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2022 HSC Chemistry Exam Paper Solutions

You've found the 2022 HSC Chemistry Exam Paper Solutions!

How do you think you went with this year’s exam? Find out with the 2022 HSC Chemistry Exam Paper Solutions, written by the Matrix Science Team.

 

2022 HSC Chemistry Exam Paper Solutions

These are the responses to the 2022 Chemistry paper, which can be found here on the NESA website.

 

Section 1: Multiple Choice Questions​

Question Answer Explanation
1 C A polymer consists of repeating monomeric units.
2 B A primary standard is a solution with an accurately known concentration. These solutions are most commonly prepared by dissolving an accurately known amount of solute into a volumetric flask and making up accurately to a final volume with a f solvent (typically water). The solute is typically weighed out using a balance.
3 D Equal concentrations of reactants and products is not a characteristic of equilibrium.
4 C Atomic absorption spectroscopy (AAS) measures the concentration of an analyte based on the amount of light absorbed by a sample. The analyte will only absorb wavelengths of light that possess an energy that corresponds to a difference between energy levels within the analyte. As a result, a lamp should be selected that emits wavelengths of light that correspond to the wavelengths that are absorbed by the analyte.
5 B The flame test is a qualitative test that is used to identify metals present in a sample. Only metals that emit wavelengths of light within the visible region of the electromagnetic spectrum can be identified using this type of analysis. Silver and magnesium do not emit wavelengths of light within the visible region. Moreover, in this type of analysis, the metal ion is first reduced to its elemental form before the electronic transitions occur. As a result, ions of the same element that possess different oxidation states cannot be differentiated using this technique.
 6 D When measuring the concentration of an analyte within a mixture, a wavelength must be selected that is only absorbed by the analyte of interest.

The spectrum shows that only analyte Q absorbs energy corresponding to 630 nm, making it an appropriate choice for selectively determining its concentration within a mixture of substances P and Q.

7 B Chemical formula diagram in response to 2022 HSC Chemistry Exam Question 7
8 B 2NaHCO3(s) ⇋ Na2CO3(s) + CO2(g) + H2O(g)
As the concentration of solids and pure liquids are usually unchanged over the course of a reaction, these are typically omitted from the equilibrium expression, giving:
Keq = [CO2][H2O]
9 A 2022 HSC Chemistry Exam Question 9 diagram
10 C A Bronsted-Lowry acid is a proton donor, while a Bronsted-Lowry base is a proton acceptor. In option C, the hydrogen carbonate ion donates a proton to ammonia.
11 B When an alcohol is treated with concentrated sulfuric acid it undergoes a dehydration reaction in which the OH group and a hydrogen atom on an adjacent carbon are eliminated to produce an alkene and water.
12 D Option D possesses only two unique chemical environments, and would result in only two signals in a 13C NMR spectrum. These environments are shown in the figure below:

13 A
14 B Decreasing the volume of the syringe will cause the equilibrium to shift right, to produce fewer moles of gas and reduce pressure (Le Chatelier’s Principle). As a result, the reactant concentration will decrease, and the product concentration will increase. Consequently, the ratio of the reactant concentration to product concentration will also decrease.
15 C HCl(aq) + NH3(aq) → NH4Cl(aq)
n(HCl) = c × V = 0.1131 × 0.025 = 2.8275 × 10-3 mol
n(NH3) = n(HCl) (1:1 stoichiometry)
In a strong acid/weak base conductivity titration curve, the end point is indicated by a sharp turning point. This occurs at a volume of 10 mL (0.010 L).
c(NH3) = n/V = 2.8275 × 10-3 mol/0.010 L = 0.283 mol L-1
16 D The absorbance of a solution can be found using the Beer-Lambert Law, which provides the expression:
A = εcl
Where A is absorbance, ε is the molar extinction coefficient, and l is the pathlength. This expression shows that the absorbance is directly proportional to the path length (A ∝ l) meaning that the longer the path length, the higher the absorbance.
17 A As stated in the question, the solubility of silver carbonate at the temperature of the experiment is 1.2 × 10-4 mol L-1. This means that only 1.2 × 10-5 mol can dissolve in 100 mL.

