Acid Base Practice Questions

Part I: Multiple Choice

1(1). An aqueous solution of which of the following reacts with magnesium metal?

A. Ammonia C. Potassium hydroxide

B. Hydrogen chloride D. Sodium hydrogencarbonate

(1)

2(2). Which is a buffer solution?

I. 0.01 mol dm−3 HCl, 0.01 mol dm–3 NaCl

II. 0.01 mol dm−3 CH3COOH, 0.01 mol dm–3 CH3COONa

A. I only

B. II only

C. Both I and II

D. Neither I nor II

(1)

3(8). Which one of the following species can act as both a Brønsted-Lowry acid and base in aqueous solution?

A. CH3COOH

B.–3NO

C. H2

–4PO

D. OH−

(1)

4(9). The Ka value for an acid is 1.0 × 10–2. What is the Kb value for its conjugate base?

A. 1.0 × 10−2

B. 1.0 × 10−6

C. 1.0 × 10−10

D. 1.0 × 10−12

1

(1)

5(16). Which of the following is/are formed when a metal oxide reacts with a dilute acid?

I. A metal salt

II. Water

III. Hydrogen gas

A. I only

B. I and II only

C. II and III only

D. I, II and III

(1)

6(17). Four aqueous solutions, I, II, III and IV, are listed below.

I. 0.100 mol dm–3 HCl

II. 0.010 mol dm–3 HCl

III. 0.100 mol dm–3 NaOH

IV. 0.010 mol dm–3 NaOH

What is the correct order of increasing pH of these solutions?

A. I, II, III, IV

B. I, II, IV, III

C. II, I, III, IV

D. II, I, IV, III

(1)

7(21). Which is a conjugate acid-base pair in the following reaction?

HNO3 +H2SO4 H2

3NO +H

–4SO

A. HNO3 and H2SO4

B. HNO3 and H2

3NO

C. HNO3 and H–4SO

D. H2NO3 and H–4SO

2

(1)

8(22). Which equation represents an acid-base reaction according to the Lewis theory but not the Brønsted-Lowry theory?

A. NH3 + HCl NH4Cl

B. 2H2O H3O+ + OH–

C. NaOH + HCl NaCl + H2O

D. CrCl3 + 6NH3 [Cr(NH3)6]3+ + 3Cl–

(1)

9(30). What is the concentration of OH− ions (in mol dm−3) in an aqueous solution in which

[H+] = 2.0 × 10−3 mol dm−3? (KW = 1.0 × 10−14 mol2 dm–6)

A.2.0 × 10−3 C. 5.0 ×10−12

B.4.0 × 10−6 D. 2.0 × 10−17

(1)

10(31). What is the relationship between Ka and PKa?

A. pKa = −log Ka

B. pKa = a

14100.1K

C. pKa = log Ka

D. pKa = a

0.1K

(1)

11(34). The pH of a solution is 2. If its pH is increased to 6, how many times greater is the [H+] of the original solution?

A. 3 C. 1000

B. 4 D. 10 000

(1)

12(37). The acid dissociation constant of a weak acid HA has a value of 1.0 × 10–5 mol dm−3.

What is the pH of a 0.10 mol dm−3 aqueous solution of HA?

A. 2 B. 3 C. 5 D. 6

3

(1)

13(38). Which mixture would produce a buffer solution when dissolved in 1.0 dm3 of water?

A. 0.50 mol of CH3COOH and 0.50 mol of NaOH

B. 0.50 mol of CH3COOH and 0.25 mol of NaOH

C. 0.50 mol of CH3COOH and 1.00 mol of NaOH

D. 0.50 mol of CH3COOH and 0.25 mol of Ba(OH)2

(1)14(39). Which compound, when dissolved in aqueous solution, has the highest pH?

A. NaCl C. NH4Cl

B. Na2CO3 D. NH4NO3

(1)

15(40). In which reaction is H2PO 4

–

(aq) acting as a Brønsted-Lowry base?

