Properties of Acids
- Change the colour of indicators. Eg. Acids turn blue litmus red.
- Are electrolytes.
- Have a sour taste.
- React with reactive metals (such as K, Ca, Mg, Zn, Fe) to produce hydrogen gas.
- React with carbonates/hydrogen carbonates to form carbon dioxide gas.
- React with metal oxides to produce a salt and water.
- React with metal hydroxides to produce a salt and water.
Properties of Bases
- Change the colour of indicators Eg. bases turn red litmus blue.
- Are electrolytes and conduct an electric current.
- React with acids to produce a salt and water
- React with amphoteric metals (such as Al, Zn) to produce hydrogen gas.
- Dissolve amphoteric metal hydroxides (such as Al(OH)3, and Zn(OH)2 )
1.2 Identify that indicators such as litmus, phenolphthalein, methyl orange and bromothymol blue can be used to determine the acidic or basic nature of a material over a range, and that the range is identified by change in indicator colour
- An indicator is a natural or synthetic substance that, in solution, changes colour depending on the acidity/basicity of the solution
- An indicator is a solution containing a weak acid in equilibrium with its conjugate base:
- It has one colour (1) in acidic form (HIn) and colour 2 (In–) in its conjugate base form. o The colour will change depending on the concentration of H3O+
- The pH ranges for colour changes in common indicators are as follows:
1.3 Identify and describe some everyday uses of indicators including the testing of soil acidity/basicity
Testing soil pH
- Some plants can only grow within narrow pH ranges – soil pH affects nutrient availability which affects plant growth. Hence indicator is used to check if soil is at a suitable pH for optimal plant growth/maximum crop yield.
- When testing soil pH, the soil colour can hide the indicator colour change. Hence, barium sulfate, an insoluble white powder, is added to the damp soil (before adding indicator) to provide contrast. It absorbs the soil moisture allowing any colour change to be visible.
- Checking swimming pool pH
- The pH of swimming pool water needs to be kept close to pH of 7.4 so as to prevent irritation of skin, eyes and mucous membranes. Hence, indicator is used to check if pH is close to 7.4.
- pH adjustments can be made if required:
o pH >> 7.4 → acid (eg. HCl or NaHSO4) is added
- pH << 7.4 → base (eg. NaCO3) is added
- Pool pH test kits commonly use phenol red (pH range 6.8–8.2) or bromothymol blue (pH range pH 6.0–7.6). Bromothymol blue is a good choice of indicator because it changes colour around pH 7.4 (green at pH 6.0–7.6), whilst other indicators have wider pH ranges when at pH 7.4 (methyl orange is yellow for pH > 4.4, phenolphthalein is colourless for pH < 8.3).
Testing aquarium water
- Most fish are pH-sensitive and can only thrive within a narrow pH range (Eg. Saltwater fish need a pH of ~8.5). Hence indicators can be used to check if adjustments need to be made.
- Aquarium water must be monitored to prevent proliferation of unwanted algae and bacteria which can lead to illness or death of its fish.
1.4 Solve problems by applying information about the colour changes of indicators to classify some household substances as acidic, neutral or basic
2.1 Identify oxides of non-metals which act as acids and describe the conditions under which they act as acids
Acidic & Basic Oxides
- Non-metal oxides are acidic:
o Reacts with water to form acids ie. Non-metal oxide + water à acid (Eg. SO2(s) + H2O(l) à H2SO3(aq))
o Reacts with bases to form salt and water ie. Non-metal oxide + base à salt + water (Eg. SO2(s) + 2NaOH(aq) à H2O(l) +
- Metal oxide are basic:
o React with water to form a basic solution ie. Metal oxide + water à base (Eg. MgO(s) + H2O(l) à Mg(OH)2(aq))
o Reacts with acids to form salt and water ie. Metal oxide + acid à salt + water (Eg. MgO(s) + 2HCl(aq) à H2O(l) + MgCl2(aq))
o The inert gases in group 8 do not form oxides – Xenon is the only noble gas that forms a relatively stable acidic oxide. o Some oxides are amphoteric, that is, they reacts with acids and bases, such as Be, Zn, Al, Pb, Sn –
(Eg. reacting with acid: ZnO(s) + 2HCl(aq) à H2O(l) + ZnCl2(aq) and reacting with base: ZnO(s) + 2NaOH(aq)à H2O(l) + Na2ZnO2(aq) )
- Some oxides are neutral such as CO, NO, N2
2.2 Analyse the position of these non-metals in the Periodic Table and outline the relationship between position of elements in the Periodic Table and acidity/basicity of oxides
Periodic Trends and Oxides
- Generally the more electronegative the element, the more acidic the oxide.
