Acids Unit Midterm Practice Test

1. Consider the following:

I H₂CO₃ + F^- D HCO₃^- + HF

II HCO₃^- + HC₂O₄^- D H₂CO₃ + C₂O₄^2-

III HCO₃^- + H₂C₆H₆O₇^- D H₂CO₃ + HC₆H₅O₇^2-

The HCO₃^- is a base in

A. I only

B. I and II only

C. II and III only

D. I, II, and III

2. Consider the following equilibrium for an indicator:

HInd + H₂O D Ind^- + H₃O⁺

When a few drops of indicator methyl red are added to 1.0 M HCl, the colour of the resulting solution is

A. red and the products are favoured

B. red and the reactants are favoured

C. yellow and the products are favoured

D. yellow and the reactants are favoured

3. The volume of 0.200 M Sr(OH)₂ needed to neutralize 50.0 mL of 0.200 M HI is

A. 10.0 mL

B. 25.0 mL

C. 50.0 mL

D. 100.0 mL

4. The pOH of 0.050 M HCl is

A. 0.050

B. 1.30

C. 12.70

D. 13.70

5. The volume of 0.450 M HCl needed to neutralize 40.0 mL of Sr(OH)₂ is

A. 18.0 mL

B. 20.0 mL

C. 40.0 mL

D. 80.0 mL

6. Consider the following

I H₃PO₄ II H₂PO₄^- III HPO₄^2- IV PO₄^3-

Which of the following solutions will have the largest [H₃0⁺]?

A. I and II only

B. II and III only

C. I, II, and III only

D. II, III, and IV only

7. Which of the following solutions will have the largest [H₃O⁺]?

A. 1.0 M HNO₂

B. 1.0 M H₃BO₃

C. 1.0 M H₂C₂O₄

D. 1.0 M HCOOH

8. Consider the following: H₂O + 57 kJ D H₃O⁺ + OH^-

When the temperature of the system is increased, the equilibrium shifts

A. left and the Kw increases

B. left and the Kw decreases

C. right and the Kw increases

D. right and the Kw decreases

9. Normal rainwater has a pH of approximately 6 as a result of dissolved

A. oxygen

B. carbon dioxide

C. sulphur dioxide

D. nitrogen dioxide

10. A 1.0 M solution of sodium dihydrogen phosphate is

A. acidic and the pH < 7.00

B. acidic and the pH > 7.00

C. basic and the pH < 7.00

D. basic and the pH > 7.00

11. Consider the following equilibrium for an indicator:

HInd + H₂O D Ind^- + H₃O⁺

When a few drops of indicator chlorophenol red are added to a colourless solution of pH 4.0, the resulting solution is

A. red as [HInd] < [Ind^-]

B. red as [HInd] > [Ind^-]

C. yellow as [HInd] < [Ind^-]

D. yellow as [HInd] > [Ind^-]

12. A Bronsted-Lowry base is defined as a chemical species that

A. accepts protons

B. neutralizes acids

C. donated electrons

D. produces hydroxides ions in solution

13. Which of the following solutions will have the greatest electrical conductivity?

A. 1.0 M HCN

B. 1.0 M H₂SO₄

C. 1.0 M H₃PO₄

D. 1.0 M CH₃COOH

14. Consider the following equilibrium: HC₆H₅O₇^{2- +} HIO₃ D H₂C₆H₅O₇^- + IO₃^-

The order of Bronsted-Lowry acids and bases is

A. acid, base, acid, base

B. acid, base, base, acid

C. base, acid, acid, base

D. base, acid, base acid

15. Consider the following: H₂O_(l) D H⁺ + OH^-

When a small amount of 1.0 M KOH is added to the above system, the equilibrium

A. shifts left and [H⁺] decreases

B. shifts left and [H⁺] increases

C. shifts right and [H⁺] decreases

D. shifts right and [H⁺] increases

16. Which of the following has the highest pH?

A. 1.0 M NaIO₃

B. 1.0 M Na₂CO₃

C. 1.0 M Na₃PO₄

D. 1.0 M Na₂SO₄

17. In a 100.0 mL sample of 0.0800 M NaOH the [H₃O⁺] is

A. 1.25 x 10^-13 M

B. 1.25 x 10^-12 M

C. 8.00 x 10^-3 M

D. 8.00 x 10^-2 M

18. Consider the following:

I ammonium nitrate II calcium nitrate III iron III nitrate

When dissolved in water, which of these salts would form a neutral solution?