The number of moles of silver carbonate added to the water is:

n(Ag2CO3) = m/MM = 2.0/275.81 = 7.2514 × 10-3 mol

As a result, the amount of silver carbonate added to 100 mL would result in a saturated solution and have undissolved solid. If an additional 100 mL of water was added, an additional 1.2 × 10-5 mol will dissolve, but this is still less than the original quantity added (7.2514 × 10-3 mol) so the solution will still be saturated with some undissolved solid. As a result, the concentration of silver ions would remain unchanged before and after dilution, giving a 1:1 ratio.

18 D The biopolymer shown is a polyester. Polyesters are a type of condensation polymer that can be formed from difunctional monomers that possess both a carboxylic acid and an alcohol functional group. During the reaction, the carboxylic acid and alcohol combine to form an ester with the simultaneous elimination of a small molecule like water.

Moreover, the question states that the desired molecular weight of the polymer is 2900 ± 100 g mol-1. Option D possesses both required functional groups, has a molar mass of 90.078 g mol-1, and consists of 40 units. When these monomers combine, 39 water molecules, with a molar mass of 18.016 g mol-1 each, will be eliminated. The overall molecular weight will therefore be 90.078 × 40 – 18.016 × 39 = 2900.496 g mol-1, which is within the desired specification.

19 A Fe(OH)2(s) ⇋ Fe2+(aq) + 2OH(aq)

Ksp = [Fe2+][OH]2 = 4.87 × 10-17 (from Data Sheet)

Let 𝑥 be the change in concentration of Fe2+

 

Ksp = [Fe2+][OH]2 = 𝑥(2𝑥)2 = 4.87 × 10-17

4𝑥3 = 4.87 × 10-17 → 𝑥 = 2.3 × 10-6 mol L-1

20 C HIO(aq) ⇋ H+(aq) +IO(aq)

pKa = 10.64 → Ka = 10-10.64

Let 𝑥 be the change in concentration of HIO

 

Ka = [H+][IO]/[HIO] = 𝑥2/(0.75 – 𝑥)

Since Ka is small assume [0.75 – 𝑥] ≈ [0.75]
Ka = 𝑥2/0.75 = 10-10.64 → 𝑥 = 4.14506 × 10-6 = [H+]Equilibrium
pH = – log10[H+] = – log10(4.14506 × 10-6) = 5.38

Based on the graph, the major species in the solution at this pH is HCy-, which is purple.

 

Section 2: Long Response Questions

 

Question 21

 

Question 22

Base: PO43−(aq)
Conjugate acid: HPO42−(aq)

OR

Base: F(aq)
Conjugate acid: HF(aq)

 

Question 23(a)

Amount of NO(g) decreases when temperature increases.

 

Question 23(b)

A catalyst provides an alternate reaction pathway of lower activation energy. The activation energy of both the forward and reverse reactions are decreased equally, hence the rate of both the forward and reverse reactions increase equally. Thus, there is no disturbance to the equilibrium, resulting in no shift in equilibrium and no change in equilibrium position.

 

Question 23(c)

Removal of H2O(g) results in a decrease in the rate of the reverse reaction due to fewer collisions between NO(g) and H2O(g) molecules.

Therefore, the rate of the forward reaction is greater than the reverse reaction so the concentrations of NH3(g) and O2(g) gradually decrease. As this occurs, the concentrations of NO(g) and H2O(g) gradually increase.

The rate of the reverse reaction subsequently increases while the rate of the forward reaction subsequently decreases until at some point the rate of the reverse reaction will be the same as the rate of the forward reaction, at which point equilibrium is established and all concentrations become constant.

 

Question 24

The boiling points reflect the thermal energy required to break the intermolecular forces between the molecules to cause boiling. Stronger intermolecular forces result in higher boiling points.

