NESA Chemistry Properties of Acids and Bases

15 sample questions with marking guides and sample answers · Avg. score: 59.5%

Q20
2024
QCAA
Paper 1
1 mark
Q20
1 mark

Identify the polyprotic acid.

A

NH3_3(aq)

B

H3_3PO4_4(aq)

C

(NH4_4)3_3PO4_4(aq)

D

CH3_3COOH(aq)

Reveal Answer
A

NH3_3(aq)

Ammonia (NH3NH_3) acts as a weak base because it accepts protons, rather than donating multiple protons like a polyprotic acid.

B

H3_3PO4_4(aq)

Correct Answer

Phosphoric acid (H3PO4H_3PO_4) is a polyprotic (specifically triprotic) acid because it contains three ionizable hydrogen atoms that can be donated in sequential steps.

C

(NH4_4)3_3PO4_4(aq)

Ammonium phosphate is an ionic salt composed of ammonium and phosphate ions, not a polyprotic acid molecule.

D

CH3_3COOH(aq)

Acetic acid (CH3COOHCH_3COOH) is a monoprotic acid because only the single hydrogen in the carboxyl group (COOH-COOH) is ionizable.

Q33
2023
SCSA
7 marks
Q33

A barium hydroxide solution is titrated against an ammonium chloride solution to produce barium chloride, ammonia and water.

Q33a
2 marks

Write a balanced ionic equation for this reaction.

Reveal Answer

OH(aq)+NH4+(aq)NH3(aq)+H2O(l)\text{OH}^-\text{(aq)} + \text{NH}_4^+\text{(aq)} \rightarrow \text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)}

Marking Criteria
DescriptorMarks

Correct species

1

Correct balancing

1
Q33b
5 marks

Consider the following indicators

IndicatorpH change rangeColour change
Methyl orange3.1–4.4Red to yellow
Bromothymol blue6.2–7.6Yellow to blue
Phenolphthalein8.3–10.0Colourless to pink

Identify the most appropriate indicator for this titration and justify your choice, using an equation to support your answer.

Reveal Answer

Phenolphthalein is the most appropriate indicator.

At the equivalence point, the solution is basic because ammonia is produced, and ammonia hydrolyses in water to form hydroxide ions:

NH3(aq)+H2O(l)NH4+(aq)+OH(aq)\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}

This means that at the equivalence point, [OH]>[H+]\mathrm{[OH^-] > [H^+]}, so the pH is greater than 7.

Therefore, the best indicator is phenolphthalein, because its colour change range is pH 8.310.0\mathrm{pH\ 8.3 - 10.0}, which is in the basic range and matches the equivalence point of this titration.

Marking Criteria
DescriptorMarks

Phenolphthalein

1

Recognition that (at the equivalence point) the ammonia hydrolyses to produce OH-

1

Recognition that at (equivalence point) [OH-]>[H+] (and solution is basic)

1

Recognition that indicator changes in the basic range/indicator colour change/end point is a similar pH to the equivalence point

1

Appropriate equation NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

1
Q33
2025
NESA
7 marks
Q33
7 marks

Chalk is predominantly calcium carbonate. Different brands of chalk vary in their calcium carbonate composition.

The table shows the composition of three different brands of chalk.

 Brand XBrand YBrand Z
CaCO3\text{CaCO}_3 (%)85.583.982.4

The following procedure was used to determine the calcium carbonate composition of a chalk sample.

  • A sample of chalk was crushed in a mortar and pestle.
  • A 3.00 g sample of the crushed chalk was placed in a conical flask.
  • 100.0 mL of 0.550 mol L1 HCl(aq)0.550 \text{ mol L}^{-1} \text{ HCl}(aq) was added to the sample and left to react completely, resulting in a clear solution.
  • Four 20 mL aliquots of this mixture were then titrated with 0.10 mol L1 KOH0.10 \text{ mol L}^{-1} \text{ KOH}.

The results of the titrations are recorded.

Burette volume (mL)Trial 1Trial 2Trial 3Trial 4
Final7.8014.9022.1029.25
Initial0.007.8014.9022.10
Total used7.807.107.207.15

Determine the brand of the chalk sample. Include a relevant chemical equation in your answer.

Reveal Answer

Exclude the outlier.