A. II only

B. III only

C. I and III only

D. I, II, and III

19. Consider the following: SO₄^{2- +} HNO₂ D HSO₄^- + NO₂^-

Equilibrium would favour the

A. the products since HSO₄^- is a weaker acid than HNO₂

B. the reactants since HSO₄^- is a weaker acid than HNO₂

C. the products since HSO₄^- is a stronger acid than HNO₂

D. the reactants since HSO₄^- is a stronger acid than HNO₂

20. The net ionic equation for the hydrolysis of Na₂CO₃ is

A. H₂O + Na⁺ D NaOH + H⁺

B. H₂O + 2Na⁺ D Na₂O + 2H⁺

C. H₂O + CO₃^2- D H₂CO₃ + O^2-

D. H₂O + CO₃^2- D HCO₃^- + OH^-

21. Consider the following equilibrium: 2H₂O_(l) D H₃O⁺ + OH^-

A few drops of 1.0 M HCl are added to the above system. When equilibrium is

re-established, the

A. [H₃O⁺] has increased and the [OH^-] has decreased

B. [H₃O⁺] has increased and the [OH^-] has increased

C. [H₃O⁺] has decreased and the [OH^-] has increased

D. [H₃O⁺] has decreased and the [OH^-] has decreased

22. A basic solution

A. tastes sour

B. feels slippery

C. does not conduct electricity

D. reacts with metals to release oxygen gas

23. The balanced formula equation for the neutralization of H₂SO₄ by KOH is

A. H₂SO₄ + KOH → KSO₄ + H₂O

B. H₂SO₄ + KOH → K₂SO₄ + H₂O

C. H₂SO₄ + 2KOH → K₂SO₄ + H₂O

D. H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

24. An Arrhenius base is defined as a substance which

A. donates protons

B. donates electrons

C. produces H⁺ in solution

D. produces OH^- in solution

25. Consider the following equilibrium: HS^- + H₃PO₄ D H₂S + H₂PO₄^-

The order of Bronsted-Lowry acids and bases is

A. acid, base, acid, base.

B. acid, base, base, acid

C. base, acid, acid, base

D. base, acid, base, acid

26. The equation representing the reaction of ethanoic acid with water is

A. CH₃COO^- + H₂O D CH₃COOH + OH^-

B. CH₃COO^- + H₂O D CH₃COO^2- + H₃O⁺

C. CH₃COOH + H₂O D CH₃COO^- + H₃O⁺

D. CH₃COOH + H₂O D CH₃COOH₂⁺ + OH^-

27. Consider the following equilibrium: 2H₂O + 57kJ D H₃O⁺ + OH^-

When the temperature is decreased, the water

A. stays neutral and the [H₃O⁺] increases

B. stays neutral and the [H₃O⁺] decreases

C. becomes basic and [H₃O⁺] decreases

D. becomes acidic and [H₃O⁺] increases

28. The equation for the reaction of Cl₂O with water is

A. Cl₂O + H₂O D 2HClO

B. Cl₂O + H₂O D 2ClO + H₂

C. Cl₂O + H₂O D Cl₂ + H₂O₂

D. Cl₂O + H₂O D Cl₂ + O₂ + H₂

29. The conjugate acid of C₆H₅0^- is

