HSC Chemistry 2021 Exam Solutions (1 Viewer)

CM_Tutor · Dec 8, 2021

My answer to Q36 would be:

H₂SO₃ (aq) + 2 H₂O (l) <---in equilibrium---> 2 H₃O⁺ (aq) + SO₃^2- (aq)

$\begin{align*} K &= \cfrac{\left[\text{H}_3\text{O}^+\right]^2\left[\text{SO}_3^{2-}\right]}{\left[\text{H}_2\text{SO}_3\right]} \\ &= \cfrac{\left[\text{H}_3\text{O}^+\right]}{\left[\text{H}_2\text{SO}_3\right]} \times \left[\text{H}_3\text{O}^+\right]\left[\text{SO}_3^{2-}\right] \\ &= \cfrac{\left[\text{H}_3\text{O}^+\right]\left[\text{HSO}_3^-\right]}{\left[\text{H}_2\text{SO}_3\right]} \times \cfrac{\left[\text{H}_3\text{O}^+\right]\left[\text{SO}_3^{2-}\right]}{\left[\text{HSO}_3^-\right]} \\ &= K_{\text{a}}\left(\text{H}_2\text{SO}_3\right) \times K_{\text{a}}\left(\text{HSO}_3^-\right) \\ &= 10^{-1.82} \times 10^{-7.17} \\ &= 10^{-8.99} \\ &= 1.0232... \times 10^{-9} \\ &\approx 1.0 \times 10^{-9}\ \text{(2 sig. fig.)} \end{align*}$

Even giving a mark for appropriate significant figures (which is 2, because the p''K''_a values are given to 2 decimal places), I struggle to give 5 marks for this without being overly generous, but I guess it could be:

1 mark for the expression for K
1 mark for rearranging to show the component K_a expressions
1 mark for substituting the K_a values from 10^-pK_a
1 mark for the numerical answer (ECD from expression given)
1 mark for answering to 2 sig. fig.

Life'sHard · Dec 8, 2021

CM_Tutor said:
My answer to Q36 would be:

H₂SO₃ (aq) + 2 H₂O (l) <---in equilibrium---> 2 H₃O⁺ (aq) + SO₃^2- (aq)

$\begin{align*} K &= \cfrac{\left[\text{H}_3\text{O}^+\right]^2\left[\text{SO}_3^{2-}\right]}{\left[\text{H}_2\text{SO}_3\right]} \\ &= \cfrac{\left[\text{H}_3\text{O}^+\right]}{\left[\text{H}_2\text{SO}_3\right]} \times \left[\text{H}_3\text{O}^+\right]\left[\text{SO}_3^{2-}\right] \\ &= \cfrac{\left[\text{H}_3\text{O}^+\right]\left[\text{HSO}_3^-\right]}{\left[\text{H}_2\text{SO}_3\right]} \times \cfrac{\left[\text{H}_3\text{O}^+\right]\left[\text{SO}_3^{2-}\right]}{\left[\text{HSO}_3^-\right]} \\ &= K_{\text{a}}\left(\text{H}_2\text{SO}_3\right) \times K_{\text{a}}\left(\text{HSO}_3^-\right) \\ &= 10^{-1.82} \times 10^{-7.17} \\ &= 10^{-8.99} \\ &= 1.0232... \times 10^{-9} \\ &\approx 1.0 \times 10^{-9}\ \text{(2 sig. fig.)} \end{align*}$

Even giving a mark for appropriate significant figures (which is 2, because the p''K''_a values are given to 2 decimal places), I struggle to give 5 marks for this without being overly generous, but I guess it could be:

1 mark for the expression for K

1 mark for rearranging to show the component K_a expressions

1 mark for substituting the K_a values from 10^-pK_a

1 mark for the numerical answer (ECD from expression given)

1 mark for answering to 2 sig. fig.

I can’t for the life of me remember if I left my answer as 10^-8.99 or 1.0x10^-9. If I did leave it as 10^-8.99 do you think this is sufficient as an answer? It still has 2 sig figs.

CM_Tutor · Dec 9, 2021

Life'sHard said:
I can’t for the life of me remember if I left my answer as 10^-8.99 or 1.0x10^-9. If I did leave it as 10^-8.99 do you think this is sufficient as an answer? It still has 2 sig figs.

The wording was "calculate the equilibrium constant" so I think they will expect an answer in scientific notation.

Further, 10^-8.99 is (as a number) an exact value and my answer is a 2sf approximation to it. I am not inclined, on first thinking, to call 10^-8.99 an answer to 2sf, though I might in a case like 10^-8.990347... = 10^-8.99 (2sf)... though even then, what if I said 10^-8.99 = 10^-9+0.01 = 10^0.01 x 10^-9 and claimed this is 1 sf?

Given the data, I don't think that rounding the answer to K = 10^-9 is justified, either.

jazz519 · Dec 9, 2021

Life'sHard said:
I can’t for the life of me remember if I left my answer as 10^-8.99 or 1.0x10^-9. If I did leave it as 10^-8.99 do you think this is sufficient as an answer? It still has 2 sig figs.

Not sure if this is correct but back when I did HSC, my tutor was a HSC marker and she said something about only 1 question in the exam being chosen for significant figure marking. So if it's that question they select which may be the case if it's 5 marks then you will lose a mark. If they don't choose that you can probably get away with it

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HSC Chemistry 2021 Exam Solutions (1 Viewer)

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