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CHEMISTRY, M6 2024 HSC 28

Iodic acid and sulfamic acid are monoprotic acids. A 0.100 mol L\(^{-1}\) solution of iodic acid has a pH of 1.151, as does a 0.120 mol L\(^{-1}\) solution of sulfamic acid.

Show that neither iodic acid nor sulfamic acid dissociates completely in water, and determine which is the stronger acid.   (3 marks)

Show Answers Only

  • Calculating the concentration of hydronium ions in solution for a pH of 1.151.
  •   \(\ce{pH}=-\log_{10}\ce{[H3O+]}\)
  •   \(\ce{[H3O+]}=10^{-\ce{pH}}=10^{-1.151}=0.0706\) 
  • As this is less than the concentration of both of the acids, neither acid completely dissociates in water.
  • A smaller concentration of iodic acid (0.100 mol/L compared to 0.120 mol/L sulfamic acid) produces the same pH level. Iodic acid must have a greater extent of ionisation compared to sulfamic acid.
  • Therefore iodic acid is a stronger acid than sulfamic acid.

Show Worked Solution

  • Calculating the concentration of hydronium ions in solution for a pH of 1.151.
  •   \(\ce{pH}=-\log_{10}\ce{[H3O+]}\)
  •   \(\ce{[H3O+]}=10^{-\ce{pH}}=10^{-1.151}=0.0706\) 
  • As this is less than the concentration of both of the acids, neither acid completely dissociates in water.
  • A smaller concentration of iodic acid (0.100 mol/L compared to 0.120 mol/L sulfamic acid) produces the same pH level. Iodic acid must have a greater extent of ionisation compared to sulfamic acid.
  • Therefore iodic acid is a stronger acid than sulfamic acid.

CHEMISTRY, M6 2016 VCE 20 MC

How does diluting a 0.1 M solution of lactic acid, \(\ce{HC3H5O3}\), change its pH and percentage ionisation?

  pH Percentage ionisation
A.   increase decrease
B. increase increase
C. decrease increase
D. decrease decrease
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\(B\)

Show Worked Solution
  • Ionisation of lactic acid:  \(\ce{HC3H5O3(aq) + H2O(l) \rightleftharpoons C3H5O3-(aq) + H3O+(aq)} \)
  • Adding water decreases \(\ce{[H3O+]}\). The equilibrium then shifts to partially compensate for this change, favouring the forward reaction.
  • \(\ce{[H3O+]}\) increases in this process although the new equilibrium \(\ce{[H3O+]}\) is lower (pH higher) than the original system.
  • Overall, the pH increases and the percentage ionisation increases.

\(\Rightarrow B\)

♦♦ Mean mark 34%.

CHEMISTRY, M6 EQ-Bank 23

Propanoic acid dissociation in water can is represented in the following equation:

\(\ce{CH3CH2COOH($aq$) + H2O($l$) \rightleftharpoons CH3CH2COO^-($aq$) + H3O^{+}($aq$)}\)

Explain how the pH of the propanoic acid solution would change if it was diluted.   (3 marks)

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  • Propanoic acid only partially ionises in solution and is defined as a weak acid.
  • Any dilution of the acid will result in a decrease of the concentration of all species (including the hydronium ion).
  • According to Le Chatelier, the decreasing concentrations of dissolved species will cause the equilibrium to shift to the right.
  • While this effect causes an increase in ionisation, it is not sufficient to counter the decrease in hydronium ion concentration caused by the original dilution.
  • Since the net effect causes the hydronium ion concentration to decrease, the solution will become less acidic and the pH will increase.
Show Worked Solution
  • Propanoic acid only partially ionises in solution and is defined as a weak acid.
  • Any dilution of the acid will result in a decrease of the concentration of all species (including the hydronium ion).
  • According to Le Chatelier, the decreasing concentrations of dissolved species will cause the equilibrium to shift to the right.
  • While this effect causes an increase in ionisation, it is not sufficient to counter the decrease in hydronium ion concentration caused by the original dilution.
  • Since the net effect causes the hydronium ion concentration to decrease, the solution will become less acidic and the pH will increase.

CHEMISTRY, M6 2015 HSC 24

  1. Explain why the salt, sodium acetate, forms a basic solution when dissolved in water. Include an equation in your answer.   (2 marks)

  2. A solution is prepared by using equal volumes and concentrations of acetic acid and sodium acetate.
  3. Explain how the pH of this solution would be affected by the addition of a small amount of sodium hydroxide solution. Include an equation in your answer.   (3 marks)

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a.   \(\ce{CH3COO-(aq) + H2O(l) \rightleftharpoons CH3COOH(aq) + OH-(aq)}\)

  • Sodium acetate is a basic salt.
  • Acetate is a strong base that accepts a proton, producing hydroxide.
  • The presence of \(\ce{OH-}\) ions produced by the hydrolysis of \(\ce{CH3COO-}\) increases the pH, producing a basic solution. 

b.  \(\ce{CH3COO-(aq) + H3O+(l) \rightleftharpoons CH3COOH(aq) + H2O(l)}\)

  • The \(\ce{OH-}\) ions introduced into the solution will react with the \(\ce{H3O+}\) ions, reducing their concentration in the equilibrium mixture.
  • By Le Chatelier’s principle, this will subsequently move the reaction to the left to increase the \(\ce{H3O+}\) ions, thus minimising any change in pH.
Show Worked Solution

a.   \(\ce{CH3COO-(aq) + H2O(l) \rightleftharpoons CH3COOH(aq) + OH-(aq)}\)

  • Sodium acetate is a basic salt.
  • Acetate is a strong base that accepts a proton, producing hydroxide.
  • The presence of \(\ce{OH-}\) ions produced by the hydrolysis of \(\ce{CH3COO-}\) increases the pH, producing a basic solution.  