Average volume KOH used excluding Trial 1 = 7.10+7.20+7.153=0.00715\frac{7.10 + 7.20 + 7.15}{3} = 0.00715 L

HCl(aqaq) + KOH(aqaq) \rightarrow KCl(aqaq) + H2_2O(ll)

moles KOH = 0.10×0.00715=0.0007150.10 \times 0.00715 = 0.000715 mol KOH

Ratio HCl : KOH = 1 : 1
\therefore 0.000715 mol HCl for each sample
\therefore 0.000715×5=0.0035750.000715 \times 5 = 0.003575 mol total in sampled solution

Initial nnHCl = 0.550×0.1000=0.05500.550 \times 0.1000 = 0.0550 mol HCl

nnHCl that reacted with CaCO3_3 = 0.05500.0035750.0550 - 0.003575 = 0.051425 mol HCl

2HCl(aqaq) + CaCO3_3(ss) \rightarrow CaCl2_2(aqaq) + H2_2O(ll) + CO2_2(gg)

Ratio HCl : CaCO3_3 = 2 : 1
\therefore 0.0514252=0.0257125\frac{0.051425}{2} = 0.0257125 mol CaCO3_3

Mass CaCO3_3 = 0.0257125×MM CaCO3(100.09)0.0257125 \times MM \text{ CaCO}_3 \, (100.09) = 2.5735641 g

% CaCO3_3 = 2.57356413.00×100=85.7854%85.8%\frac{2.5735641}{3.00} \times 100 = 85.7854\% \approx 85.8\%

Chalk sample has to be Brand X.

Marking Criteria
DescriptorMarks

Correctly identifies the brand AND Provides all correct calculations AND Provides a balanced chemical equation including states

7

Provides substantially correct calculations AND Provides a balanced chemical equation

6

Provides the main steps in the calculation AND Provides a balanced chemical equation

5

Provides the main steps in the calculation AND Provides a balanced chemical equation

4

Provides some steps in the calculation

3

Provides some steps in the calculation

2

Provides some relevant information

1

None of the above

0
Q12
2024
SCSA
1 mark
Q12
1 mark

One limitation of the Brønsted-Lowry Theory of acids and bases is that it

A

does not explain reactions between acidic and basic oxides, such as SO3(g)+MgO(s)MgSO4(s)SO_3(g) + MgO(s) \rightarrow MgSO_4(s), as they do not involve the transfer of protons.

B

does not explain the production of a neutral salt solution resulting from the reaction between a strong acid and strong base.

C

links acids and bases into conjugate acid-base pairs rather than accounting for the transfer of protons.

D

cannot explain the acidity and basicity of acidic and basic salts.

Reveal Answer
A

does not explain reactions between acidic and basic oxides, such as SO3(g)+MgO(s)MgSO4(s)SO_3(g) + MgO(s) \rightarrow MgSO_4(s), as they do not involve the transfer of protons.

Correct Answer

The Brønsted-Lowry theory defines acids and bases strictly in terms of proton (H+H^+) transfer. Therefore, it cannot explain acid-base reactions where no protons are exchanged, such as the reaction between Lewis acids and bases like SO3SO_3 and MgOMgO.

B

does not explain the production of a neutral salt solution resulting from the reaction between a strong acid and strong base.

The Brønsted-Lowry theory successfully explains neutralization reactions between strong acids and strong bases through the transfer of a proton from the acid (or hydronium ion) to the base (or hydroxide ion).

C

links acids and bases into conjugate acid-base pairs rather than accounting for the transfer of protons.

The concept of conjugate acid-base pairs in the Brønsted-Lowry theory is actually based entirely on the transfer of protons, rather than being an alternative to it.

D

cannot explain the acidity and basicity of acidic and basic salts.

The Brønsted-Lowry theory effectively explains the acidity and basicity of salts through hydrolysis, where the constituent ions of the salt act as proton donors or acceptors when interacting with water.

Q17
2020
SCSA
1 mark
Q17
1 mark

Acid-base indicators

A

are oxidising or reducing agents.

B

change colour at a specific pH value.

C

are strong acids or bases.

D

are weak acids or bases.

Reveal Answer
A

are oxidising or reducing agents.

Acid-base indicators respond to changes in hydrogen ion concentration, not electron transfer. Indicators that act as oxidising or reducing agents are known as redox indicators.

B

change colour at a specific pH value.

Indicators change colour over a specific pH range (typically spanning about 2 pH units), rather than at one exact pH value.

C

are strong acids or bases.

If indicators were strong acids or bases, they would completely dissociate in solution, preventing the reversible equilibrium shift needed to produce a colour change.

D

are weak acids or bases.

