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CHEMISTRY, M6 2025 HSC 26

A hydrogen atom on the methyl group of ethanoic acid can be replaced with a single halogen atom. A general formula for these haloethanoic acids is shown.

\(\ce{X-CH2COOH \quad (X = F,Cl,Br or I)}\)

The \(p K_a\) values of the four haloethanoic acids are given in the table.

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Acid} \rule[-1ex]{0pt}{0pt}& \quad \quad \ce{X} \quad \quad & \quad \quad p K_a \quad \quad\\
\hline
\rule{0pt}{2.5ex}\text{Fluoroethanoic acid } \quad \rule[-1ex]{0pt}{0pt}& \ce{F} & 2.6 \\
\hline
\rule{0pt}{2.5ex}\text{Chloroethanoic acid } \rule[-1ex]{0pt}{0pt}& \ce{Cl} & 2.9 \\
\hline
\rule{0pt}{2.5ex}\text{Bromoethanoic acid } \rule[-1ex]{0pt}{0pt}& \ce{Br} & 2.9 \\
\hline
\rule{0pt}{2.5ex}\text{Iodoethanoic acid } \rule[-1ex]{0pt}{0pt}& \ce{I}& 3.2 \\
\hline
\end{array}

  1. Construct an appropriate graph for the four haloethanoic acids, showing their \(p K_a\) values and the identity of the halogen \(\ce{X}\) in each molecule, in the order provided in the table.   (3 marks)
     

  1. Describe the trend in the relative strengths of the haloethanoic acids.   (2 marks)

Show Answers Only

a. 
     

b.    The strength of the haloethanoic acids can be explained by the relationship between \(pK_a\), \(K_a\) and the degree of ionisation.

  • Since \(pK_a\) is inversely related to \(K_a\) (\(pK_a = -\log_{10}(K_a)\)), a lower \(pK_a\) corresponds to a higher \(K_a\) and therefore greater ionisation in water.
  • Fluoroethanoic acid, with the lowest \(pK_a\) value, has the highest \(K_a\) and ionises the most, making it the strongest acid.
  • As you move down the halogen group from \(\ce{F}\) to \(\ce{I}\), \(pK_a\) increases and \(K_a\) decreases, so the acids ionise less and become weaker, with iodoethanoic acid being the weakest.
Show Worked Solution

a.    
         

b.    The strength of the haloethanoic acids can be explained by the relationship between \(pK_a\), \(K_a\) and the degree of ionisation.

  • Since \(pK_a\) is inversely related to \(K_a\) (\(pK_a = -\log_{10}(K_a)\)), a lower \(pK_a\) corresponds to a higher \(K_a\) and therefore greater ionisation in water.
  • Fluoroethanoic acid, with the lowest \(pK_a\) value, has the highest \(K_a\) and ionises the most, making it the strongest acid.
  • As you move down the halogen group from \(\ce{F}\) to \(\ce{I}\), \(pK_a\) increases and \(K_a\) decreases, so the acids ionise less and become weaker, with iodoethanoic acid being the weakest.

CHEMISTRY, M6 2025 HSC 19 MC

0.1 mol of solid sodium acetate is dissolved in 500 mL of 0.1 mol L\(^{-1}\) \(\ce{HCl}\) in a beaker. This solution has a pH of 4.8 .

500 mL of distilled water is then added to the beaker.

What is the pH of the final solution?