The fewer the oxygen atoms associated with each element, the less acidic is the molecule. Eg. SO3=strong, SO2=weak acid
- Elements to the left of the PT are metals and tend to form basic oxides. These have low electronegativity. Eg. MgO dissolves
in water to form a basic solution of Mg(OH)2: MgO(s) + H2O(l) à Mg(OH)2(aq)
- Elements to the right of the PT are non-metals and tend to form acidic oxides. These have high electronegativity. Eg. CO2(s) +
H2O(l) à H2CO3(aq). Exceptions include the noble gases in group 8 which do not form stable oxides (only noble gas Xenon forms a stable acidic oxide), and some neutral oxides such as CO, NO, N2O.
- Elements near the right of the transition metal block (at borderline between metals and non-metals) form amphoteric oxides. These have moderate electronegativity. They can react with both acids and bases: Eg. Reacting with acid: ZnO(s) + 2HCl(aq) à H2O(l) + ZnCl2(aq) and reacting with base: ZnO(s) + 2NaOH(aq)à H2O(l) + Na2ZnO2(aq)
- Le Chatelier’s principle= When a system at equilibrium is disturbed, the equilibrium will shift so as to minimize the effect of the disturbance.
- Chemical/Dynamic equilibrium= A closed system where:
o Forward reaction rate = reverse reaction rate o [Reactants] and [products] are constant
o Macroscopic properties (eg. colour) are constant.
Effects of Changing Factors on Equilibrium
- ↑[ ] = shift to the opposite side (to remove some of added component)
- ↓[ ] = shift to the same side (to replace some of removed component)
- ↑P/↓V = shift to side with less gas moles (to reduce the pressure)
- ↓P/↑V = shift to side with more gas moles (to increase the pressure)
- ↑ T = shift in the endothermic direction (to remove some of added heat) (eg. Haber process)
- ↓ T = shift in the exothermic direction (to replace some of removed heat)
2.5 Describe the solubility of carbon dioxide in water under various conditions as an equilibrium process and explain in terms of Le Chatelier’s principle
Effect of Changes in Pressure & Temperature on the Solubility of CO2
- It involves the following equilibria:
|CO2(g) ⇌ CO2(aq) + heat||ΔH = -19kJ||…(1)|
|CO2(aq) + H2O(l) ⇌ HCO3–(aq) + H+||(aq) + heat ΔH = -50kJ||…(2)|
- When pressure is increased (i.e. volume is decreased), (1) will shift to the RHS (with lower number of gas moles) to reduce the pressure, that is, to remove some gaseous CO2 by dissolving it. This will affect (2) such that the [H+(aq)] eventually increases and the solution becomes more acidic. \Sharper/more sour.
- When pressure is decreased (i.e. volume is increased), (1) will shifts to the LHS (side with higher number of gas moles) to increase the pressure. This will affect (2) such that the [H+(aq)] eventually decreases and the solution becomes less acidic. \ Neutral/less sour.
- When temperature is increased (by heating/un-refrigerating), (1) will shift to the LHS (endothermic direction) to remove/absorb some of the applied heat, decreasing [CO2(aq)] (ie. decreasing solubility of CO2 and acidity of solution).
- When temperature is decreased (by refrigerating), (1) will shift in RHS (exothermic direction) to replace some of the lost heat, increasing [CO2(aq)] (ie. increasing CO2 solubility and acidity of solution).
Summary of Effects of Changing Factors on CO2 Solubility
- ↑[H+(aq)] = shift to LHS in (2) and shift to LHS in (1) = decreased CO2 solubility
- ↓[H+(aq)] = shift to RHS in (2) and shift to RHS in (1) = increased CO2 solubility
- ↑P/↓V = shift to RHS in (1) = increased CO2 solubility
- ↓P/↑V = shift to LHS in (1) = decreased CO2 solubility
- ↑ T = shift to LHS (endothermic direction) in (1) = decreased CO2 solubility
- ↓ T = shift to RHS (exothermic direction) in (1) = increased CO2 solubility
Analogies for Effects of Changing Factors on CO2 Solubility
- Opening bottle of carbonated soda = ↓P = ↑[CO2(g)] = decreased CO2 solubility
- Taking soda out of fridge/Heating soda = ↑ T = goes flat quicker ie. ↑ [CO2(g)] = decreased CO2 solubility
2.6 Identify natural and industrial sources of sulfur dioxide and oxides of nitrogen (PLUS) 2.7 Describe, using equations, examples of chemical reactions which release sulfur dioxide and chemical reactions which release oxides of nitrogen
2.8 Assess the evidence which indicates increases in atmosphere concentration of oxides of sulfur and nitrogen
2.12 Analyse information from secondary sources to summarise the industrial origins of sulfur dioxide and oxides of nitrogen and evaluate reasons for concern about their release into the environment
- Natural unpolluted rain: pH≈ 5.6 (because of the natural CO2/H2CO3 equilibrium)
- Acid rain: pH< 5.6
Formation of acid rain
- Acidic oxides formed mainly enter atmosphere: (See equation for combustion of fossil fuel and oxidation)
- Acidic oxides of sulfur, nitrogen (and carbon) react with moisture in the air forming acids:
SO2 (g) + H2O(l) ⇌ H2SO3 (aq) SO3 (g) + H2O(l) → H2SO4 (aq)
2NO2 (g) + H2O(l) → HNO2 (aq) + HNO3 (aq)
- These forms a dilute solution of acid which is incorporated into rainclouds (‘acidified clouds’)
- This is precipitated as acid rain.