A. H2PO 4

–

(aq) + NH3(aq) → HPO 4

–2

(aq) + NH 4

(aq)

B. H2PO 4

–

(aq) + OH−(aq) → HPO 4

–2

(aq) + H2O(l)

C. H2PO 4

–

(aq) + C2H5NH2(aq) → HPO 4

–2

(aq) + C2H5NH 3

(aq)

D. H2PO 4

–

(aq) + CH3COOH(aq) → H3PO4(aq) + CH3COO−(aq)

(1)16(44). Which methods will distinguish between equimolar solutions of a strong base and a strong

acid?

I. Add magnesium to each solution and look for the formation of gas bubbles.

II. Add aqueous sodium hydroxide to each solution and measure the temperature change.

III. Use each solution in a circuit with a battery and lamp and see how bright the lamp glows.

A. I and II only

B. I and III only

C. II and III only

4

D. I, II and III

(1)

17(46). Which values are correct for a solution of NaOH of concentration 0.010 mol dm−3 at 298 K?

(Kw = 1.0 × 10−14 mol2 dm–6 at 298 K)

A. [H+] = 1.0 × 10−2 mol dm−3 and pH = 2.00

B. [OH–] = 1.0 × 10−2 mol dm−3 and pH = 12.00

C. [H+] = 1.0 × 10−12 mol dm−3 and pOH = 12.00

D. [OH− ] = 1.0 × 10−12 mol dm−3 and pOH = 2.00

(1)

18(47). Which solution, of concentration 0.10 mol dm−3, has the highest pH value?

A. HCl(aq) C. NaCl(aq)

B. MgCl2(aq) D. AlCl3(aq)

(1)

19(10). Separate 20.0 cm3

solutions of a weak acid and a strong acid of the same concentration are titrated with NaOH solution. Which will be the same for these two titrations?

I. Initial pH

II. pH at equivalence point

III. Volume of NaOH required to reach the equivalence point

A. I only

B. III only

C. I and II only

D. II and III only

(1)

20(24). An acid-base indicator, HIn, dissociates according to the following equation.

In(aq) H+

(aq) + In–

(aq)

colour A colour B

Which statement about this indicator is correct?

I. In a strongly acidic solution colour B would be seen.

II. In a neutral solution the concentrations of HIn–

(aq) and In (aq) must be equal.

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III. It is suitable for use in titrations involving weak acids and weak bases.

A. I only C. III only

B. II only D. None of the above

(1)

21(32). Which curve is produced by the titration of a 0.1 mol dm−3

weak base with 0.1 mol dm−3

strong acid?

A .

C .

B .

D .

p H

p H

p H

p H

1 2

8

4

1 2

8

4

1 2

8

4

1 2

8

4

Vo lu m e o f titran t

Vo lu m e o f titran t

Vo lu m e o f titran t

Vo lu m e o f titran t

22(33). Consider the following equilibria in 0.10 mol dm−3

carbonic acid.

HCO3(aq) H+

(aq) + HCO 3

–

(aq)

HCO 3

–

(aq) H+

(aq) + CO 3

–2

(aq)

Which species is present in the highest concentration?

A. H2CO3(aq)

B. H+

(aq)

C. HCO 3

–

(aq)

D. CO 3

–2

(aq)

(1)

6

23(48). Which statement about indicators is always correct?

A. The mid-point of an indicator‘s colour change is at pH = 7.

B. The pH range is greater for indicators with higher pK4 values.

C. The colour red indicates an acidic solution.

D. The pK4 value of the indicator is within its pH range.

(1)

7

Part II: Open Response

1(3). Define the terms strong acid and weak acid. Using hydrochloric and ethanoic acid as examples, write equations to show the dissociation of each acid in aqueous solution.

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(Total 4 marks)

2(4). (i) Calcium carbonate is added to separate solutions of hydrochloric acid and ethanoic acid of the same concentration. State one similarity and one difference in the observations you could make.

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(2)(ii) Write an equation for the reaction between hydrochloric acid and calcium carbonate.

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(2)

(iii) Determine the volume of 1.50 mol dm–3 of hydrochloric acid that would react with exactly 1.25 g of calcium carbonate.

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(3)(iv) Calculate the volume of carbon dioxide, measured at 273 K and

1.01 × 105 Pa, which would be produced when 1.25 g of calcium carbonate reacts completely with the hydrochloric acid.