Correct Answer

Acid-base indicators are typically weak organic acids or bases that exist in equilibrium with their conjugate forms, where the un-ionised and ionised forms have different colours.

Q27
2020
SCSA
8 marks
Q27
8 marks

Write balanced equations for any reactions occurring between the following substances and describe the observation(s).

If there is no reaction, write 'no reaction' for the equation and if there is no change observed, write 'no visible reaction' for the observations. Where applicable, use the colours stated in the Chemistry Data Booklet.

  • Iron filings and dilute hydrochloric acid

  • Chromium(III) nitrate solution and magnesium ribbon

  • Potassium chloride solution and bromine water

Reveal Answer

Iron filings and dilute hydrochloric acid.
Equation
Fe(s)+2H+(aq)Fe2+(aq)+H2(g)Fe(s) + 2 H^+(aq) \rightarrow Fe^{2+}(aq) + H_2(g)
or
Fe(s)+2HCl(aq)FeCl2(aq)+H2(g)Fe(s) + 2 HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)

Observations (Any two of the following):

  • colourless, (odourless) bubbles /effervescence/gas
  • silver/grey solid dissolves
  • (pale) green solution formed

Chromium(III) nitrate solution and magnesium ribbon.
Equation
2Cr3+(aq)+3Mg(s)2Cr(s)+3Mg2+(aq)2 Cr^{3+}(aq) + 3 Mg(s) \rightarrow 2 Cr(s) + 3 Mg^{2+}(aq)
or
2Cr(NO3)3(aq)+3Mg(s)2Cr(s)+3Mg(NO3)2(aq)2 Cr(NO_3)_3(aq) + 3 Mg(s) \rightarrow 2 Cr(s) + 3 Mg(NO_3)_2(aq)

Observations (Any two of the following):

  • silver/grey solid dissolves
  • blackish/grey/silver solid forms
  • solution becomes less green/colourless

Potassium chloride solution and bromine water.
Equation: No reaction
Observations: No visible reaction

Marking Criteria

Iron filings and dilute hydrochloric acid - Equation

DescriptorMarks

correct species

1

correct balancing

1

Iron filings and dilute hydrochloric acid - Observations

DescriptorMarks

Any two of the following: colourless, (odourless) bubbles /effervescence/gas; silver/grey solid dissolves; (pale) green solution formed

1

Chromium(III) nitrate solution and magnesium ribbon - Equation

DescriptorMarks

correct species

1

correct balancing

1

Chromium(III) nitrate solution and magnesium ribbon - Observations

DescriptorMarks

Any two of the following: silver/grey solid dissolves; blackish/grey/silver solid forms; solution becomes less green/colourless

1

Potassium chloride solution and bromine water

DescriptorMarks

Equation: a statement indicating 'no reaction'

1

Observations: a statement indicating 'no visible reaction'

1
Q11
2024
QCAA
Paper 1
1 mark
Q11
1 mark

A Brønsted–Lowry acid

A

accepts a proton to form its base.

B

donates a proton to form its base.

C

accepts a proton to form its conjugate base.

D

donates a proton to form its conjugate base.

Reveal Answer
A

accepts a proton to form its base.

This describes a Brønsted–Lowry base, not an acid. Acids donate protons, whereas bases accept them.

B

donates a proton to form its base.

While acids do donate protons, the specific term for the species formed after the loss of a proton is the "conjugate base," making Option D the more precise and standard definition.

C

accepts a proton to form its conjugate base.

Acids donate protons rather than accepting them. Additionally, when a species accepts a proton, it forms a conjugate acid, not a conjugate base.

D

donates a proton to form its conjugate base.

Correct Answer

By definition, a Brønsted–Lowry acid is a proton (H+H^+) donor. Once it donates a proton, the remaining species is called its conjugate base.

Q10
2022
QCAA
Paper 1
1 mark
Q10
1 mark

The midpoint of the colour change of a weak acid indicator occurs when

A

[In]=[H+][\text{In}^-] = [\text{H}^+]

B

[In]=[HIn][\text{In}^-] = [\text{HIn}]

C

[H+]=[OH][\text{H}^+] = [\text{OH}^-]

D

[HIn]=[OH][\text{HIn}] = [\text{OH}^-]

Reveal Answer
A

[In]=[H+][\text{In}^-] = [\text{H}^+]

This condition implies the concentration of the indicator anion equals the hydrogen ion concentration, which is not the general definition for the color change midpoint.