2.10 Calculate volumes of gases given masses of some substances in reactions, and calculate masses of substances given gaseous volumes, in reactions involving gases at 0˚C and 100kPa or 25˚C and 100kPa
2.11 Identify data, plan and perform a first-hand investigation to decarbonate soft drink and gather data to measure the mass changes involved and calculate the volume of gas released at 25˚C and 100kPa
3.2 Identify acids including acetic (ethanoic), citric (2-hydroxypropane-1,2,3-tricarboxylic), hydrochloric and sulfuric acid
3.3 Describe the use of the pH scale in comparing acids and bases (PLUS) 3.5 Identify pH as -log10(H+) and explain that a change in pH of 1 means a ten-fold change in (H+)
- pH (potential of hydrogen) is a measure of the acidity/basicity of a solution, which is dependent on the hydrogen ion concentration.
- pH is the negative of the logarithm (to base 10) of the hydrogen/ hydronium ion concentration:
o pH = -log10[H+]
- [H+] = 10-pH (∵ [H+] = antilog(-pH))
- pH + pOH = 14
- #(dp.) for pH = #(sf.) for [H+] (usually 2dp. for pH)
- Because pH is a log scale, each one pH unit increase/decrease corresponds to a ten-fold decrease/increase respectively in [H+(aq)] concentration. Eg. pH decrease by 1 indicates that the ten-fold increase in [H+]: pH drop of 1= [H+] x 101
- pH can be measured by using indicators/pH meters/probes. pH meters/probes measure the voltage, which is dependent on the hydrogen ion concentration, and convert it to a pH reading. They are accurate but expensive.
- pH depends on:
o acid strength
- number of ionisable protons
- Ion product constant of water: Kw = [H3O+] [OH–] = 1.0 x 10-14 (at 25°C).
- If [OH–(aq)] increases, then [H3O+(aq)] must decrease so that the product of the ion concentrations remains constant.
- neutral solutions: [H3O+(aq)] = 10-7 molL-1 (at 25°C).
- basic solutions: [H3O+(aq)] < 10-7 molL-1
o acidic solutions: [H3O+(aq)] > 10-7 molL-1
- Using pH = -log10[H+]:
- neutral solutions: pH = 7
- acidic solutions: pH < 7
- basic solutions: pH > 7
3.4 Describe acids and their solutions with the appropriate use of the terms strong, weak, concentrated and dilute (PLUS) 3.7 Describe the difference between a strong and a weak acid in terms of an equilibrium between the intact molecule and its ions
Strong Vs Weak Acids
o Acid strength ∝ degree of ionisation
o There are degrees of weakness of weak acids, but no degrees of strongness for strong acids.
3.6 Compare the relative strengths of equal concentrations of citric, acetic and hydrochloric acids and explain in terms of the degree of ionisation of their molecules
3.8 Solve problems and perform a first-hand investigation to use pH meters/probes and indicators to distinguish between acidic, basic and neutral chemicals
- Methyl orange is only red in strongly acidic solutions, so even a very weak base would have been able to elicit a colour change.
- A wide range of substances was used to portray the wide range of possible bases. This showed that bases are not only limited to metal hydroxides.
- A pH meter gave instantaneous accurate results.
- The methyl orange test was not able to distinguish basic and neutral substances, as both caused no colour change from the red.
- Using a pH meter or probe is a non-destructive way of testing whether a chemical solution is acidic, basic or neutral. Provided the pH meter electrode or probe is washed well with distilled water between measurements, the solutions tested should be unaffected, but the indicator will contaminate the portion of solution tested.
3.9 Plan and perform a first-hand investigation to measure the pH of identical concentrations of strong and weak acids
3.10 Gather and process information from secondary sources to write ionic equations to represent the ionisation of acids
3.11 Use available evidence to model the molecular nature of acids and simulate the ionisation of strong and weak …….acids
3.12 Gather and process information from secondary sources to explain the use of acids as food additives
3.13 Identify data, gather and process information from secondary sources to identify examples of naturally occurring acids and bases and their chemical composition
3.14 Process information from secondary sources to calculate pH of strong acids given appropriate hydrogen ion concentrations
4.1 Outline the historical development of ideas about acids including those of Lavoisier, Davy and Arrhenius (PLUS) 4.10 Gather and process information from secondary sources to trace developments in understanding and describing acid/base reactions
|·||A acid is a PROTON donor (ie. gives protons to a base)|
|·||A base is a PROTON acceptor (ie. accepts protons from an acid)|
4.4 Identify a range of salts which form acidic, basic or neutral solutions and explain their acidic, neutral or basic nature
- The salt or conjugate acids/bases can act as B-L acids/bases. The nature of a salt depends on the strength of the reactants: o strong acid + strong base → neutral salt
o strong acid + weak base → moderately acidic salt o weak acid + strong base → moderately basic salt
o weak acid + weak base → neutral OR slightly acidic/basic salt (depending on their degrees of weakness)
- The acidity/alkalinity of a salt must be proved by hydrolysis (reaction with water to produce OH–/ H3O+)
4.3 Describe the relationship between an acid and its conjugate base and a base and its conjugate acid (PLUS) 4.5 Identify conjugate acid/base pairs
4.6 Identify amphiprotic substances and construct equations to describe their behaviour in acidic and basic solutions
- Neutralisation reactions are reactions between acids and bases.