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(2)(Total 9 marks)

3(6). (i) Define the term buffer solution.

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(2)(ii) State a suitable mixture that can act as a buffer solution.

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(2)(Total 4 marks)

4(7). Propanoic acid, CH3CH2COOH is a weak acid.

(a) Give the equation for the ionization of propanoic acid in water and deduce the expression for the ionization constant, Ka, of propanoic acid.

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(2)(b) Calculate the Ka value of propanoic acid using the pKa value in the Data Booklet.

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(1)

(c) Use your answer from (b) to calculate the [H+] in an aqueous solution of propanoic acid of

concentration 0.0500 mol dm–3, and hence the pH of this solution.

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(3)(Total 6 marks)

5(11). Predict whether each of the following solutions would be acidic, alkaline or neutral. In each case explain your reasoning.

(i) 0.1 mol dm–3 FeCl3 (aq)

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(ii) 0.1 mol dm–3 NaNO3 (aq)

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(iii) 0.1 mol dm–3 Na2CO3 (aq)

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(Total 6 marks)

6(15).Use the Data Booklet to find the pKa values of the following acids. State how the presence of substituents in carboxylic acids affects their acid strengths. For each pair, explain the difference in acid strength by referring to the substituents.

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Ethanoic acid and propanoic acid

Chloroethanoic acid and dichloroethanoic acid

Chloroethanoic acid and fluoroethanoic acid

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(Total 5 marks)

7(25). (a) (i) Calculate the Ka value of methanoic acid, HCOOH, using table 16 in the Data Booklet.

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(1)(ii) Based on its Ka value, state and explain whether methanoic acid is a strong or weak

acid.

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(2)

(iii) Calculate the hydrogen ion concentration and the pH of a 0.010 mol dm−3 methanoic acid solution. State one assumption made in arriving at your answer.

(4)...........................................................................................................................

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(b) Explain how you would prepare a buffer solution of pH3.75 starting with methanoic acid.

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(3)(Total 10 marks)

8(51). Choosing suitable examples from the following:

NH3, O2− , Cu2+, OH–, NH 2

, H2O

explain, using a different equation in each case, the meaning of the terms below.

(i) Brønsted-Lowry acid

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(2)(ii) Lewis acid

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(2)

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(iii) conjugate acid-base pair (Identify each member of both acid-base pairs.)

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(3)

13

Acid Base Practice Test – Focus on Titrations and Indicators

9(5). The graph below shows the change in pH when aqueous sodium hydroxide is added to 20 cm3

of aqueous hydrochloric acid.

1 2

11

1 0

9

8

7

6

5

4

3

2

1

0

p H

0 5 1 0 1 5 2 0 2 5 3 0

Vo lu m e o f N aO H / cm 3

By reference to the graph

(i) state the [H+] before any alkali is added.

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(1)

(ii) state by how much the [H+] changes after the addition of 20 cm3

of aqueous sodium hydroxide.

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(1)

(iii) determine the volume of the same sodium hydroxide solution needed to neutralize 20 cm3

of aqueous ethanoic acid of the same concentration as the hydrochloric acid.

14

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(1)(Total 3 marks)

10(12). The following graph shows how the pH changes during the titration of 10 cm3

of a solution of a

weak acid (HA) with 0.10 mol dm–3

NaOH.

1 4

1 3

1 2

11

1 0

9

8

7

6

5

4

3

2

1

00 1 2 3 4 5 6 7 8 9 1 0 11 1 2 1 3 1 4

p H

Vo lu m e o f 0 .1 0 m o l d m N aO H / c m– 3 3

(i) State the pH at the equivalence point and explain why the pH changes rapidly in this region.

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(2)

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(ii) Calculate the initial concentration of the acid (HA).

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(3)

(iii) Calculate the [H+

] of the acid before any sodium hydroxide is added. Use this value to determine the Ka value and the pKa value of the acid.

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(5)(Total 10 marks)

16

10(41). The indicator bromophenol blue, HIn(aq), has a form that is yellow and an In–

(aq) form that is blue.