B

[In]=[HIn][\text{In}^-] = [\text{HIn}]

Correct Answer

The color of an indicator depends on the ratio of its acid form ([HIn][\text{HIn}]) to its conjugate base form ([In][\text{In}^-]). The midpoint of the color change occurs when these concentrations are equal, resulting in pH=pKIn\text{pH} = \text{pK}_{\text{In}}.

C

[H+]=[OH][\text{H}^+] = [\text{OH}^-]

This condition defines a neutral solution (pH=7\text{pH} = 7 at 25C25^\circ\text{C}), whereas indicators change color at specific pH ranges determined by their pKIn\text{pK}_{\text{In}} values, not necessarily at neutrality.

D

[HIn]=[OH][\text{HIn}] = [\text{OH}^-]

There is no theoretical relationship equating the concentration of the undissociated indicator to the hydroxide ion concentration that defines the midpoint of the color transition.

Q16
2021
QCAA
Paper 1
1 mark
Q16
1 mark

Calculate the percentage yield of magnesium ethanoate when 8.0 moles of ethanoic acid reacts with 6.0 moles of magnesium carbonate, producing 3.5 moles of magnesium ethanoate as shown in the equation.

2CH3COOH(aq)+MgCO3(s)(CH3COO)2Mg(aq)+CO2(g)+H2O(l)2CH_3COOH(aq) + MgCO_3(s) \rightarrow (CH_3COO)_2Mg(aq) + CO_2(g) + H_2O(l)

A

44%

B

58%

C

88%

D

100%

Reveal Answer
A

44%

This value is obtained by dividing the actual yield (3.5 mol) by the initial moles of ethanoic acid (8.0 mol), neglecting the 2:1 stoichiometric ratio which dictates the theoretical yield is half the moles of the acid.

B

58%

This calculation incorrectly identifies magnesium carbonate as the limiting reactant. Since magnesium carbonate is in excess, the theoretical yield should be based on ethanoic acid.

C

88%

Correct Answer

Ethanoic acid is the limiting reactant (8.0 mol/2=4.0 mol8.0 \text{ mol} / 2 = 4.0 \text{ mol} theoretical yield). The percentage yield is calculated as 3.54.0×100=87.5%\frac{3.5}{4.0} \times 100 = 87.5\%, which rounds to 88%.

D

100%

This option implies the actual yield equals the theoretical yield, but 3.5 moles is less than the calculated theoretical yield of 4.0 moles.

Q3
2024
SCSA
1 mark
Q3
1 mark

Which of the following represents an acid-base reaction in which the underlined species is acting as a Brønsted-Lowry acid?

A

2CrO42+2HCO3Cr2O72+2CO32+H2O2 CrO_4^{2-} + 2 \underline{HCO_3^-} \rightleftharpoons Cr_2O_7^{2-} + 2 CO_3^{2-} + H_2O

B

CH3CH2CH2COOH+NaOHCH3CH2CH2COONa+H2OCH_3CH_2CH_2COOH + \underline{NaOH} \rightleftharpoons CH_3CH_2CH_2COONa + H_2O

C

NH3+O2NH2+OH\underline{NH_3} + O^{2-} \rightleftharpoons NH_2^- + OH^-

D

HClO3+CH3NH2ClO3+CH3NH3+HClO_3 + \underline{CH_3NH_2} \rightleftharpoons ClO_3^- + CH_3NH_3^+

Reveal Answer
A

2CrO42+2HCO3Cr2O72+2CO32+H2O2 CrO_4^{2-} + 2 \underline{HCO_3^-} \rightleftharpoons Cr_2O_7^{2-} + 2 CO_3^{2-} + H_2O

While HCO3HCO_3^- does donate a proton, the overall process involves the condensation of chromate into dichromate, making it a more complex reaction rather than a simple Brønsted-Lowry acid-base proton transfer.

B

CH3CH2CH2COOH+NaOHCH3CH2CH2COONa+H2OCH_3CH_2CH_2COOH + \underline{NaOH} \rightleftharpoons CH_3CH_2CH_2COONa + H_2O

NaOHNaOH provides the hydroxide ion (OHOH^-), which accepts a proton from the carboxylic acid. Therefore, it acts as a Brønsted-Lowry base, not an acid.

C

NH3+O2NH2+OH\underline{NH_3} + O^{2-} \rightleftharpoons NH_2^- + OH^-

Correct Answer

A Brønsted-Lowry acid is defined as a proton (H+H^+) donor. In this reaction, NH3NH_3 donates a proton to O2O^{2-} to become NH2NH_2^-, perfectly fitting the definition.

