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CHEMISTRY, M6 EQ-Bank 25

The graph shows changes in pH for the titrations of equal volumes of solutions of two monoprotic acids, Acid 1 and Acid 2.
 

Explain the differences between Acid 1 and Acid 2 in terms of their relative strengths and concentrations.   (3 marks)

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→ Acid 1 is a strong acid. Its initial pH = 1 and its equivalence point is at pH = 7.

→ Acid 2 is a weaker acid. Its initial pH ~ 2 and its equivalence point is >7.

→ Acid 2 has a higher concentration than Acid 1 as it doesn’t take that much more \(\ce{KOH}\) to neutralise it.

Show Worked Solution

→ Acid 1 is a strong acid. Its initial pH = 1 and its equivalence point is at pH = 7.

→ Acid 2 is a weaker acid. Its initial pH ~ 2 and its equivalence point is >7.

→ Acid 2 has a higher concentration than Acid 1 as it doesn’t take that much more \(\ce{KOH}\) to neutralise it.

CHEMISTRY, M6 2016 HSC 6 MC

Which combination of equimolar solutions would produce the most basic mixture?

  1. Acetic acid and barium hydroxide
  2. Acetic acid and sodium carbonate
  3. Sulfuric acid and barium hydroxide
  4. Sulfuric acid and sodium carbonate
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`A`

Show Worked Solution

→ Acetic acid \(\ce{(CH3COOH)}\) is a monoprotic acid, which when mixed with barium hydroxide \(\ce{(Ba(OH)2)}\), leaves excess hydroxide ions creating a basic solution.

→ When acetic acid reacts with sodium carbonate \(\ce{(Na2CO3)}\), it leaves excess sodium carbonate in the solution. Although this is the same concentration as the hydroxide ions above, it is a weaker base.

→ Sulfuric acid \(\ce{(H2SO4)}\) is a strong diprotic acid and the mixtures in C and D will produce more neutral solutions.

`=>A`

CHEMISTRY, M6 2015 HSC 13 MC

Which of the following solutions has the highest pH?

  1. `1.0 \ text{mol L}^(-1)` acetic acid
  2. `0.10 \ text{mol L}^(-1)` acetic acid
  3. `1.0 \ text{mol L}^(-1)` hydrochloric acid
  4. `0.10 \ text{mol L}^(-1)` hydrochloric acid
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`B`

Show Worked Solution

By Elimination:

→ Acetic acid is weaker than hydrochloric acid and therefore has a higher pH (eliminate C and D).

→ A more dilute acid has a higher pH (eliminate A).

`=>B`


Mean mark 55%.

CHEMISTRY, M6 2020 HSC 34

The effect of concentration on the pH of acrylic acid `(text{C}_2 text{H}_3 text{COOH})` and hydrochloric acid `(text{HCl})` solutions is shown in the graph. Both of these acids are monoprotic.
 

 

Explain the trends in the graph. Include relevant chemical equations in your answer.   (4 marks)

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\(\ce{HCl(aq) -> H+ (aq) + Cl– (aq)}\)

\(\ce{C2H3COOH(aq) \rightleftharpoons H+ (aq) + C2H3COO– (aq)}\)

→ \(\ce{HCl}\) is a strong acid that fully dissociates in water, resulting in a high concentration of \(\ce{H+}\) ions and a low pH.

→ Acrylic acid, on the other hand, is a weak acid that only partially dissociates in water, resulting in a lower concentration of \(\ce{H+}\) ions and a higher pH.

→ When the concentration of \(\ce{HCl}\) decreases by a factor of 10, its pH increases by 1 due to the decrease in \(\ce{H+}\).

→ However, when the concentration of acrylic acid decreases by a factor of 10, its pH increases by less than 1.

→ This is due to the change in pH causing the equilibrium to shift right, producing more \(\ce{H+}\) ions in response to the dilution, resulting in a smaller change in the concentration of \(\ce{H+}\), and thus smaller change in pH.

→ At very dilute concentrations, the degree of dissociation of acrylic acid approaches 100% and its pH converges closely to that of \(\ce{HCl}\).

Show Worked Solution

\(\ce{HCl(aq) -> H+ (aq) + Cl– (aq)}\)

\(\ce{C2H3COOH(aq) \rightleftharpoons H+ (aq) + C2H3COO– (aq)}\)

→ \(\ce{HCl}\) is a strong acid that fully dissociates in water, resulting in a high concentration of \(\ce{H+}\) ions and a low pH.