(a) Write an equation to show how bromophenol blue acts as an indicator.

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(1)

(b) State and explain the colour of bromophenol blue

(i) on the addition of a strong acid.

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(ii) at the equivalence point of a titration.

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(3)

(Total 4 marks)

11(49). An experiment was carried out to determine the concentration of an aqueous solution of

ammonia by titrating it with a solution of sulfuric acid of concentration 0.150 mol dm−3

. It was

found that 25.0 cm3

of the ammonia solution required 20.1 cm3

of the sulfuric acid solution for neutralization.

(a) Write the equation for the reaction and calculate the concentration, in mol dm−3

, of the ammonia solution.

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(4)

(b) Several acid-base indicators are listed in Table 17 of the Data Booklet. State and explain which one of the following indicators should be used for this experiment: bromocresol green, phenol red or phenolphthalein.

17

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(3)

(c) Determine the pOH of a solution with an ammonia concentration of 0.121 mol dm−3

. (pKb of ammonia is 4.75.)

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(4)(Total 11 marks)

18

Acids and Bases – Practice Test for IB Seniors Answer Key – Part I - Multiple Choice

1(1). B

2(2). B

3(8). C

4(9). D

5(16). B

6(17). B

7(21). B

8(22). D

9(30). C10.(31). A11(34). D

12(37). B

13(38). B

14(39). B

15(40). D

16(44). A

17(46). B

18(47). C

19(10). B

20(24). D

21(32). C22(33). A

23(48). D

Acids and Bases – Practice Test for IB Seniors – Part II – Open Response - KEY

1(3). strong acid completely dissociated / ionized;weak acid only partially dissociated / ionized;

HCl(aq) H+(aq) + Cl+(aq);

CH3COOH(aq) CH3COO–(aq) + H+(aq); 4Insist on both arrows as shown, state symbols not needed.

Also accept H2O(1) and H3O+ (aq) in equations

19

2(4). (i) bubbling / effervescence / dissolving of / gas given off CaCO3

(do not accept CO2 produced);more vigorous reaction with HCl / OWTTE; 2

(ii) 2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(1); 2[1] for correct formulas, [1] for balanced, state symbols not essential.

(iii) amount of CaCO3 = 09.1001.25

(no penalty for use of 100);amount of HCl = 2 × 0.0125 = 0.0250 mol (allow ECF);

volume of HCl = 0.0167 dm3 / 16.7 cm3 (allow ECF); 3

(iv) 1:1 ratio of CaCO3 to CO2 to / use 0.0125 moles CO2 (allow ECF);

(0.0125 × 22.4) = 0.28 dm3 / 280 cm3 / 2.8 × 10–4 m3 (allow ECF); 1Accept calculation using pV=nRT.

3(6). (i) a solution that resists pH change / maintains a (nearly) constant pH;when small amounts of acid or alkali are added; 2

(ii) Any suitable example, [1] for each correct component.e.g. ammonia solution and ammonium chloride;

ethanoic acid and sodium ethanoate; 2weak acid salt of weak acid / weak base salt of weak base [1 max] blood [1 max]

4(7). (a) CH3CH2COOH CH3CH2COO– + H+ (no penalty for );

Accept equation with H2O and H2O+. 2

Ka = COOH]CH[CH]][HCOOCH[CH

23

–23

;

(b) 1.35 × 10–5 (mol dm–3) 1Units not needed. No penalty for too many sig figs.

(c) [H+] = COOH]CHCH[ 23aK ;

050.010135 5–

= 8.21 or 8.22 × 10–4 (mol dm–3); 3Units not needed, ECF from (b). No penalty for too many sig figs.(ECF) penalize more than 4 sig figs;

pH (= –log[H+] = –log (8.22 ×10–4) = 3.1 or 3.09 (ECF) penalize more than 4 sig figs;

If expression for [H+] missing but both answers correct, award [3], if one answer correct, award [2].

5(11). (i) acidic;

20

Fe(H2O) 6

3

is a weak acid / Fe3+ reacts with OH– / equation to show

formation of HCl or H+;“FeCl3 is acidic” is not acceptable. 2

(ii) neutral;NaNO3/ sodium nitrate is formed from strong base and strongacid / ions do not hydrolyse; 2

(iii) alkaline;

As CO 3

–2

is weak base / combines with H+/ equation showing

formation of OH–; 2

Acidic, neutral, alkali mark in each case is independent of reason.

