Additional Answer Options: 7.2.3

2023 - May/June - Paper 2 - Question 7

7.1.1

A substance whose aqueous solution contains ions.
A substance that dissolves in water to give a solution that conducts electricity.
A substance that forms ions in water/forms ions when molten.

Tips:

It is essential that your answer has the idea that the solution contains ions that can move unless it is mentioned that an ionic substance is molten or that it forms in water/is aqueous.
Either point will be awarded the marks.

2 marks

(1 x 2 marks)

7.1.2

H₂SO₄ is diprotic/donates more than one mole of H⁺ ions per mole of acid (and both acids are of the same concentration)/H₂SO₄ has a higher Ka value.

It ionises to produce more than one mole of protons/H⁺ions for each mole of H₂SO₄/ H₂SO₄ has a higher Ka value.

Tips:

The bulb is the brightest in the solution which has the highest concentration of ions.
In this case, both HNO₃ and H₂SO₄ are strong acids and will ionise completely in the solution to generate a high concentration of ions. H₂SO₄ produces more ions than HNO₃ because it is diprotic and produces three ions for every molecule of HNO₃ that ionises.
Remember to only refer to A and B.

2 marks

(2 x 1 mark)

7.1.3

Stronger acid/ionises completely (and both acids are of the same concentration)/ HNO₃ has a higher Ka value.

CH₃COOH/C is a weaker acid/ionises incompletely.

Tips:

The bulb is the brightest in the solution with the highest concentration of ions.
HNO₃ is a strong acid, therefore, it will ionise completely to generate a higher concentration of hydronium ions in the solution.
Both the H₃O⁺ ions and the NO₃^- ions that form, will conduct electricity.

2 marks

(2 x 1 mark)

7.2.1

How to approach this question:

The equation in option one can only be used when the reaction is neutralized.
The n_a and n_b refer to the stoichiometric coefficients of the acids and bases in the balanced equation.

Option 1

$\frac{c_{a}V_{a}}{c_{b}V_{b}}=\frac{n_{a}}{n_{b}}$

$\frac{c_{a}(19,5)}{(0,04)(25)}=\frac21$ ^{(1 + 2)}

$\mathrm{C}_{\mathrm{a}}=0,10\mathrm{mol}\cdot\mathrm{dm}^{-3}(0,103)$ ⁽³⁾

The equation above is one of the only equations where you can leave volumes in cm³ or convert them to dm³ without changing the answer.

Option 2

$\mathrm{n}\left(\mathrm{Na}_2\mathrm{CO}_3\right)=\mathrm{cV}$

$=0,04\times0,025$ ⁽¹⁾

$=0,001\mathrm{~mol}$

$n(\mathrm{HCl})=2\mathrm{n}\left(\mathrm{Na}_2\mathrm{CO}_3\right)$

$=0,002\mathrm{~mol}$

$[\mathrm{HCl}]=\frac{\mathrm{n}}{\mathrm{V}}$

$=\frac{0,002}{0,0195}$ ⁽²⁾

$=0,10\mathrm{mol}\cdot\mathrm{dm}^{-3}(0,103)$ ⁽³⁾

3 marks

(3 x 1 mark)

Marking criteria:

⁽¹⁾Substitute 0,04 mol∙dm^-3 and 25 x 10^-3 dm³ (25 cm3) and 19,5 x 10^-3 dm³ (19,5 cm³)
⁽²⁾Use mol ratio: n(Na₂CO₃) : n(HCℓ) = 1:2
⁽³⁾Final answer: 0,10 to 0,103 mol·dm^-3

7.2.2

Greater than

The few drops of water will dilute the HCℓ, therefore greater volume of acid will be needed to neutralise the base.

Tips:

If there are drops of water in the burette, it will dilute the HCℓ which means that a greater volume of HCℓ will be needed to reach the same number of moles required for the endpoint.
If drops of water are left in the conical flask before the Na₂CO₃ is added, this will not have any impact on the titration as the number of moles of Na₂CO₃ pipetted into the conical flask will remain the same.