- As acids are proton-donors & bases are proton-acceptors, neutralisation reactions between acids and bases are proton-transfer reaction
Features of a Primary Standard
- Very high purity solid – so that its weight is truly representative of the number of moles of substance contained so as to produce accurate results. Impurities are of unknown composition and cannot be taken into account. Solid provides ease of weighing.
- Non-hygrosopic and non-efflorescent – so that it does not absorb water to prevent inaccurate mass (part of it=water). Hygroscopic would results in a higher no. of moles calculated (b/c n=m/M) and thusly higher concentration that the actual concentration. Non-efflorescence means that it does not readily lose its water in its chemical composition to the atmosphere which would result in inaccurate mass. ie. It absorbs water from air (hence inaccurate mass) and reacts with carbon dioxide in the air (inaccurate concentration). [NOTE: NaOH is hygroscopic, so it readily absorbs water from the air. This means part of the mass will water and impurities. NaOH also reacts with CO2, reducing the concentration. The calculated concentration for NaOH using the mass of pellets will be less that actual concentration.]
- High solubility – so that it can to completely dissolve to make aqueous solution.
- High molecular weight – to ensure accuracy as it is less prone to weighing errors.
Preparation of Standard Solutions
- Calculate amount of primary standard required (eg. to create 250 mL of the 0.05 M standard solution of sodium carbonate) [EG CALCULATIONS: n = c × v = 0.05 x 0.25 = 0.0125 mol; m = 0.125 x (2×22.99 + 12.01+ 3×16) = 1.325g]
- Weigh primary standard on watch glass.
- Transfer into beaker.
- Add distilled water and stir until completely dissolved.
- Transfer into volumetric flask.
- Rinse the funnel, beaker and stirring rod using distilled water to ensure entire primary standard solid is transferred into the flask. (ie. Quantitative transfer)
- Fill the flask until the bottom of the meniscus lies on the calibration line. Use a dropper to add the last few millilitres.
- Place lid on top and invert the flask to ensure uniform mixing
- Prepared a standard solution
- Rinse burette with ~10mL of solution to be transferred (Solution A of known conc.). Discard rinsings. (NOTE: Rinsing with the solution removes any water which would reduce the volume, and thus the number of moles. Rinsing with water= lower base conc.= less acid required for endpoint= higher acid conc. cal.)
- Fill burette above the 0.00mL mark with Solution A. (NOTE: Burettes have 0.1 mL divisions which provide accuracy of ±0.5% whereas a measuring cylinder gives accuracy of ±5%)
- Open the tap and run out the solution so that there are no air bubbles (between the stopcock & tip) and until the bottom of meniscus lies on the 0.00mL mark (Burette solution= TITRANT. Volume of liquid delivered by burette= TITRE).
- Clamp the burette vertically using a burette clamp.
- Record initial volume in burette.
- Rinse conical flask several times with distilled water. Discard rinsings. (NOTE: The number of moles of solution to be used has already been measured out by the pipette: Any water left in the flask will not change the number of moles. It would be incorrect to rinse the flask with the solution as this would change the number of moles of reactant present).
- Rinse pipette several times with the solution to be transferred (Solution B of UNKNOWN concentration). Swirl and discard rinsings. (Volume of liquid delivered by pipette= ALIQUOT). (NOTE: Rinsing with the solution removes any water which would reduce the volume, and thus the number of moles.)
- Fill the pipette with Solution B. Adjust volume until the bottom of meniscus lies on the calibration line. (NOTE: Squeeze A and compress bulb. Release A. Release bulb. Put tip into solution. Press S. Press E to eject)
- Deliver the accurate volume of aliquot (Solution B) to the conical flask. Hold tip is against the inside wall of the flask for draining time (~20s). (NOTE: Pipette is calibrated to take into account the small amount of solution left in the tip – do not blow/shake it).