→ Acrylic acid, on the other hand, is a weak acid that only partially dissociates in water, resulting in a lower concentration of \(\ce{H+}\) ions and a higher pH.

→ When the concentration of \(\ce{HCl}\) decreases by a factor of 10, its pH increases by 1 due to the decrease in \(\ce{H+}\).

→ However, when the concentration of acrylic acid decreases by a factor of 10, its pH increases by less than 1.

→ This is due to the change in pH causing the equilibrium to shift right, producing more \(\ce{H+}\) ions in response to the dilution, resulting in a smaller change in the concentration of \(\ce{H+}\), and thus smaller change in pH.

→ At very dilute concentrations, the degree of dissociation of acrylic acid approaches 100% and its pH converges closely to that of \(\ce{HCl}\).

CHEMISTRY, M6 2020 HSC 18 MC

An aqueous solution of sodium hydrogen carbonate has a pH greater than 7 .

Which statement best explains this observation?

  1. `text{H}_(2) text{O}(l)` is a stronger acid than `text{HCO}_(3)^(\ -)(aq)`.
  2. `text{HCO}_(3)^(\ -)(aq)` is a weaker acid than `text{H}_(2) text{CO}_(3)(aq)`.
  3. `text{Na}^(+)(aq)` reacts with water to produce the strong base `text{NaOH}(aq)`.
  4. The conjugate acid of `text{HCO}_(3)^(\ -)(aq)` is a stronger acid than `text{H}_(2)text{O}(l)`.
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`A`

Show Worked Solution

→ \(\ce{HCO3-}\) can act as an acid or a base (amphiprotic).

→ \(\ce{H2O}\) is a stronger acid than \(\ce{HCO3-}\), so \(\ce{H2O}\) donates a proton to \(\ce{HCO3-}\).

→ As a result, this produces \(\ce{OH-}\) ions in solution, causing the pH to be greater than 7.

`=> A`


♦♦♦ Mean mark 23%.

CHEMISTRY, M6 2022 HSC 25

The pH of two aqueous solutions was compared.
 

Explain why the `\text{HCN}(aq)` solution has a higher pH than the `\text{HCl}(aq)` solution. Include a relevant chemical equation for the `\text{HCN}(aq)` solution.   (3 marks)

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→ \( \ce{HCl} \) is a strong acid, ie it completely ionises in water to form \( \ce{H+} \) ions.

→ On the other hand, \( \ce{HCN} \) is a weak acid, ie it partially ionises in water to form \( \ce{H+} \) ions.

\( \ce{HCl(aq) -> H+ (aq) + Cl- (aq)} \)

\( \ce{HCN(aq) \rightleftharpoons H+ (aq) + CN– (aq)}\)

→ As \( \ce{[H+]} \) decreases, pH increases  (\( \ce{\text{pH} = – log [H+]} \))

→ Therefore, at the same 0.2M, the \( \ce{HCN} \) solution would have a lower \( \ce{[H+]} \) and thus would have a higher pH than \( \ce{HCl} \).

Show Worked Solution

→ \( \ce{HCl} \) is a strong acid, ie it completely ionises in water to form \( \ce{H+} \) ions.

→ On the other hand, \( \ce{HCN} \) is a weak acid, ie it partially ionises in water to form \( \ce{H+} \) ions.

\( \ce{HCl(aq) -> H+ (aq) + Cl- (aq)} \)

\( \ce{HCN(aq) \rightleftharpoons H+ (aq) + CN– (aq)}\)

→ As \( \ce{[H+]} \) decreases, pH increases  (\( \ce{\text{pH} = – log [H+]} \))

→ Therefore, at the same 0.2M, the \( \ce{HCN} \) solution would have a lower \( \ce{[H+]} \) and thus would have a higher pH than \( \ce{HCl} \).


Mean mark 57%.

CHEMISTRY, M6 2021 HSC 32

The molar enthalpies of neutralisation of three reactions are given.