A. C₆H₄O^-

B. C₆H₅OH

C. C₆H₄O^2-

D. C₆H₅OH⁺

30. Which of the following solutions will have the greatest electrical conductivity?

A. 1.0 M HCl

B. 1.0 M HNO₂

C. 1.0 M H₃BO₃

D. 1.0 M HCOOH

31. A solution of 1.0 M HF has

A. a lower pH than a solution of 1.0 M HCl

B. a higher pOH than a solution of 1.0 M HCl

C. a higher [OH^-] than a solution of 1.0 M HCl

D. a higher [H₃O⁺] than a solution of 1.0 M HCl

32. Which of the following is the weakest acid

A. HIO₃

B. HCN

C. HNO₃

D. C₆H₅COOH

33. Considering the following data

H₃AsO₄ Ka = 5.0 x 10^-5

H₂AsO₄^- Ka = 8.0 x 10^-8

HAsO₄^2- Ka = 6.0 x 10^-10

The Kb value for H₂AsO₄^- is

A. 2.0 x 10^-10

B. 8.0 x 10^-8

C. 1.2 x 10^-7

D. 1.7 x 10^-5

34. In a solution at 25^oC, the [H₃O⁺] is 3.5 x 10^-6 M. The [OH^-] is

A. 3.5 x 10^-20 M

B. 2.9 x 10^-9 M

C. 1.0 x 10^-7 M

D. 3.5 x 10^-6 M

35. In a solution with a pOH of 4.22, the [OH-] is

A. 1.7 x 10^-10 M

B. 6.0 x 10^-5 M

C. 6.3 x 10^-1 M

D. 1.7 x 10⁴ M

36. An aqueous solution of NH₄CN is

A. basic because Ka < Kb

B. basic because Ka > Kb

C. acidic because Ka < Kb

D. acidic because Ka > Kb

37. The net ionic equation for the predominant hydrolysis reaction of KHSO₄ is

A. HSO₄^- + H₂O D SO₄^2- + H₃O⁺

B. HSO₄^- + H₂O D H₂SO₄ + OH^-

C. KHSO₄ + H₂O D K⁺ + SO₄^2- + H₃O⁺

D. KHSO₄ + H₂O D K⁺ + H₂SO₄ + OH^-

38. The [OH^-] in an aqueous solution always equals

A. Kw x [H₃O⁺]

B. Kw - [H₃O⁺]

C. Kw/[H₃O⁺]

D. [H₃O⁺]/Kw

39. The [H₃O⁺] in a solution with pH of 0.253 is

A. 5.58 x 10^-15 M

B. 1.79 x 10^-14 M

C. 5.58 x 10^-1 M

D. 5.97 x 10^-1 M

40. The equilibrium expression for the hydrolysis reaction of 1.0 M K₂HPO₄ is

A. [H₂PO₄^-][OH^-] B. [H₃PO₄][OH^-]

[HPO₄^-] [H₂PO₄^-]

C. [K⁺] [KHPO₄^-] D. [K⁺]² [HPO₄^2-]

[K₂HPO₄] [K₂HPO₄]

41. The solution with the highest pH is

A. 1.0 M NaCl

B. 1.0 M NaCN

C. 1.0 M NaIO₃

D. 1.0 M Na₂SO₄

42. The pH of 100.0 mL of 0.0050 M NaOH is

A. 2.30

B. 3.30

C. 10.70

D. 11.70

43. Consider the following equilibrium for an indicator: HInd + H₂O D Ind^- + H₃O⁺

At the transition point,

A. [HInd] > [Ind^-]

B. [HInd] = [Ind^-]

C. [HInd] < [Ind^-]

D. [HInd] = [H₃O⁺]

Acids Unit Midterm Practice Test Subjective

1. a) Write the net ionic equation for the reaction between NaHSO₃ and NaHC₂O₄.

HSO₃^- + HC₂O₄^- D H₂SO₃ + C₂O₄^2-

b) Explain why the reactants are favoured in the above reaction.