D

HClO3+CH3NH2ClO3+CH3NH3+HClO_3 + \underline{CH_3NH_2} \rightleftharpoons ClO_3^- + CH_3NH_3^+

CH3NH2CH_3NH_2 accepts a proton from HClO3HClO_3 to form CH3NH3+CH_3NH_3^+. Because it accepts a proton, it is acting as a Brønsted-Lowry base.

Q27
2022
QCAA
Paper 1
5 marks
Q27
5 marks

Five colourless 0.1 M solutions of NH3_3, HCl, KOH, H2_2SO4_4 and CH3_3CH2_2COOH have lost their labels. The substances are randomly relabelled A, B, C, D and E. The conductivity of each solution and the colour of the solution when phenol red was added are shown.

SolutionConductivity (S/m)Colour with phenol red
A4.1yellow
B0.14red
C0.08yellow
D6.7yellow
E4.9red

Identify the five solutions. Explain your reasoning.

Reveal Answer

Solutions A, C and D are acids and solutions B and E are bases, because phenol red is yellow when pH < 6.8 and red when pH > 8.4.
Solution C is a weak electrolyte (from low conductivity), therefore C is propanoic acid.
Solution D has higher conductivity than solution A, therefore D is sulfuric acid and solution A is HCl, because H2SO4H_2SO_4 is diprotic acid and will give a higher concentration of ions in solution than monoprotic HCl.
Solution E is KOH as it has a higher conductivity and therefore is a strong base.
Solution B has a lower conductivity and therefore is ammonia, a weak base.

Marking Criteria
DescriptorMarks

Identifies all five solutions

1

Uses indicator data to identify acids and bases

1

Uses conductivity data to identify relative strength of bases

1

Uses conductivity data to identify relative strength of acids

1

Identifies the diprotic acid is more conductive than monoprotic acid

1
Q7
2020
SCSA
1 mark
Q7
1 mark

The following equation shows the reaction between copper and concentrated nitric acid:

4 HNO3()+Cu(s)Cu(NO3)2(aq)+2 NO2(g)+2 H2O()4 \text{ HNO}_3(\ell) + \text{Cu(s)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2 \text{ NO}_2\text{(g)} + 2 \text{ H}_2\text{O}(\ell)

Observable changes associated with this reaction are the dissolving of the copper, the formation of a deep blue solution and the evolution of a pungent brown gas.

Which of the following are some of the atomic/molecular scale events needed for these observable changes to occur?

(i) collisions between HNO3\text{HNO}_3 molecules and Cu atoms
(ii) donation and acceptance of protons
(iii) reduction of copper atoms

A

i only

B

ii only

C

i and iii only

D

i, ii and iii

Reveal Answer
A

i only

Correct Answer

This is correct because for any chemical reaction to occur, the reactant particles (in this case, HNO3\text{HNO}_3 molecules and Cu atoms) must collide. Statements ii and iii are incorrect.

B

ii only

This is incorrect because the reaction is a redox process, not a simple acid-base proton transfer. Furthermore, reactant collisions (statement i) are essential.

C

i and iii only

This is incorrect because copper atoms are oxidized (losing electrons to go from an oxidation state of 0 to +2), not reduced.

D

i, ii and iii

This is incorrect because copper is oxidized rather than reduced, making statement iii false. Additionally, the primary mechanism is redox, not proton donation/acceptance.

Q22
2023
NESA
4 marks
Q22
4 marks

Explain how the following substances would be classified under the Arrhenius and Brønsted–Lowry definitions of acids. Support your answer with relevant equations.

  • HCl(aq)\text{HCl}(aq)
  • NH4Cl(aq)\text{NH}_4\text{Cl}(aq)
Reveal Answer

According to Arrhenius, acids are hydrogen-containing compounds that dissociate in water to give H+\text{H}^+ ions. HCl(aq)\text{HCl}(aq) would be considered an acid by Arrhenius as it produces H+\text{H}^+ ions in water.

HCl(aq)H+(aq)+Cl(aq)\text{HCl}(aq) \rightarrow \text{H}^+(aq) + \text{Cl}^-(aq)

Arrhenius would not recognise the salt NH4Cl\text{NH}_4\text{Cl} as an acid, as the predominant ions present in aqueous solution are ammonium and chloride.

In Brønsted–Lowry theory, acids are defined as proton donors. HCl(aq)\text{HCl}(aq) is a proton donor and therefore a Br€™nsted–Lowry acid.