1-chloroalkanes are polar molecules that exhibit dipole-dipole forces and dispersion forces. The strength of the dipole-dipole force is similar in all the 1-chloroalkanes due to the molecules having the same functional group. However, dispersion force strength increases with an increasing number of electrons which is proportional to molar mass. Hence, the boiling points of 1-chloroalkanes increase with increasing molar mass.

However, with increasing molar mass, the increase in boiling point of successive 1-chloroalkanes becomes smaller. This is because the incremental increase in dispersion force strength becomes a smaller and smaller percentage of the total intermolecular force.

 

Question 25

HCl is a strong acid and undergoes complete ionisation:
HCl(aq) + H2O(l) → H3O+(aq) + Cl(aq)

Therefore [H3O+] is 0.2 M, and pH = −log10[H3O+] = 0.70.

However, HCN is a weak acid and only partially ionises:
HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN(aq)

Therefore, the concentration of H3O+ ions is less than 0.2 M, and pH > 0.70.

 

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Question 26(a)

Average initial temperature: 21.1 °C

q(solution)=mc∆T
= (100.7 + 102.0)× 10-3 × 4.18 × 103 × (24.4 – 21.1)
= 2796.0438 J = 2800 J (2 s.f.)


Question 26(b)

\begin{align*}
∆H &=  \frac{-q}{n} \\
&= \frac{-2.6 \text{kJ}}{0.10 \times 0.50 \text{ mol}} \\
&= 52 \text{ kJ mol}^{-1}
\end{align*}

 

Question 27(a)

 

Question 27(b)

The isomers can be identified by 13C NMR spectroscopy as they will show a different number of signals in the 13C spectrum.

As seen in the diagram below, propan-1-ol has 3 unique carbon environments, thus it will exhibit three signals in the 13C spectrum. Propan-2-ol on the other hand is symmetrical and only has 2 unique carbon environments, hence will only exhibit 2 signals in the 13C spectrum.

 

Question 27(c)

 

Question 28(a)

Iron(III) hydroxide (Fe(OH)3)

 

Question 28(b)

\begin{align*}
&Fe_{(s)} → FeCl_{3(s)} → Fe(OH)_{3(s)} → Fe_2 O_{3(s)} \\
&n(Fe_2 O_3 ) = \frac{m}{MM} = \frac{4.21}{55.85 × 2 + 16.00 × 3} = 0.0263619… \text{ mol} \\
&n(Fe) = 2×n(Fe_2 O_3 )=0.0527238 … \text{ mol} \\
&m(Fe)=n×MM=0.0527238 ×55.85=2.944627 \ g \\
& \text{%} \frac{m}{m}=\frac{m(Fe)}{m(Sample)} ×100=2.944627/4.32×100=68.2 \text{% (3 sig. fig.)}\\
\end{align*}


Question 29(a)

Enthalpy of combustion of butan-1-ol = −2150 kJ mol−1

 

Question 29(b)

Heat loss to the surroundings is inevitable in the experiment. Since the heat loss is unaccounted for, the measured temperature change is smaller, and the experimentally determined enthalpy of combustion will be smaller in magnitude.

OR

Incomplete combustion, wherein less heat is released per mole of fuel, resulting in the temperature change being smaller and therefore the experimentally determined enthalpy of combustion being smaller in magnitude.

 

Question 30

IR spectrum:

The IR spectrum shows absorptions around 3000 cm-1 which are characteristic of C–H bonds and a strong absorption at 1700 cm-1, which is consistent with a carbonyl group.

The absence of a broad absorption band in the region from 2500–3550 cm-1 indicates the absence of hydroxyl and carboxyl groups. This is consistent with a ketone.

 

Mass spectrum:

The mass spectrum has a parent ion peak at 86, indicating that the compound must have a molar mass of around 86 g mol−1. This is consistent with the formula C5H10O.

 

13C NMR:
4 signals indicate 4 unique carbon environments. This is consistent with the proposed structure.