- Add 3 drops of SUITABLE indicator into flask and mix. (Indicator itself is weakly acidic/basic, so minimise no. of drop)
- Place conical flask on a white tile under the burette
- Run the standard solution slowly into the conical flask until the indicator JUST changes colour permanently. Squirt the sides of the conical flask with distilled water if any reactant splashed onto the sides. When close to the end point, add titrant drop by drop, swirling between each added drop. (NOTE: This signals that the volume delivered has gone JUST beyond the equivalence point, that is, the end-point has been reached. Swirl the reaction mixture continuously.
- Record final volume in burette.
- Repeat steps 6-12 three times for three titres, excluding the first titre (rough titre). (For reliability, these titres must be concordant and must lie within narrow margin not exceeding 0.1-0.2mL)
- Average the values, excluding any outliers.
- Write a balanced equation, including states.
- Calculate the number of moles of KNOWN standard solution using n(s) =c x V (DO NOT ROUND OFF EARLY)
- Calculate the number of moles of UNKNOWN solution using n(u) = x n(s), where is determined by the mole ratio.
- Calculate the concentration of the UNKNOWN solution using c(u) = n(u)/V(u)
4.9 Qualitatively describe the effect of buffers with reference to a specific example in a natural system
- Buffers = mixture of weak acid and its conjugate base in equimolar concentration that minimises pH changes when small amounts of acid/base are added which is therefore able to maintain an approximately constant pH.
- H2CO3/HCO3– buffer system in the blood that maintains blood pH at ~ 7.4. This is important ∵ the efficiency of biochemical reactions in the blood relies on carefully controlled pH.
- H2CO3(aq) + H2O(l) ⇌ HCO3–(aq) + H3O+(aq) …(1)
- A pH increase in the blood acid (drop in [H3O+]), disturbs equilibrium. By LCP, the equilibrium will shift to LHS to produce more of the reduced H3O+. OR An increase in [OH–] will cause the reaction: H3O+(aq) + OH-(aq) → 2H2O(l) , decreasing [H3O+] in (1). By LCP, the equilibrium will shift to LHS to produce more of the reduced H3O+.
- A pH decrease in the blood acid (increase in [H3O+]) causes the equilibrium to shift to LHS to use up some of the added H3O+.
- This keeps the blood pH relatively constant.
4.11 Choose equipment and perform a first-hand investigation to identify the pH of a range of salt solutions
4.12 Perform a first-hand investigation and solve problems using titrations and including the preparation of standard solutions, and use available evidence to quantitatively and qualitatively describe the reaction between selected acids and bases
- Misjudging the colour of the indicator near the end point – Not only colour change is often very delicate and slow, but different people have different sensitivity to colours. Parallax error
- Misreading the volume – parallax error as measurement may not have been at eye level to the increment.
- Not transferring all solid/liquid when preparing samples – it may happen that part of the solid was left in the funnel during
transferring it into flask
- Rinsing of burette/pipette in solution to be used for the titration + rinsing of the flask with distilled water.
- Alignment of eye level with the bottom of the meniscus prevents parallax error.
- Use of burette is marked with 0.1 mL divisions which allow accurate readings of the volume. (If volumes half-way between 1 divisions are accounted for, the limit of accuracy is 0.05mL.)
- Elimination of air bubble below tap ensures accurate volume.
- Use of white tile directly behind burette helps read accurately the meniscus and see the colour change in the reaction
- Use of a pH meter/data logger over indicator to accurately determine the end point.
- Repetition of titration and subsequent evaluation of the concordance of results across the three titres. By repeating it multiple times, outliers can be to identified and excluded. Titres must be concordant to each other, not exceeding narrow margin not exceeding 0.1-0.2mL.
- Taking averaging of the results.
- Ensuring variables are controlled including: Volume of unknown solution, Type of indicator, Concentration of standard solution, Equipment used
- Use of a control, that is, the aliquot with indicator untitrated. This would act as the ‘baseline’ which allows us to establish that the independent variable (volume of titrant/titre) is the only variable acting on the dependent variable (colour of flask solution). ie. It allow valid determination of the endpoint, thus making the investigation valid.
Justification of Indicator
- Explain why ___ is not a suitable indicator.
The equivalence point lies outside the suitable range
- Explain ___ would be a suitable indicator.
____changes colour at pH ___ to ___, which is on the vertical section of each titration curve. ie. the pH of the equivalence point lies within the pH range of the colour change for phenolphthalein
End point VS Equivalence Point
- Equivalence point — the point where the no. of moles of acid (H+) and the no. of moles of base (OH-) are stoichiometrically equivalent.
- End point — the point where a permanent colour change is observed (that is, when the no. of moles of acid/base just exceed no. of moles of base/acid in the conical flask)
- “If the indicator is chosen correctly, these two points will coincide, the indicator changing colour when the reaction is complete”
- Inflexion point/midpt of vertical section= pH of equivalence point= volume of titrant
- Strong acid and/or base= sharp increase in pH
4.13 Perform a first-hand investigation to determine the concentration of a domestic acidic substance using computer-based technologies
To prepare 250 of a 0.05 solution of (oxalic acid) and determine the concentration of acetic acid in vinegar.
