CHEMISTRY, M6 2023 HSC 35
- A 0.2000 mol L\(^{-1} \) solution of dichloroacetic acid \( \ce{(CHCl2COOH)} \) has a pH of 1.107. Dichloroacetic acid is monoprotic.
- Calculate the \( K_a \) for dichloroacetic acid. (3 marks)
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- The following data apply to the ionisation of acetic acid \( \ce{(CH3COOH)} \) and trichloroacetic acid \( \ce{(CCl3COOH)} \).
\( \ce{CH3 COOH }\) | \( \ce{CCl3COOH} \) | |
\( p K_a \) | \(4.76\) | \(0.51\) |
\( \Delta H° \text{(kJ mol}^{-1})\) | \(-0.1\) | \(+1.2\) |
\(\Delta S° \text{(J K}^{-1} \text{ mol}^{-1})\) | \(-91.6\) | \(-5.8\) |
\( -T \Delta S° \text{(kJ mol}^{-1}) \) | \(+27.3\) | \(+1.7\) |
\( \Delta G° \text{(kJ mol}^{-1}) \) | \(+27.2\) | \(+2.9\) |
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Explain the relative strength of these acids with reference to the data. (3 marks)
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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?
- \(\ce{HCl}\)
- \(\ce{H3PO4}\)
- \(\ce{CH3COOH}\)
- \(\ce{HCN}\)
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?
- Acetic acid as it is less likely to lose a hydrogen ion
- Acetic acid as it is more likely to lose a hydrogen ion
- Trichloroacetic acid as it is less likely to lose a hydrogen ion
- Trichloroacetic acid as it is more likely to lose a hydrogen ion
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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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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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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