The 2 signals between 2-50 ppm are consistent for carbons adjacent to a carbonyl carbon (carbons A and C) and the signal below 20 ppm is consistent for an aliphatic carbon (carbons D).The downfield signal above 200 ppm is consistent with a carbonyl carbon of a ketone (carbon B) which has the characteristic chemical shift in the region of 190-220 ppm.

 

1H NMR:

3 signals indicate 3 unique proton environments. This is consistent with the proposed structure.

The hydrogen generating the septet around 2.5 ppm is consistent with hydrogen labelled b. It is adjacent to 6 non-equivalent hydrogens.

The 3 hydrogens generating the singlet around 2 ppm is consistent with the 3 hydrogens labelled a. They have no neighbouring hydrogens.

The 6 hydrogens generating the doublet around 1 ppm is consistent with the 6 hydrogens labelled c. They are adjacent to 1 non-equivalent hydrogen.

 

Question 31(a)

Adding NaI(aq) will precipitate Ag+(aq) as AgI(s) according to the following equilibrium:

AgI(s) ⇌ Ag+(aq) + I(aq)
Ksp = 8.52 × 10–17

Since Ksp is very small, this equilibrium lies very far to the left, which indicates that AgI(s) is highly insoluble. NaI(aq) is thus a good choice of reagent to precipitate all Ag+(aq) ions from the solution as very few Ag+(aq) ions will remain dissolved. The total [Ag+] can therefore be assessed accurately and validly.

However, the objective is to determine the concentration of free Ag+(aq); that is, Ag+(aq) ions that are not part of the [Ag(NH3)2]+(aq) complex. This complex forms according to the following equilibrium:

Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)2]+(aq)
Keq = 1.6 × 107

Removing Ag+(aq) ions by precipitating them as AgI(s) disturbs this equilibrium. According to Le Châtelier’s principle, the system shifts left to replace the lost Ag+(aq) and minimise the disturbance. Consequently, some Ag+(aq) ions that were part of the [Ag(NH3)2]+(aq) complex dissociate and become free Ag+(aq) ions.

Additional NaI(aq) titrant will be needed to precipitate these additional Ag+(aq) ions, which will lead to an inaccurate calculation of the original concentration of free Ag+(aq) ions in the solution.

Therefore, this method is not suitable for determining the concentration of free Ag+(aq) ions in the solution. A more valid method would be to use atomic absorption spectroscopy to determine the concentration of free Ag+(aq) ions by comparing the solution’s absorption of specific wavelengths of light to standard solutions of Ag+(aq). This method avoids disturbing the equilibrium and causing changes to the concentration of Ag+(aq).

 

Question 31(b)

\(K_{eq} = \frac{[[Ag(NH_3 )_2 ]^+]}{[Ag^+ ][NH_3 ]^2}=1.6×10^7\)

If 0.010% of the total silver ions in solution are present as Ag+(aq) at equilibrium, then 99.99% of the total silver ions in solution are [Ag(NH3)2]+(aq). Thus:

\([Ag^+ ]= \frac{0.010}{99.99} × [[Ag(NH_3 )_2 ]^+]\)

Therefore:

\begin{align*}
1.6×10^7 = \frac{[[Ag(NH_3 )_2 ]^+]}{(\frac{0.010}{99.99}×[[Ag(NH_3 )_2 ]^+])[NH_3 ]^2}= \frac{1}{ \frac{0.010}{99.99} × [NH_3 ]^2 } \\
[NH_3 ]^2= \frac{1}{ \frac{0.010}{99.99} ×1.6×10^7 } \\
[NH_3 ]= \sqrt{ \frac{1}{ \frac{0.010}{99.99} ×1.6 × 10^7 } } =0.025 \text{ mol L}^{-1} \text{ (2 sig. fig.)} \\
\end{align*}


Question 32(a)