- Part A – Preparation of a Dilute Vinegar Solution:
- Transfer 25mL of vinegar to a rinsed 250mL volumetric flask using a pipette.
- Dilute with distilled water to the mark using a water-rinsed glass funnel.
- Stopper the flask and invert many times.
- Part B – Titration:
- Transfer about 75mL of vinegar solution to a 100mL beaker rinsed with vinegar solution.
- Transfer 25mL of vinegar solution to a 250mL conical flask using a pipette.
- Transfer about 75mL of 0.1M NaOH solution to a 100 beaker rinsed with NaOH.
- Fill the NaOH-rinsed burette with the NaOH solution, ensuring there are no air bubbles.
- Record the initial reading of the burette.
- Add 2 drops of phenolphthalein indicator into the conical flask and place it under the burette.
- Add the solution rapidly at first while swirling the conical flask. When the pink colour disappears more slowly, add the NaOH drop wise. The titration is complete when the solution is permanently light pink.
- Repeat steps 1-7 for at least 3 concordant titres.
- Calculate the concentration of acetic acid in the undiluted vinegar
- Find n(NaOH) using n=c x Vavg of titre 1,2,3
- Find n(CH3COOH) using mole ratio
- Find c(CH3COOH)f using c= n/V(aliquot)
- Find [CH3COOH]i using civi= cfvf
- Discuss whether or not the use of a pH probe and data logger has produced a more accurate result than using an indicator.
More accurate, because eliminate the possibility of human errors including misjudgement of the colour of the indicator at the end point and parallax error. Also, provides more dps.
- Explain why the pH at the equivalence point was not equal to 7.
This is due to the basic nature of the salt formed. When NaCH3COO dissociates to Na+ and CH3COO- , CH3COO- hydrolyses to produce OH-, which in turn increases pH ie. pH > 7.
4.14 Analyse information from secondary sources to assess the use of neutralisation reactions as a safety measure or to minimise damage in accidents or chemical spills
- Neutralisation produces salt, and water, all of which are relatively harmless compared to the high toxicity of strong acids and bases which can cause: chemical burns, damage to waterways, aquatic life and structures.
- Na2CO3 is useful for acid spills ∵ cheap, powder form (easy to store/handle).
NaHCO3 is useful because it is an amphiprotic, fizzing, solid, weak base → acid and base spills, signal, easy to store/handle, buffer.
- Due to exothermic nature of neutralisation reactions, use of strong acid/base to neutralize can potentially cause will release even more heat to be and intensify the burn.
Although there are negatives associated with its use, negatives can be solved. Hence, positives outweigh negatives as there are very significant benefits to preventing harm to society and the environment.
Use of NaHCO3 for chemical spills
- Non-toxic solid (easy to store and handle) and produces harmless produces salt, H2O, CO2.
- Amphiprotic nature allows it to neutralise both acid and base spills: Acting as a base: NaHCO3 + HCl → NaCl + H2O + CO2
Acting as an acid: NaHCO3 + NaOH → Na2CO3 + H2O
- It is a weak base, which can ‘buffer’ itself if slightly too much is added
- It reacts with the acid to form CO2(g) – which can be observed as effervescence. Hence, a lack of fizzing will provide a signal for the completion of neutralisation, making harder to add too much.
Use of NaOH for chemical spills
- Not available in solid form → hard to store and handle
- Strong base → cause further damage if too much is added as it is caustic.
5.1 Describe the differences between the alkanol and alkanoic acid functional groups in carbon compounds
- Alkanols – Organic compounds containing the hydroxyl group (-OH), attach to an alkyl group. general formula: ROH, where R stands for a saturated alkyl group.
- Alkanoic Acids – Organic compounds containing the carboxyl group (-COOH), attached to an alkyl group. Their general formula is RCOOH, where R stands for a saturated alkyl group.
5.2 Explain the difference in melting point and boiling point caused by straight-chained alkanoic acid and straight-chained primary alkanol structures
Melting Point and Boiling Point Trends
- For the same number of carbons, the order of boiling points and melting points are in the ascending order: alkanes < alkanols < alkanoic acids
- The strength of intermolecular forces between molecules determines the melting and boiling point. The stronger the intermolecular force, the more energy is needed to overcome these forces, thus the higher melting/boiling point.
- For the same molecular mass, the strength of dispersion forces between molecules of each compound is relatively the same and thus the differences in melting and boiling points are due to the differences in polarity.
- Also, the longer the carbon chains, the stronger the attractions between molecules due to dispersion forces (the polar functional groups become less important in comparison) and thus the higher the boiling and melting points are, but less soluble they become.
5.3 Identify esterification as the reaction between an acid and an alkanol and describe, using equations, examples of esterification
- Esters are organic compounds containing the ester functional group (–COO–) formed via condensation reaction (between an alkanol and alkanoic acids.