H₂SO₃ is a stronger acid than HC₂O₄^-

2. What is the [H₃O⁺] in a solution formed by adding 60.0 mL of water to 40.0 mL of 0.400 mL of KOH?

KOH → K⁺ + OH^-

40 x 0.40 M 0.16 M 0.16 M

100

[H⁺][OH^-] = 1.00 x 10^-14 [H⁺] = 6.25 x 10^-14 M

3. A solution of 0.100 M HOCN has a pH of 2.24. Calculate the Ka value for the acid.

Ka = 10^-2.24 = 0.005754 M

HOCN D H⁺ + OCN^-

I 0.100 M 0 0

C 0.005754 M 0.005754 M 0.005754 M

E 0.094245 M 0.005754 M 0.005754 M

Ka = (0.005754 )² = 3.5 x 10^-4

0.094245

4. Calculate the pH in 100.0 mL M H₃BO₃.

H₃B0₃ D H⁺ + H₂BO₃^-

I 0.400 M 0 0

C x x x

E 0.400 - x x x

0 small Ka approximation

(x)² = 7.3 x 10^-10

0.400

x = 1.709 x 10^-5 M

pH = 4.77

5. Calculate the pH of the solution formed by mixing 20.0 mL of 0.500 M HCl with 30.0 mL of 0.300 M NaOH.

HCl + NaOH → NaCl + H₂O

0.0200 L x 0.500 mole = 0.01000 mole HCl 0.0300 L x 0.500 mole = 0.01000 mole NaOH

L L

I 0.00100 mole 0.000900 mole

C 0.000900 mole 0.000900 mole

E 0.00010 mole 0

Note the loss of significant figures! Total new volume = 20.0 mL + 30.0 mL = 50.0 mL = 0.0500 L

[H⁺] = 0.00010 moles = 0.0020 M

0.0500 L

pH = 1.70 M

6. a) Write the balanced equation representing the reaction of HF with H₂O.

HF + H₂O D H₃O⁺ + F^-

b) Identify the Bronsted-Lowry bases in the above equation.

H₂O and F^-

7. Consider the following data:

Barbituric acid HC₄H₃N₂O₃ Ka = 9.8 x 10^-5

Sodium propanoate NaC₃H₅O₂ Kb = 7.5 x 10^-10

Propanoic acid HC₃H₅O₂ Ka = ?

Which is the stronger acid, propanoic acid or babituric acid? Explain using calculations.

Ka (HC₄H₃N₂O₃) = Kw = 1.0 x 10^-14 = 1.3 x 10^-5

Kb(C₃H₅O₂^-) 7.5 x 10^-10

Barbituric acid is a stronger acid because it has a larger Ka.

8. A solution of 0.0100 M lactic acid, HC₃H₅O₃, has a pH of 2.95. Calculate the Ka value.

HC₃H₅O₃ D H⁺ + C₃H₅O₃^-

I 0.0100 M 0 0

C 0.001122 0.001122 0.001122

E 0.008878 0.001122 0.001122

Ka = (0.001122)² = 1.4 x 10^-4

0.008878

9. a) Write equations showing the amphiprotic nature of water as it reacts with HCO₃^-.

HCO₃^- + H₂O D H₃O⁺ + CO₃^2-

HCO₃^- + H₂O D H₂CO₃ + OH^-

b) Calculate the Kb for HCO₃^-.

Kb (HCO₃^-) = Kw = 1.0 x 10^-14 = 2.3 x 10^-8

Ka(H₂CO₃) 4.3 x 10^-7

10. Calculate the [H₃O⁺] in 0.550 M C₆H₅COOH.

C₆H₅COOH D H⁺ + C₆H₅COO^-

I 0.550 M 0 0

C x x x

E 0.550 - x x x

0 small Ka approximation

(x)² = 6.5 x 10^-5

0.550

x = [H₃O⁺] = 6.0 x 10^-3 M