For the standardisation of NaOH by titration with KHP:

\begin{align*}
& n(KHP)= \frac{m}{MM} = \frac{4.989}{204.22} =0.02442… \text{ mol} \\
& [KHP]= \frac{n}{V} = \frac{0.02442…}{0.1} =0.2442… \text{M} \\
& n(KHP)_{titration}=c×V=0.2442…×0.025=0.006107… \text{mol}\\
& KHP_{(aq)}+NaOH_{(aq)}→NaKP_{(aq)}+H_2 O_{(l)} \\
& n(NaOH)=n(KHP)_{titration}=0.006107… \text{mol}\\
& V_{average} (NaOH)= \frac{0.0274+0.0272+0.0276}{3}=0.0274 \text{ L (trial 1 is an outlier)} \\
& [NaOH]= \frac{n}{V} = \frac{0.006107…}{0.0274} = 0.2228… \text{M} \\
\end{align*}

For the titration of NaOH against citric acid (H3A):

\begin{align*}
& n(NaOH)= c×V=0.2228…×0.0131=0.002919… \text{mol} \\
& H_3 A_{(aq)} +3NaOH_{(aq)} →Na_3 A_{(aq)}+3H_2 O_{(l)} \\
& n(H_3 A)= \frac{1}{3} × n(NaOH)= \frac{1}{3} × 0.002919…=0.0009733… \text{mol} \\
& [H_3 A]_{dilute} = \frac{n}{V} = \frac{0.0009733…}{0.025} =0.03893… \text{M}\\
& [H_3 A]_{original} = \frac{[H_3 A]_{dilute} × V_{dilute}}{V_{original}} = \frac{0.03893…×0.25}{0.075} =0.1298 \text{ M (4 sig. fig.)} \\
\end{align*}

 

Question 32(b)

Carbon dioxide, CO2(g) is sparingly soluble in water. In solution, CO2(aq) can react with water to produce carbonic acid, H2CO3(aq) according to the following equilibrium:

CO2(g) ⇌ CO2(aq)+H2O(l) ⇌ H2CO3(aq)

H2CO3(aq) can react with NaOH(aq) in a neutralisation reaction:

H2CO3(aq) + 2NaOH(aq) → Na2CO3(aq) + 2H2O(l)

If CO2(aq) was not removed from the soft drink before titrating it against NaOH(aq), additional NaOH(aq) titrant would be required to neutralise the CO2(aq) as well as the citric acid, H3A(aq). This would result in the calculation of a greater n(NaOH), a greater n(H3A), a greater [H3A]dilute, and finally a greater [H3A]original than in part (a).

 

Question 33

Esterification is a chemical synthesis process that may be undertaken in a school laboratory. In this process, an ester is produced by a condensation reaction in which an alcohol and a carboxylic acid are reacted together in the presence of an acid catalyst. The reaction proceeds according to the general equilibrium:

A typical procedure for this process to produce ethyl acetate is as follows:

  1. Place 10 mL glacial acetic acid, 10 mL ethanol, and 3 mL 32% (v/v) sulfuric acid in a 100 mL pear-shaped flask with boiling chips.
  2. Heat mixture under reflux for 1 hour.
  3. Allow the mixture to cool, then transfer to a 250 mL separating funnel and wash with water (3 × 100 mL).
  4. Discard the aqueous layers and wash the organic layer with 2.5 M sodium carbonate solution (3 × 100 mL), venting the separating funnel regularly.
  5. Dry the organic layer over anhydrous sodium sulfate, then filter to remove solids.
  6. The product, ethyl acetate, may be further purified by fractional distillation.

 

Selection of reagents

  • Acetic acid is a good choice for the carboxylic acid as it is readily available and is not exceedingly hazardous.
  • Ethanol is a good choice for the alcohol as it is also readily available and is less toxic than most other alcohols.
  • Concentrated sulfuric acid is the best choice for the acid catalyst as it is a strong acid that will provide a high concentration of H+ ions to catalyse the reaction. It also has a useful secondary utility as a dehydrating agent.