- General formula: RCOOR’
- Carboxylate esters are formed when a carboxyl functional group (–COOH) of an alkanoic acid reacts with the hydroxyl functional group (–OH) of an alkanol.
- Esterification is a condensation reaction. It is the carboxylic acid (–COOH) that supplies the OH, and the alkanol (-OH) that supplies an H to produce water. It is a slow equilibrium reaction in which lies much to the left at room temperature.
|·||Concentrated sulfuric acid added during esterification serves two main purposes:|
|1.||Sulfuric acid increases the rate of reaction by lowering the activation energy, and thus the rates of the forward and|
|reverse reactions will be increased equally allowing the point of equilibrium to be reached faster.|
|2.||Sulfuric acid increases the yield of the reaction. It does this by acting as a dehydrating agent; it absorbs the water,|
|encouraging the forward reaction, shifting equilibrium to the right according to Le Chatelier’s principle.|
- Refluxing is a technique involving the condensation of vapors and the return of this condensate to the system from which it originated, this is done by a refluxing apparatus which is basically a condenser placed vertically onto a boiling flask; it cools any vapours that boil off so that they drip back into the flask.
- The need for it includes the fact that the natural production of ester is a slow process (esp. at normal SLC) and a high un-yielding process and dangerous.
- Heating the reaction flask has two main benefits:
- The higher the temperature (more kinetic energy), the faster the rate of reaction; equilibrium can be reached much faster than if it was left at room temperature.
- Also, esterification is an endothermic reaction; increasing the heat of the flask encourages the forward reaction, creating more ester.
- \Faster with greater yields. Also the products and reactants of an esterification reaction (acids, alcohols, esters) are volatile substances, and thus would escape from the container as they were heated, presenting a hazard to students and loss of reactant and produces. Reflux enables safe technique and more viable reaction.
5.6 Identify the IUPAC nomenclature for describing the esters produced by reactions of straight-chained alkanoic acids from C1 to C8 and straight-chained primary alkanols from C1 to C8
- Count the number of carbons, take the parent alkane name, drop the ‘e’ and add on an ‘-ol’.
Naming alkanoic acids
- Count the number of carbons; taking the name of the parent alkane with the same number of carbons, drop the ‘e’ and add on ‘-oic acid’.
- Exceptions include the IUPAC-preferred name for the alkanoic acids for C1 & C2: C1 ¹ methanoic acid = formic acid, C2 ¹ ethanoic acids = acetic acid
- Given the alkanol and the alkanoic acid:
- Replace ‘-anol’ with ‘-yl’. replace ‘-oic acid’ with ‘-oate’: Alkanol + alkanoic acid → alkyl alkanoate
- Given the ester molecular formula:
- Identify # of carbons in the alkanol and alkanoic acid
- Split the ester along the -COOC- bond. The side with the unsaturated CO bond is the acid.
- Replace ‘-anol’ with ‘-yl’. replace ‘-oic acid’ with ‘-oate’.
Note: Use numbering to indicate which carbon of the alcohol the ester bond is attached to, if necessary.
o Esters occur commonly in nature.
- Usually, it is a mixture of esters that creates the characteristic smells or tastes found in nature, such as those of flowers or fruit.
o Sometimes, a single ester can be identified as the main smell of a plant: o Eg. A typical ripe pineapple with contain 120 mg/Kg of ethyl ethanoate.
Production & Uses
o Many esters are industrially produced for many reasons.
o Domestic uses of esters include artificial flavourings for foods, scents for perfumes and as nail polish remover.
- Short esters such as ethyl acetate are used as industrial solvents, whereas larger esters are used as plasticisers to soften hard plastics (like PVC).
5.8 Process information from secondary sources to identify and describe the uses of esters as flavours and perfumes in processed foods and cosmetics
- Esters are industrially produced to mimic flavours and scents found in nature
- These are for use in processed foods, or mixed to produce unique perfumes.
- Domestic food flavourings are often esters dissolved in a solvent such as ethanol
- Many processed foods are flavoured artificially, e.g. banana-flavoured milk is flavoured with the ester iso-pentyl acetate.
- Cosmetics contain esters as scents, such as perfumes, with are comprised almost exclusively of a mixture of esters in a solvent, or to give soaps, hand-lotions or other cosmetics a pleasant smell.
- Other cosmetic uses of esters include as solvents for other products, such as nail polish removers.
5.9 Identify data, plan, select equipment and perform a firsthand investigation to prepare an ester using reflux
To prepare the ester butyl ethanoate by reflux from 1-butanol and ethanoic acid.