Reaction 1:

\(\ce{HCl($aq$) + KOH($aq$) -> KCl($aq$) + H2O($l$)}\)                \(\ce{Δ$H$}\) \(\pu{=-57.6 kJ mol-1}\)

Reaction 2:

\(\ce{HNO3($aq$) + KOH($aq$) -> KNO3($aq$) + H2O($l$)}\)       \(\ce{Δ$H$}\) \(\pu{=-57.6 kJ mol-1}\)

Reaction 3:

\(\ce{HCN($aq$) + KOH($aq$) -> KCN($aq$) + H2O($l$)}\)            \(\ce{Δ$H$}\) \(\pu{=-12.0 kJ mol-1}\)

 
Explain why the first two reactions have the same enthalpy value but the third reaction has a different value.   (4 marks)

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→ Reaction 1 and reaction 2 are both neutralisation reactions between strong acids and strong bases. These reactions completely ionise in solution when added to water.

→ Both reactions have the same net ionic equation:

`text{H}^+ (aq) + text{OH}^-  (aq) →  text{H}_2 text{O} (l)`

→ Therefore, the enthalpy values obtained are the same for both reactions.

→ In reaction 3, `text{HCN}` is a weak acid that only partially ionises in an equilibrium reaction with water.

`text{HCN} (aq) + text{H}_2 text{O} (l)  ⇋  text{CN}^-  (aq) + text{H}_3 text{O}^+ (aq).`

→ As the reaction continues, `text{HCN}` will further ionise as the equilibrium shifts to the right.

→ The bond-breaking is an endothermic process and thus will consume energy to break the bonds. As a result, the overall reaction is less exothermic than reaction 1 and reaction 2.

Show Worked Solution

→ Reaction 1 and reaction 2 are both neutralisation reactions between strong acids and strong bases. These reactions completely ionise in solution when added to water.

→ Both reactions have the same net ionic equation:

`text{H}^+ (aq) + text{OH}^-  (aq) →  text{H}_2 text{O} (l)`

→ Therefore, the enthalpy values obtained are the same for both reactions.

→ In reaction 3, `text{HCN}` is a weak acid that only partially ionises in an equilibrium reaction with water.

`text{HCN} (aq) + text{H}_2 text{O} (l)  ⇋  text{CN}^-  (aq) + text{H}_3 text{O}^+ (aq).`

→ As the reaction continues, `text{HCN}` will further ionise as the equilibrium shifts to the right.

→ The bond-breaking is an endothermic process and thus will consume energy to break the bonds. As a result, the overall reaction is less exothermic than reaction 1 and reaction 2.


♦ Mean mark 44%.

CHEMISTRY, M6 2021 HSC 23

Methanoic acid reacts with aqueous potassium hydroxide. A salt is produced in this reaction.

  1. Write a balanced chemical equation for this reaction.   (2 marks)

  2. Is the salt acidic, basic or neutral? Justify your answer.  (2 marks)

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a.   `text{HCOOH}_((aq)) + text{KOH}_((aq))  -> text{KHCOO}_((aq)) + text{H}_2text{O}_((l))`, or

`text{HCOOH}_((aq)) + text{KOH}_((aq))  -> text{KHCOO}_((s)) + text{H}_2text{O}_((l))`

 

b.   Potassium methanoate is a basic salt.

This is because the `text{HCOO}^-` is a moderately strong conjugate base as it comes from a weak acid `text{HCOOH}`.

Show Worked Solution

a.   `text{HCOOH}_((aq)) + text{KOH}_((aq))  -> text{KHCOO}_((aq)) + text{H}_2text{O}_((l))`, or

`text{HCOOH}_((aq)) + text{KOH}_((aq))  -> text{KHCOO}_((s)) + text{H}_2text{O}_((l))`

 

b.   Potassium methanoate is a basic salt.

This is because the `text{HCOO}^-` is a moderately strong conjugate base as it comes from a weak acid `text{HCOOH}`.

CHEMISTRY, M6 2021 HSC 6 MC

Which row of the table describes what happens when a solution of a weak acid is diluted? (Assume constant temperature.)
 

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`C`

Show Worked Solution

A weak acid has the following equilibrium:

`text{HA} (aq) + text{H}_2 text{O} (l) ⇋ text{A}^(-) (aq) + text{H}_3 text{O}^(+) (aq)`
 

`text{K}_a =  [[text{A}^(-)][text{H}_3text{O}^(+)] ]/[[text{HA}]]`

→ The value of  `text{K}_a` is only affected by temperature, and thus the value of `text{K}_a` will remain the same.

→ When the solution is diluted, water is added. According to Le Chatelier’s Principle, the equilibrium will shift to the right to counteract the change.

→ Thus, the equilibrium will shift to the right and increase the extent of ionisation.

`=>C`


♦♦♦ Mean mark 25%.