 

Reaction conditions (rate)

  • Esterification is a slow equilibrium reaction at room temperature, so conditions must be optimised to maximise the rate of the process.
    The reaction is carried out under reflux, which involves heating the reactants with a condenser attached to prevent the loss of volatile reactants and products by condensing them back to liquids. This allows the reaction to proceed at higher temperatures and facilitates superior mixing of reactants, which increases the rate of the reaction.
  • Boiling chips are included in the reaction mixture to promote even boiling of reactants, which also contributes to increasing the rate of the reaction.
  • Concentrated sulfuric acid is used as a catalyst to further increase the rate of the reaction.

 

Hazards and safety precautions

Hazard Risk Control
Acetic acid and concentrated sulfuric acid are corrosive May cause severe skin burns and eye damage Wear safety glasses, lab coat, and gloves when handling
Ethanol and sodium carbonate are irritants May cause serious eye irritation Wear safety glasses and lab coat when handling
Acetic acid and ethanol are flammable as liquid or vapour May cause fire if ignited Use a hot plate or heating mantle to carry out reaction, remove all ignition sources
Reflux heating May cause pressure build-up inside apparatus (volatile reactants and products) Ensure the top of the reflux condenser is open to avoid dangerous pressure build-up
Separating funnel with sodium carbonate solution May cause pressure build-up inside apparatus (carbon dioxide gas produced) Vent separating funnel regularly to avoid dangerous pressure build-up

 

Yield and Purity

Since the reaction is a dynamic equilibrium, it does not go to completion. Using concentrated sulfuric acid improves the yield as it is a dehydrating agent and will remove water from the system.

According to Le Châtelier’s principle this drives the equilibrium further forward to replace the water that is removed, resulting in a greater yield of ethyl acetate.

Once the system has reached equilibrium, some unreacted alcohol and carboxylic acid will remain in the reaction mixture, which also includes the desired ester, water as a by-product, and leftover sulfuric acid catalyst. Steps 4–6 in the procedure are necessary to separate the ester from the other components in the mixture.

  • Washing with water removes the water produced from the esterification and any water-soluble substances, which includes the sulfuric acid catalyst and (possibly) some unreacted alcohol and carboxylic acid
  • Washing with sodium carbonate solution removes the remaining carboxylic acid, RCOOH(aq), as it reacts with the sodium carbonate to form a sodium carboxylate salt, RCOONa(aq), which is water-soluble, and water according to the neutralisation reaction:
    2RCOOH(aq) + Na2CO3(aq) → 2RCOONa(aq) + CO2(g) + H2O(l)
  • Drying the organic layer over anhydrous sodium sulfate removes any remaining water droplets in the organic layer left over from separation.
  • Since some unreacted alcohol may remain in the organic layer at this point, fractional distillation may be necessary to separate this from the desired ester based on their different boiling points. Alcohols can form Hydrogen bonds to each other whereas esters cannot, and consequently esters have lower boiling points than alcohols of similar size.

 

Question 34

\(K_b=\frac{[HOCl][OH^-]}{[OCl^-]}\)

If at equilibrium pH = 7.5, then pOH = 6.5, and [OH–] = 10–6.5 M. Also, [HOCl] = 1.3 × 10–4 M, and Kb = 3.33 × 10–7. Therefore:

\begin{align*}
3.33 × 10^{-7}= \frac{1.3×10^{-4}×10^{-6.5}}{[OCl^-]_{eqm}}\\
[OCl^- ]_{eqm}=\frac{1.3×10^{-4}×10^{-6.5}}{3.33×10^{-7}} =1.23… × 10^{-4} \text{ M} \\
\end{align*}

This is the [OCl] when the system has come to equilibrium after the NaOCl(aq) solution is added to the pool. To calculate the volume of NaOCl(aq) required, the initial [OCl] must be determined using the equilibrium concentrations and an ICE table:

conc. (M) OCl–(aq) H2O(l) HOCl(aq) OH(aq)
Initial 2.53… × 10–4 0
Change – 1.3 × 10–4 + 1.3 × 10-4
Equilibrium 1.23… × 10–4 1.3 × 10-4 10-6.5

Thus, the initial [OCl]pool = 2.53… × 10–4 M.

