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CHEMISTRY, M6 2024 HSC 28

Iodic acid and sulfamic acid are monoprotic acids. A 0.100 mol L solution of iodic acid has a pH of 1.151, as does a 0.120 mol L 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)

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→ Calculating the concentration of hydronium ions in solution for a pH of 1.151.

 
 

 

→ 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.

 
 

 

→ 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 2015 VCE 8

Hydrogen sulfide, in solution, is a diprotic acid and ionises in two stages.

A student made two assumptions when estimating the pH of a solution of :

Assumption 1: The pH can be estimated by considering only the first ionisation reaction.

Assumption 2: The concentration of at equilibrium is approximately equal to .

  1. Explain why these two assumptions are justified.  (2 marks)

  2. Use the two assumptions given above to calculate the pH of a 0.01 M solution of .  (3 marks)

  3. Some solid sodium hydrogen sulfide, , is added to a 0.01 M solution of .
  4. Predict the effect of this addition on the pH of the hydrogen sulfide solution. Justify your prediction.  (2 marks)

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a.   1st assumption:

→ is significantly smaller than the first ionisation , making its impact on the / pH level negligible.

2nd assumption:

→ is very small, making the extent of the ionisation of very small and hence a minimal change in results.

b.   

c.    Adding :

→ Increases the .

→ This increase causes the 1st ionisation equilibrium back to the left.

→ This left shift in the equilibrium decreases the and the pH will therefore increase.

Show Worked Solution

a.   1st assumption:

→ is significantly smaller than the first ionisation , making its impact on the / pH level negligible.

2nd assumption:

→ is very small, making the extent of the ionisation of very small and hence a minimal change in results.
 

♦♦ Mean mark (a) 27%.
b.   
 

 
 
   

 

  

c.    Adding :

→ Increases the .

→ This increase causes the 1st ionisation equilibrium back to the left.

→ This left shift in the equilibrium decreases the and the pH will therefore increase.

♦♦ Mean mark (c) 35%.

CHEMISTRY, M6 2023 HSC 6 MC

The pH of a solution changes from 8 to 5.

What happens to the concentration of hydrogen ions during this change of pH?

  1. It increases by a factor of 3.
  2. It decreases by a factor of 3.
  3. It increases by a factor of 1000.
  4. It decreases by a factor of 1000.
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→ Each increase/decrease of pH by a magnitude of 1 represents a change in of a factor of 10.

→ Therefore, concentration change when pH moves from 8 to 5 = 10 × 3 = 1000 (increase).

CHEMISTRY, M6 2016 HSC 12 MC

Which of the following could be added to 100 mL of 0.01 mol L¯1 hydrochloric acid solution to change its pH to 4?

  1. 900 mL of water
  2. 900 mL of 0.01 mol L¯1 hydrochloric acid
  3. 9900 mL of water
  4. 9900 mL of 0.01 mol L¯1 hydrochloric acid
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By Elimination:

→ If 0.01 mol L¯1 hydrochloric acid is added, remains unchanged and hence pH is unchanged (eliminate B and D)

→ If 900 mL of water added, dilutes to 0.001 mol L¯1, with a pH = 3 (eliminate A)

→ If 9900 mL of water added, dilutes to 0.0001 mol L¯1, with a pH = 4


Mean mark 59%.

CHEMISTRY, M6 2019 HSC 27

The relationship between the acid dissociation constant, , and the corresponding conjugate base dissociation constant, , is given by:

Assume that the temperature for part (a) and part (b) is 25°C.

  1. The of hypochlorous acid is  .
  2. Show that the of the hypochlorite ion, , is  .   (1 mark)

  3. The conjugate base dissociation constant, , is the equilibrium constant for the following equation:
  4.      
  5. Calculate the pH of a 0.20 mol L¯1 solution of sodium hypochlorite .   (4 mark)

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a.   

 
   

 

b.   
 

 

Assume    because is negligible:

 
 
   

 


♦ Mean mark (b) 45%.

CHEMISTRY, M6 2020 HSC 33

Excess solid calcium hydroxide is added to a beaker containing 0.100 L of 2.00 mol L¯1 hydrochloric acid and the mixture is allowed to come to equilibrium.

  1. Show that the amount (in mol) of calcium hydroxide that reacts with the hydrochloric acid is 0.100 mol.   (2 marks)

  2. It is valid in this instance to make the simplifying assumption that the amount of calcium ions present at equilibrium is equal to the amount generated in the reaction in part (a).
  3. Calculate the pH of the resulting solution.  (4 marks)

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a.   

 

b.  

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a.   

 

b.   


 

 
 
 
   
 
   

 


♦♦♦ Mean mark (b) 20%.

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  .
  3. Calculate the concentration of hydroxide ions in this solution.   (3 marks)

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a.   

   

b.   

 

Assume because is negligible:

 
 
   

 

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a.   

   


♦ Mean mark (a) 48%.

b.   

 

Assume because is negligible:

 
 
   

 


Mean mark (b) 51%.

CHEMISTRY, M6 2022 HSC 34

Sodium hypochlorite is the active ingredient in pool chlorine. It completely dissolves in water to produce the hypochlorite ion , which undergoes hydrolysis according to the following equilibrium.

The equilibrium constant for this reaction at 25°C is .

For pool chlorine to be effective the pH is maintained by a different buffer at 7.5 and the hypochlorous acid concentration should be  mol L ¯1.

Calculate the volume of 2.0 mol L ¯1 sodium hypochlorite solution that needs to be added to a 1.00 × 104 L pool to meet the required conditions.   (4 marks)

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♦ Mean mark 46%.

CHEMISTRY, M6 2022 HSC 25

The pH of two aqueous solutions was compared.
 

Explain why the solution has a higher pH than the solution. Include a relevant chemical equation for the solution.   (3 marks)

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→ is a strong acid, ie it completely ionises in water to form ions.

→ On the other hand, is a weak acid, ie it partially ionises in water to form ions.

→ As decreases, pH increases  ()

→ Therefore, at the same 0.2M, the solution would have a lower and thus would have a higher pH than .

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→ is a strong acid, ie it completely ionises in water to form ions.

→ On the other hand, is a weak acid, ie it partially ionises in water to form ions.

→ As decreases, pH increases  ()

→ Therefore, at the same 0.2M, the solution would have a lower and thus would have a higher pH than .


Mean mark 57%.

CHEMISTRY, M6 2022 HSC 20 MC

Cyanidin is a plant pigment that may be used as a pH indicator. It has four levels of protonation, each with a different colour, represented by these equilibria:
 

The following graph shows the relative amount of each species present at different pH values.
 

What colour would the indicator be if added to a 0.75 mol L solution of hypoiodous acid, ?

  1. Red
  2. Colourless
  3. Purple
  4. Blue
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Show Worked Solution

 


♦ Mean mark 46%.