Ammonium chloride (NH4Cl\text{NH}_4\text{Cl}) is classified as a Br€™nsted–Lowry acid as the ammonium ion donates a proton to water and forms a hydronium ion.

NH4+(aq)+H2O(l)NH3(aq)+H3O+(aq)\text{NH}_4^+(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_3(aq) + \text{H}_3\text{O}^+(aq)
Marking Criteria
DescriptorMarks
  • Classifies both substances with respect to Arrhenius and Brønsted–Lowry theories
  • Provides two relevant chemical equations
4
  • Classifies one substance with respect to Arrhenius and Brønsted–Lowry theories
  • Provides a relevant chemical equation
    OR
  • Classifies both substances with respect to Arrhenius OR Brønsted–Lowry theory
  • Provides a relevant chemical equation
    OR
  • Classifies both substances with respect to Arrhenius and Brønsted–Lowry theories
3
  • Classifies one substance with respect to Arrhenius and Brønsted–Lowry theories
    OR
  • Classifies both substances with respect to Arrhenius OR Brønsted–Lowry theory
2

• Provides some relevant information

1

None of the above

0
Q1
2022
SCSA
1 mark
Q1
1 mark

Consider the following statements about acid-base indicators. Acid-base indicators
(i) change colour as the concentration of hydrogen ions changes.
(ii) are weak acids or bases.
(iii) must not react with the reactants or products in a titration.
(iv) must be used in large volumes for the best results.

Which of the above statements is/are correct?

A

i only

B

i and ii

C

i and iii

D

i, ii, iii and iv

Reveal Answer
A

i only

Incorrect. While statement (i) is true, statement (ii) is also correct because acid-base indicators are typically weak acids or weak bases.

B

i and ii

Correct Answer

Correct. Acid-base indicators are weak acids or bases that change color depending on the hydrogen ion concentration ([H+][H^+]) in the solution.

C

i and iii

Incorrect. Statement (iii) is false because indicators must react with the acid or base in the solution to change color, though they are used in small amounts.

D

i, ii, iii and iv

Incorrect. Statements (iii) and (iv) are false. Indicators do react with the reactants, and they must be used in very small volumes so they do not consume a significant amount of the titrant.

Q15
2021
SCSA
1 mark
Q15
1 mark

Which of the following processes does not contribute to the building of weaker seashells through ocean acidification?

A

HCO3(aq)+H2O()CO32(aq)+H3O+(aq)HCO_3^-(aq) + H_2O(\ell) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)

B

2H+(aq)+CaCO3(s)Ca2+(aq)+CO2(aq)+H2O()2 H^+(aq) + CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_2(aq) + H_2O(\ell)

C

CO2(g)+H2O()H2CO3(aq)CO_2(g) + H_2O(\ell) \rightleftharpoons H_2CO_3(aq)

D

H2CO3(aq)+H2O()HCO3(aq)+H3O+(aq)H_2CO_3(aq) + H_2O(\ell) \rightleftharpoons HCO_3^-(aq) + H_3O^+(aq)

Reveal Answer
A

HCO3(aq)+H2O()CO32(aq)+H3O+(aq)HCO_3^-(aq) + H_2O(\ell) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)

This equilibrium determines the concentration of carbonate ions (CO32CO_3^{2-}). Increased ocean acidity shifts this reaction to the left, reducing the carbonate available for marine organisms to build strong shells.

B

2H+(aq)+CaCO3(s)Ca2+(aq)+CO2(aq)+H2O()2 H^+(aq) + CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_2(aq) + H_2O(\ell)

Correct Answer

This reaction represents the dissolution of already-formed calcium carbonate (CaCO3CaCO_3) shells by increased acidity. While it destroys existing shells, it is not a process that inhibits the initial building of new shells.

C

CO2(g)+H2O()H2CO3(aq)CO_2(g) + H_2O(\ell) \rightleftharpoons H_2CO_3(aq)

This reaction shows carbon dioxide dissolving in water to form carbonic acid, which is the primary driver of ocean acidification that ultimately impairs shell building.

D

H2CO3(aq)+H2O()HCO3(aq)+H3O+(aq)H_2CO_3(aq) + H_2O(\ell) \rightleftharpoons HCO_3^-(aq) + H_3O^+(aq)

This reaction shows the dissociation of carbonic acid into bicarbonate and hydronium ions, which increases ocean acidity and leads to a reduction in the carbonate ions needed for shell building.

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