When in solution, NaOCl(aq) dissociates completely according to the following reaction:

NaOCl(aq) → Na+(aq) + OCl(aq)
so [OCl] = [NaOCl]. Therefore:

\(V(NaOCl)_{required}= \frac{[OCl^- ]_{pool}×V_{pool}}{[NaOCl]}= \frac{2.53…×10^{-4}×10^4}{2}=1.3 \text{ L (2 sig. fig.)} \)

 

Question 35

\begin{align*}
& n(Sr(NO_3)_2) = c × V = 1.50 × 0.080 = 0.12 \text{ mol} = n(Sr^{2+})_{initial} \\
& n(NaOH) = c × V = 0.855 × 0.080 = 0.0684 \text{ mol} = n(OH^-)_{initial} \\
& n(Sr(OH)_2) = \frac{m}{MM} = \frac{3.93}{121.63} = 0.0323111 \text{ mol} \\
& n(Sr^{2+})_{used} = n(Sr(OH)_2) = 0.0323111 \text{ mol} \\
& n(OH^-)_{used} = 2 × n(Sr(OH)_2) = 2 × 0.0323111 \text{ mol} = 0.0646222 \text{ mol} \\
& n(Sr^{2+} )_{remaining} = n(Sr^{2+} )_{initial}-n(Sr^{2+} )_{used}= 0.12-0.0323111 = 0.08768889 \text{ mol} \\
& [Sr^{2+}]_{remaining} = \frac{n}{ V_{total}} = \frac{0.08768889}{0.160} = 0.548055578 \text{ mol L} ^{-1} \\
& n(OH^- )_{remaining} = n(OH^- )_{initial}  \ – \ n(OH^- )_{used} = 0.0684-0.0646222 = 3.77785×10^{-3} \\
& [OH^- ]_{remaining} = \frac{n}{V_{total}} = \frac{3.77785×10^{-3}}{0.160} = 0.02361115679 \text{ mol L} ^{-1} \\
& Sr(OH)_{2(s)} ⇌ Sr^{2+}_{(aq)} + 2OH^-_{(aq)} \\
& K_{sp} = [Sr^{2+}][OH^-]^2 = 0.548055578 × 0.023611156792 = 3.06×10^{-4} \text{ (3 s.f.)} \\
\end{align*}

 

Question 36

Since the evaporation of water results in the formation of gaseous particles that have a greater freedom of movement, the forward reaction will be associated with an increase in entropy (or positive entropy change).

The evaporation of water requires energy to be absorbed, hence the process is endothermic and thus enthalpy increases (or enthalpy change is positive).

Since the enthalpy change and entropy change is positive for the forward evaporation reaction, it can be predicted that the complete evaporation of water is spontaneous at high temperatures. At 23 °C, only a small percentage of liquid water molecules will have sufficient energy to become gaseous water.

100 mL of water held in a sealed bottle represents a closed system, so dynamic equilibrium will eventually be established. In this system, the rate of the forward and reverse reactions will be equal at dynamic equilibrium and there would be no change in the concentration of gaseous water, so the liquid water will not completely evaporate.

100 mL of water held in a beaker represents an open system, so dynamic equilibrium will not be established. The gaseous water diffuses away over time, removing it from the mixture. According to Le Châtelier’s principle, the equilibrium will shift continuously to the right to replace the gaseous water to minimise the disturbance. Therefore, all of the liquid water will be converted to gaseous water and escape, eventually resulting in complete evaporation.

Written by Matrix Science Team

The Matrix Science Team are teachers and tutors with a passion for Science and a dedication to seeing Matrix Students achieving their academic goals.

 

© Matrix Education and www.matrix.edu.au, 2018. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.

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