2 marks

(2 x 1 mark)

7.2.3

How to approach this question:

First, calculate the number of moles in the 18,7 cm³ of 0,10 mol·dm^-3 solution.
If you got the answer to question 7.2.1 wrong, you can still be awarded marks in this question for substituting correctly.

Option 1

n(HCl)=cV

$=(0,1)\left(18,7\times10^{-3}\right)$

$=1,87\times10^{-3}\mathrm{~mol}$

Using the balanced equation, NH₃ (aq) + HCℓ(aq) → NH₄⁺ (aq)+ Cℓ^- (aq), determine the mol ratio of NH₃ to HCℓ.

$\mathrm{n}\left(\mathrm{NH}_3\right)_{\text{reacted}}=\mathrm{n}(\mathrm{HCl})_{\text{reacted }}$

$=1,87\times10^{-3}\mathrm{~mol}$

$\mathrm{n}\left(\mathrm{NH}_3\right)\text{ in }22\mathrm{~cm}^3=1,87\times10^{-3}\mathrm{~mol}$

Using a ratio, determine the number of moles in the 250 cm³ volumetric flask.

$\mathrm{n}\left(\mathrm{NH}_3\right)\text{ in }250\mathrm{cm}^3=\frac{\left(1,87\times10^{-3}\right)(250)}{22^{}}$

$=0,021\mathrm{~mol}$

$\left(2,13\times10^{-2}\right)$

$\mathrm{n}\left(\mathrm{NH}_3\right)\text{ in initial }20\mathrm{~cm}^3=0,021\mathrm{~mol}$

Now that you know the number of moles in the volumetric flask, use the molar mass of ammonia to determine the mass of ammonia in 1dm³ of ChemClean.

$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

$0,021=\frac{m}{17}$

$\mathrm{m}\left(\mathrm{NH}_3\right)=0,357\mathrm{~g}\text{ in }20\mathrm{~cm}^3$

$m\left(\mathrm{NH}_3\right)=\frac{(0,357)(1000)}{20}$

$=17,85\mathrm{g}(18,06)$

There are more ways to answer this question. To view other options, click here.

7 marks

(7 x 1 mark)

Marking criteria:

Substitute 0,1 mol.dm^-3 &
18,7 x 10^-3 dm³ (18,7 cm³)
Use mole ratio: 1:1
Calculate n(NH³) / m(NH³) in
250 cm³: Substitute 0,25 dm³ (250 cm³)
Substitute 0,022 dm³ (22 cm³)
Substitute 0,02 dm³ (20 cm³) to calculate mole/mass in the initial solution
Use 17 g·mol^-1 in n = m/M
Final answer: 18,06g
Range: 17 to 19,13g.

7.2.4

Less than 7

$\mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{H}_2\mathrm{O}(\ell)\rightleftharpoons\mathrm{NH}_3(\mathrm{aq})+\mathrm{H}_3\mathrm{O}^{+}(\mathrm{aq})$

Tips:

When answering this question, some students get confused and incorrectly assume that if a titration is neutralized, the pH must be 7.
Consider a hydrolysis reaction, the salt produced is NH₄Cℓ. In this case, NH₄⁺is a strong conjugate acid produced when NH₃ acts as a weak base. Cℓ^-is a weak conjugate base formed when HCℓ acts as a strong acid.
When the ammonium ion (NH₄⁺) reacts with water, it donates a proton to form ammonia and hydronium ions.