- Assemble the refluxing apparatus as shown in the diagram. Ensure that the glassware is securely clamped and supported on a retort stand. The round bottomed flask should sit in a water bath on top of a tripod with a gauze mat
- Connect tubing to the condenser and adjust the water volume to a mild, uniform flow
- Drop in a few boiling chips through the top of the condenser to allow for smooth boiling
- Mix together 25 mL of 1-butanol, 30 mL of ethanoic acid and ten drops of concentrated sulfuric acid in a 250mL beaker. Place a funnel into the top of the condenser and pour in the reaction mixture
- Start heating using a blue Bunsen burner flame and reflux for at least 30 minutes
- Stop heating and allow it to cool. Disassemble the apparatus. Pour the refluxed mixture into a separating funnel. Add 50 mL
of water and shake. Drain off the lower aqueous layer and discard. Repeat twice with two more 50 mL portions of water. Add 15 mL of saturated NaHCO3 solution and 35 mL of water. Drain off the lower aqueous layer, so that the ester remains.
- Smell the butyl ethanoate and record observations. Optional: Purify by fractional distillation (bp. =126.50C)
Justification of Method Questions
- Write the equilibrium equation that shows the esterification of pentyl ethanoate. Use structural formulae and show required conditions
1-pentanol + ethanoic acid pentyl ethanoate + water
C5H11OH(aq) + CH5COOH(aq) C6H16COOH(l) + H2O(l)
- Why was the reaction refluxed?
A reflux water condenser is arranged vertically above the reaction flask such that volatile reagents are condensed and returned to the flask. The contents can therefore be boiled for long periods without any loss of material.
- Why does acid need to be in EXCESS of alcohol ? Explain using Le Chatelier’s principle.
According to Le Châtelier’s principle the equilibrium will move so as to counteract any changes to the system. If either of the reactants is used in excess, the equilibrium will move to the right in an attempt to counteract the presence of the excess reactant. Thus the equilibrium moves to the right in favour of the ester product. The overall effect is to increase the yield.
- Why must the concentrated sulphuric acid be added slowly and with cooling?
Reaction b/w conc. sulfuric acid and water acts as an acid-base reaction which is exothermic, that is, it releases much heat energy, which can lead to spitting/splashing if large amounts of H2SO4 is added quickly. Therefore, the acid should be added slowly with cooling to minimize too much heating from adding acid.
- The esterification reaction is catalysed by H2SO4. How does the catalyst affect: (a) the equilibrium, (b) the reaction?
Sulfuric acid acts as a catalyst. It lowers the activation energy, therefore speeds up the rate of reaction, allowing the point of equilibrium to be reached faster.
Sulfuric acid increases the yield of the reaction. It does this by acting as a dehydrating agent; it absorbs the water. According to Le Chatelier’s principle the equilibrium will move so as to counteract any changes to the system. Thus the equilibrium will move to the RIGHT (forward reaction) to produce more water (AND ESTER).
- Why must the mixture be homogeneous?
Conc. sulphuric acid is much denser than any of the other reagents. Hence, the mixture must be mixed thoroughly to ensure even distribution of sulphuric acid. If not well mixed initially, the solution is liable to get too hot and boil uncontrollably when mixing occurs later in the reaction.
- Why are boiling chips used?
Provides a surface for bubbles to form. It prevents any of the liquids in the reaction mixture heating about its boiling point vigorously and consequently prevents any sudden production of vapours, which could lead to the liquid being forced explosively out of the container. i.e. Provides a surface for bubbles to form. They also prevent ‘bumping’ due to superheating and promote steady/gentle boiling. Severe bumping can lead to explosive spillage of material.
- What is the function of the water bath?
Allows even heating around surface of reaction flask. Prevents direct contact of naked flame with flammable substances (if breakage occurs). Reactants are volatile and flammable.
- What is the function of the separating funnel?
Used to separate the ester from the rest of the reaction mixture. This is more cost effective than fractional distillation as it relies the differences in densities and water solubility of the components – All the reactants are water soluble, whilst the ester is immiscible in water.
- What chemicals are left in the round bottomed flask?
After esterification: ester (insoluble) + sulphuric acid (soluble) + unreacted reactants (both soluble) + H2O.
- Why do you wash with water?
Water washings dissolve the reactants, sulphuric acid, and water, creating two immiscible layers – an aqueous layer at the bottom and the ester at the top. This allows separation and purification of the ester
- What is the function of the sodium carbonate solution?
Neutralises the unreacted acid and some of the acid catalyst in the remaining mixture. The products form a aqueous salt, water and carbon dioxide. The CO2 is able to escape the funnel, and the water + salt forms a bottom aqueuous layer immiscible with the ester layer. This increases the purity of the ester.
- Why do you need to open the tap from time to time?
To release CO2
- Why was NaHCO3 solution used to purify the ester? (Remember that esterification is an equilibrium reaction. What will NaHCO3 react with)?
This neutralizes the acid (H2SO4 + some alkanoic acid). By reacting the NaHCO3, water, soluble salt, carbon dioxide is produced. CO2 escapes and the salt and water form an aqueous layer that is immiscible with the ester, allowing for it to be separation of the layers via separating funneling