6(15). presence of electron-releasing groups decreases acid strength;presence of electron-withdrawing groups increases acid strength;propanoic weaker than ethanoic because of extra electron-releasingCH2 group / the values are similar as the R groups have a similar effect;dichloroethanoic stronger than chloroethanoic because of extraelectron-withdrawing Cl;fluoroethanoic stronger than chloroethanoic because of greater electronegativity/electron-withdrawing power of F; 5

(The first two points may be implicit in the explanations)

7(25). (a) (i) pKa = 3.75, therefore Ka = 1.78 × 10−4 (accept 1.8 × 10−4) 1No units required.

(ii) weak acid;

less [H+] / partial dissociation / more reactants / less products /Ka << 1/ small Ka; 2

(iii) (HCOOH(aq) H+(aq) + HCOO− (aq))

010.0[HCOOH]]][HCOO[H 2–

x

;

(x2 = 1.78 ×10−6)

x = 1.33 × 10−3 mol dm−3 = [H+] (no mark without units);ECF from (a)(i).No penalty for incorrect significant figures.

pH = 2.88 / 2.9 (ECF);assume x << 0.010 / 25°C / negligible dissociation; 4

(b) add strong base / sodium hydroxide or other named alkali / salt of methanoicacid / HCOONa to methanoic acid;

in equimolar amounts / quantities / so that [HCOOH] = [HCOO–];(from Ka expression) pH = pKa (= 3.75); 3

8(51). (i) proton donor / OWTTE;

21

suitable equation; 2

(ii) electron pair acceptor / OWTTE; suitable equation; 2

(iii) two species whose formulas differ by H+ / OWTTE; suitable equation;both acid-base pairs correctly identified; 3

Examples of suitable equations:

NH3 + Η2O → NH 4

+ ΟΗ−

base1 acid2 acid1 base2

NH 2

+ Η2O → NH 4

+ ΟΗ− O2- + H2O → 2OH−

Cu2+ + 4NH3 → Cu(NH3) 4

2

NH 2

+ H2O → NH3 + OH−

Each equation must use at least one of the species in the question. Each equation must be balanced.

KEY for Acids and Bases – Focus on Titrations and Indicators

9(5). (i) 0.1 (mol dm–3

) (units not needed but penalize if incorrect); 1

(ii) to 0.01 / decreases by factor of 10 / goes down by 0.09; 1

(iii) 25 cm3

; 1

[3]

22

10(12). (i) 8.7 ± 0.7;

low [H+

] thus small addition of OH–

has great effect / OH–

increasesrapidly as NaOH is a strong base / logarithmic nature of pH; 2

(ii) volume of NaOH = 8.2 cm3

(exact);

amount of NaOH; 10002.8

× 0.1 = 0.00082 mol;

[HA] = 010.000082.0

× 0.1 = 0.082 mol dm–3

/ 0.082 M; 3

Correct answer [3], units needed for last mark.(iii) correct pH reading from graph (2.9) (allow 2.8 or 3.0);

thus [H+

] = 1.26 × 10–3

(mol dm–3

)

Ka = 082.01010 9.2–9.2–

= 1.9 × 10–5

(mol dm–3

);pKa = 4.71 5

Accept 4.7 and allow ECF from (ii).

If pH given as 2.8, Ka = 3.06 × 10–5

and pKa = 4.51

If pH given as 2.8, Ka = 1.22 × 10–5

and pKa = 4.91If half equivalence method used:

Volume = 4.1 cm3

pKa = 4.75Award [2] out of last [4].

[10]

11(41). (a) HIn(aq) H+

(aq) + In–

(aq); 1

needed for mark. State symbols not essential.

(b) (i) yellow as equilibrium shifts to left to remove (added) H+

(aq); 1Colour and explanation needed for the mark.

(ii) green / blue-yellow;

both HIn(aq) and In–

(aq) are present; 2

[4]

23

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