♦♦ Mean mark (a) 37%.

b.  \(\ce{CH3COO-(aq) + H3O+(l) \rightleftharpoons CH3COOH(aq) + H2O(l)}\)

  • The \(\ce{OH-}\) ions introduced into the solution will react with the \(\ce{H3O+}\) ions, reducing their concentration in the equilibrium mixture.
  • By Le Chatelier’s principle, this will subsequently move the reaction to the left to increase the \(\ce{H3O+}\) ions, thus minimising any change in pH.

♦♦♦ Mean mark (b) 25%.

CHEMISTRY, M6 2017 HSC 24

A solution of sodium hydroxide was titrated against a standardised solution of acetic acid which had a concentration of 0.5020 mol L¯1.

  1. The end point was reached when 19.30 mL of sodium hydroxide solution had been added to 25.00 mL of the acetic acid solution.
  2. Calculate the concentration of the sodium hydroxide solution.   (3 marks)

  3. Explain why the pH of the resulting salt solution was not 7. Include a relevant chemical equation in your answer.   (2 marks)

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a.   \(\ce{CH3COOH(aq) + NaOH(aq) -> NaCH3COO(aq) + H2O(l)}\)
 

b.   \(\ce{CH3COO-(aq) + H2O \rightleftharpoons CH3COOH(aq) + OH-(aq)}\)

  • The acetate ion is a weak base.
  • As a result, it has accepted a proton from the water resulting in production of hydroxide ions.
  • Therefore the solution has a pH > 7.
Show Worked Solution

a.   \(\ce{CH3COOH(aq) + NaOH(aq) -> NaCH3COO(aq) + H2O(l)}\)

\(\ce{n(CH3COOH) = c \times V = 0.5020 \times 0.0250 = 0.01255 mol}\)

\(\ce{n(NaOH) = n(CH3COOH) = 0.01255 mol}\)

\[\ce{[NaOH] = \frac{n}{V} = \frac{0.01255}{0.01930} = 0.6503 mol L^{-1}}\]  

b.   \(\ce{CH3COO-(aq) + H2O \rightleftharpoons CH3COOH(aq) + OH-(aq)}\)

  • The acetate ion is a weak base.
  • As a result, it has accepted a proton from the water resulting in production of hydroxide ions.
  • Therefore the solution has a pH > 7.

♦♦ Mean mark (b) 34%.

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, M5 2020 HSC 27

A student makes up a solution of propan-2-amine in water with a concentration of 1.00 mol L ¯1.

  1. Using structural formulae, complete the equation for the reaction of propan-2-amine with water.   (2 marks)
     
         
  2. The equilibrium constant for the reaction of propan-2-amine with water is  `4.37 xx10^(-4)`.
  3. Calculate the concentration of hydroxide ions in this solution.   (3 marks)

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a.   

   

b.   

\begin{array} {|l|c|c|c|}
\hline  & \ce{C3H7NH2} & \ce{C3H7NH3} & \ce{OH–} \\
\hline \text{Initial} & 1.00 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 1.00-x & x & x \\
\hline \end{array}

 
\[ K_b = \ce{\frac{[C3H7NH3+][OH– ]}{[C3H7NH2 ]}} = \frac{x^2}{(1.00-x)} \]

Assume `1.00-x=1.00` because `x` is negligible:

`4.37 xx 10^(−4)` `= x^2 / 1.00`  
`x` `=sqrt(4.37 xx 10^(−4))`  
  `= 0.0209\ text{mol L}^(–1)`  

 
`=> [text{OH}^– ] = 0.0209\ text{mol L}^(–1)`

Show Worked Solution

a.   

   


♦ Mean mark (a) 48%.

b.   

\begin{array} {|l|c|c|c|}
\hline  & \ce{C3H7NH2} & \ce{C3H7NH3} & \ce{OH–} \\
\hline \text{Initial} & 1.00 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 1.00-x & x & x \\
\hline \end{array}

 
\[ K_b = \ce{\frac{[C3H7NH3+][OH– ]}{[C3H7NH2 ]}} = \frac{x^2}{(1.00-x)} \]

Assume `1.00-x=1.00` because `x` is negligible:

`4.37 xx 10^(−4)` `= x^2 / 1.00`  
`x` `=sqrt(4.37 xx 10^(−4))`  
  `= 0.0209\ text{mol L}^(–1)`  

 
`=> [text{OH}^– ] = 0.0209\ text{mol L}^(–1)`


Mean mark (b) 51%.

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 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}`.