3 marks

(3 x 1 mark)

Option 2

n(HCl)=cV

$=(0,1)\left(18,7\times10^{-3}\right)$ ⁽¹⁾

$=1,87\times10^{-3}\mathrm{~mol}$

$\left(\mathrm{NH}_3\right)_{\text{reacted }}=\mathrm{n}(\mathrm{HCl})_{\text{reacted }}$

$=1,87\times10^{-3}\mathrm{~mol}$ ⁽²⁾

$\mathrm{n}\left(\mathrm{NH}_3\right)\text{ in }22\mathrm{~cm}^3=1,87\times10^{-3}\mathrm{~mol}$

$n=\frac{m}{M}$

$1,87\times10^{-3}=\frac{m}{17}$ ⁽⁶⁾

$\mathrm{m}\left(\mathrm{NH}_3\right)=3,72\times10^{-3}\mathrm{q}\text{ in }22\mathrm{~cm}^3$

$\mathrm{m}\left(\mathrm{NH}_3\right)\text{ in }250\mathrm{cm}^3=\frac{\left(3,72\times10^{-3}\right)(250)}{22}$ ^{(3 + 4)}

$=0,361\mathrm{~g}$

$\mathrm{m}\left(\mathrm{NH}_3\right)\text{ in initial }20\mathrm{~cm}^3=0,361\mathrm{~g}$

$\mathrm{m}\left(\mathrm{NH}_3\right)\text{ in }1000\mathrm{cm}^3=\frac{(0,361)(1000)}{20}$ ⁽⁵⁾

$=18,06\mathrm{~g}$ ⁽⁷⁾

Option 3

$\frac{c_{b}V_{b}}{c_{a}V_{a}}=\frac{n_{b}}{n_{a}}$

$\frac{c_{b}(22)}{(0,1)(18,7)}=\frac11$ ^{(1 + 2 + 4)}

$\mathrm{c}_1=0.085\mathrm{~mol}\cdot\mathrm{dm}^{-3}$

$\left[\mathrm{NH}_3\right]\text{ in }22\mathrm{~cm}^3=0,085\mathrm{~mol}\cdot\mathrm{dm}^{-3}$

$\left[\mathrm{NH}_3\right]\text{ in }250\mathrm{~cm}^3=0,085\mathrm{~mol}\cdot\mathrm{dm}^{-3}$

$\mathrm{C}_1\mathrm{~V}_1=\mathrm{C}_2\mathrm{~V}_2$

c_1(0,02)=(0,085)(0,25) ^{(3 + 5)}

$\mathrm{c}_1=1,06\mathrm{~mol}\cdot\mathrm{dm}^{-3}$

$\mathrm{m}=\mathrm{cVM}$

=(1,06)(1)(17) ⁽⁶⁾

$=18,06\mathrm{~g}$ ⁽⁷⁾

Option 4

$n(\mathrm{HCl})=\mathrm{cV}$

$=(0,1)\left(18,7\times10^{-3}\right)$ ⁽¹⁾

$=1.87\times10^{-3}\mathrm{~mol}$

$\left(\mathrm{NH}_3\right)_{\text{reacted }}=\mathrm{n}(\mathrm{HCl})_{\text{reacted }}$

$=1,87\times10^{-3}\mathrm{~mol}$ ⁽²⁾

$\mathrm{n}\left(\mathrm{NH}_3\right)\text{ in }22\mathrm{~cm}^3=1,87\times10^{-3}\mathrm{~mol}$

$\mathrm{n}\left(\mathrm{NH}_3\right)\text{ in }250\mathrm{cm}^3=\frac{\left(1,87\times10^{-3}\right)(250)}{22}$ ^{(3 + 4)}

$=0,021\mathrm{~mol}$

$c\left(20\mathrm{~cm}^3\right)=c\left(1\mathrm{dm}^3\right)$

$\frac{\mathrm{n}_1}{\mathrm{~V}_1}=\frac{\mathrm{n}_2}{\mathrm{~V}_2}$

$\mathrm{n}\left(\mathrm{NH}_3\right)\text{ in }1000\mathrm{~cm}^3=\frac{0,021\times1000}{20}$ ⁽⁵⁾

$=1,06\mathrm{~mol}$

$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

$1,06=\frac{\mathrm{m}}{17}$ ⁽⁶⁾

$\mathrm{m}\left(\mathrm{NH}_3\right)=18,06\mathrm{~g}$ ⁽⁷⁾