endothermic autoionization of pure water is represented by the chemical equation shown above. The pH of pure water is measured to be 7.00 at 25.0 °C and 6.02 at 100.0 °C. Which of the following statements best explains these observations?

Because the reaction is endothermic, increasing temperature will increase Kw. So the water dissociates more, and [H3O+] and [OH−] will always be equal in pure water.

Which of the following gives the best estimate for the pH of a 5 x 10⁻⁴ M Sr(OH)₂ solution at 25 °C?

Sr(OH)2(aq) is a strong base, and its complete dissociation produces 2 moles of OH− ions per mole of Sr(OH)2 that dissolves. As a result, [OH−]=2×(5×10−4M)=1×10−3M and pOH≈3.0, so pH≈11.0.

Which of the following gives the best estimate for the pH of a 1 x 10⁻⁵ M HClO₄(aq) solution at 25 °C?

Since HClO4 is a strong acid, it dissociates completely, producing 1 mol of H3O+ per 1 mol of HClO4(aq). Therefore, pH=−log[H3O+]=−log[HClO4]initial=−log(1×10−5)≈5.0. (In general, the negative log of 10−x is x. Therefore, the negative log of 1×10−5 is 5.)

Which of the following is the correct mathematical relationship to use to calculate the pH of a 0.10 M aqueous HBr solution?

By definition, pH=−log[H3O+]. In this case, pH=−log(1.0×10−1), since for a strong acid like HBr, it is assumed that the acid ionizes 100% and [HBr]initial=[H3O+].

Based on the information given, when the initial concentration of the weak acid increases, the percent ionization decreases and the equilibrium concentration of H3O+ increases. A higher [H3O+]eq results in a lower [OH−]eq and a higher pOH.

Access all questions and much more by creating a free account

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Science

Sr(OH)2(aq) is a strong base, and its complete dissociation produces 2 moles of OH− ions per mole of Sr(OH)2 that dissolves. As a result, [OH−]=2×(5×10⁻⁴M)=1×10⁻³M and pOH≈3.0, so pH≈11.0.

Since HClO4 is a strong acid, it dissociates completely, producing 1 mol of H3O+ per 1 mol of HClO4(aq). Therefore, pH=−log[H3O+]=−log[HClO4]_initial=−log(1×10−5)≈5.0. (In general, the negative log of 10^−x is x. Therefore, the negative log of 1×10−5 is 5.)

By definition, pH=−log[H3O+]. In this case, pH=−log(1.0×10−1), since for a strong acid like HBr, it is assumed that the acid ionizes 100% and [HBr]_initial=[H3O+].