  1. 2.4
  2. 4.5
  3. 4.8
  4. 5.1
Show Answers Only

\(C\)

Show Worked Solution
  • The initial reaction between sodium acetate and hydrochloric acid runs to completion as \(\ce{HCl}\) is a strong acid:
  •    \(\ce{CH3COO-(aq) + HCl(aq) -> CH3COOH(aq) + Cl-(aq)}\)
  • \(n(\ce{CH3COO-}) = 0.1\ \text{mol}\)
  • \(n(\ce{HCl}) = 0.5\ \text{L} \times 0.1\ \text{mol L}^{-1} = 0.05\ \text{mol}\)
  • As they react in a \(1:1\) ratio, \(\ce{HCl}\) is the limiting reagent.
  • \(n(\ce{CH3COO-_{\text{after reaction}}}) = n(\ce{CH3COO_{\text{initial}}})-n(\ce{CH3COO_{\text{reacted}}}) = 0.1-0.05 = 0.05\ \text{mol}\)
  • \(n(\ce{CH3COOH_{\text{after reaction}}}) = n(\ce{CH3COO_{\text{reacted}}}) = 0.05\ \text{mol}\)
  • The following equilibrium reaction is then established below dilution
  •    \(\ce{CH3COOH(aq) + H2O(l) \leftrightharpoons CH3COO-(aq) + H3O+(aq)}\)
  • Therefore the following ice table can be constructed:

\begin{array} {|c|c|c|c|}
\hline  & \ce{[CH3COOH]} & \ce{[CH3COO-]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.1 & 0.1 & 0 \\
\hline \text{Change} & -10^{-4.8} & +10^{-4.8} & +10^{-4.8} \\
\hline \text{Equilibrium} & 0.1-10^{-4.8} & 0.1+10^{-4.8} & +10^{-4.8} \\
\hline \end{array}

\(\therefore K_a = \dfrac{(0.1+10^{-4.8})(10^{-4.8})}{0.1-10^{-4.8}} = 1.585395 \times 10^{-5}\)

  • Then considering the dilution which would shift the equilibrium position to the right.

\begin{array} {|c|c|c|c|}
\hline  & \ce{[CH3COOH]} & \ce{[CH3COO-]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.1 \times \dfrac{500\ \text{mL}}{1000\ \text{mL}} = 0.05 & 0.1 \times \dfrac{500\ \text{mL}}{1000\ \text{mL}} = 0.05 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.05-x & 0.05+x & +x \\
\hline \end{array}

  • As \(K_{eq}\) is small, \(0.05 -x \approx 0.05\)  and  \(0.05 + x \approx 0.05\). 
  •    \(\therefore K_{eq} = \dfrac{0.05x}{0.05} = x = 1.585395 \times 10^{-5}\)
  • \(\text{pH} = -\log_{10}\ce{[H3O+]} = -\log_{10}(1.585395 \times 10^{-5}) = 4.79962 = 4.8\ \text{(1 d.p.)}\)

\(\Rightarrow C\)

CHEMISTRY, M6 2023 HSC 35

  1. A 0.2000 mol L\(^{-1} \) solution of dichloroacetic acid \( \ce{(CHCl2COOH)} \) has a pH of 1.107. Dichloroacetic acid is monoprotic.
  2. Calculate the \( K_a \) for dichloroacetic acid.   (3 marks)
  1. The following data apply to the ionisation of acetic acid \( \ce{(CH3COOH)} \) and trichloroacetic acid \( \ce{(CCl3COOH)} \).  
  \( \ce{CH3 COOH }\) \( \ce{CCl3COOH} \)
\( p K_a \) \(4.76\) \(0.51\)
\( \Delta H° \text{(kJ mol}^{-1})\) \(-0.1\) \(+1.2\)
\(\Delta S° \text{(J K}^{-1} \text{ mol}^{-1})\) \(-91.6\) \(-5.8\)
\( -T \Delta S° \text{(kJ mol}^{-1}) \) \(+27.3\) \(+1.7\)
\( \Delta G° \text{(kJ mol}^{-1}) \) \(+27.2\) \(+2.9\)
  1.  
    Explain the relative strength of these acids with reference to the data.   (3 marks)
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a.    \(K_a = 0.0501\)

b.    Relative strength of acids:

  • The \(pK_a\) of trichloroacetic acid is lower than the \(pK_a\) of acetic acid, so trichloroacetic acid is a stronger acid than acetic acid.
  • The \(\Delta S°\) term for acetic acid is a significantly lower number than for the trichloroacetic acid (noting they are both negative).
  • In both cases, this value will contribute unfavourably to each acid’s \(\Delta G°\) value, with the effect much larger for acetic acid than for trichloroacetic acid.
  • It follows from this result that the ionisation of acetic acid is less favourable than it is for trichloroacetic acid, making the latter the stronger acid.
Show Worked Solution

a.   \(\ce{CHCl2COOH(aq) \rightleftharpoons H+(aq) + CHCl2COO-(aq)} \)

\(\ce{[H+] = 10^{-\text{pH}} = 10^{-1.107} = 0.0782\ \text{mol L}^{-1}} \)

♦ Mean mark (a) 54%.

\begin{array} {|l|c|c|c|}
\hline  & \ce{CHCl2COOH(aq)} & \ce{H+(aq)} & \ce{CHCl2COO^{-}(aq)} \\
\hline \text{Initial} & \ \ \ \ 0.2000 &  0 & 0 \\
\hline \text{Change} & -0.0782 & +0.0782 & \ \ \ +0.0782 \\
\hline \text{Equilibrium} & \ \ \ \ 0.1218 & \ \ \ \ 0.0782 & \ \ \ \ \ \ 0.0782 \\
\hline \end{array}

\(K_a\) \(= \dfrac{\ce{[H+][CHCl2COO-]}}{\ce{[CHCl2COOH]}} \)  
  \(= \dfrac{0.0782 \times 0.0782}{0.1218} \)  
  \(= 0.0501\)  

 
b.    Relative strength of acids:

  • The \(pK_a\) of trichloroacetic acid is lower than the \(pK_a\) of acetic acid, so trichloroacetic acid is a stronger acid than acetic acid.
  • The \(\Delta S°\) term for acetic acid is a significantly lower number than for the trichloroacetic acid (noting they are both negative).
  • In both cases, this value will contribute unfavourably to each acid’s \(\Delta G°\) value, with the effect much larger for acetic acid than for trichloroacetic acid.
  • It follows from this result that the ionisation of acetic acid is less favourable than it is for trichloroacetic acid, making the latter the stronger acid.
♦ Mean mark (b) 52%.

CHEMISTRY, M6 EQ-Bank 14 MC

Equal volumes of four different acids are titrated with the same base at 25°.

Information about these acids is given in the table.
 

Which acid requires the greatest volume of base for complete reaction?

  1. \(\ce{HCl}\)
  2. \(\ce{H3PO4}\)
  3. \(\ce{CH3COOH}\)
  4. \(\ce{HCN}\)
Show Answers Only

`B`

Show Worked Solution
  • The volume of base required for complete reaction is independent of the strength of the acid as all neutralisation reactions go to completion.
  • Phosphoric acid (\(\ce{H3PO4}\)) has three acidic hydrogens whereas the other given bases only have one. So, phosphoric acid requires three times the amount of base as any other acid given.

`=> B`

CHEMISTRY, M6 EQ-Bank 13 MC

The pKa of trichloroacetic acid is 0.70 and the pKa of acetic acid is 4.8.

Which of the following identifies the acid with the higher pH and explain why?

  1. Acetic acid as it is less likely to lose a hydrogen ion
  2. Acetic acid as it is more likely to lose a hydrogen ion
  3. Trichloroacetic acid as it is less likely to lose a hydrogen ion
  4. Trichloroacetic acid as it is more likely to lose a hydrogen ion
Show Answers Only

`A`

Show Worked Solution
  • Higher pH corresponds to a weaker acid.
  • Higher pKa corresponds to a weaker acid (i.e. a higher pH).
  • Acetic acid the weaker acid (higher pKa), meaning it is less likely to dissociate in solution and lose a hydrogen ion.

`=> A`

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)

Show Answers Only
  • 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 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)

Show Answers Only
  • \( \ce{HCl} \) 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)

Show Answers Only
  • 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%.