aims to increases students’ understanding of: history,...
TRANSCRIPT
Aims to increases students’ understanding of:
• History, nature and practice of chemistry
• Applications and uses of chemistry
• Implications of chemistry for society and the environment
1. Definitions: Lavoisier / Davy; Arrhenius; Brønsted-Lowry; conjugate pairs; salts – acidic/basic/neutral;amphoteric substances; neutralisation; titrations;standard solutions; pH; buffers
2. Indicators: natural, common lab-based; uses of outsidethe lab; identification of acidic, basic, neutral;Arrhenius definition
3. Acids in the air: non-metal oxides (e.g. CO2, NO,NO2, SO3) as acids; equilibria; Le Chatelier’sPrinciple; natural and industrial sources, chemicalequations; assessment of evidence; environmentalconsequences
4. Metal oxides (e.g. CuO, MgO, CaO, Fe2O3) as bases;Periodic Table; solubility;
5. Acids in Biological Systems: Brønsted-Lowrydefinitions; simple organic acids; strong/weak-dilute/concentrated distinctions; degree of ionisation– weak/strong/equilibria
Antoine Lavoisier and Humphry Davy
Lavoisier thought acids contained oxygen as lots of oxides are acidic (but many are not!)
Davy thought acids contained hydrogen
(as “replaceable hydrogen” – Hydrogen that could be replaced with a metal)
2004 and 2010 exams asks for the Davy definition of an acid
Arrhenius Definition of Acids and Bases
An acid is a substance that provides H+ ions* in aqueous solution.
e.g. HCl(g) H+(aq) + Cl–(aq)
e.g. H2SO4(l) 2 H+(aq) + SO42–(aq)
*Note do not have free H+ ions as it isattached to water to form the hydronium ion H3O+
OH
H+ H + O
H
HH
+
formation of hydronium ion
A base is a substance that providesOH– (hydroxide) ions in aqueous solution.
e.g. NaOH(s) Na+(aq) + OH–(aq)
Arrhenius definition of base
The hydronium ion becomes surroundedby water.
Therefore generally refer to ionic species as“aquated” species:
e.g. HCl(g) + H2O(l) H3O+(aq) + Cl–(aq)
For convenience commonly write:
HCl(g) + H2O(l) H+(aq) + Cl–(aq)
This applies to all ions in waterNaCl in water looks like:
The Arrhenius definition explains the properties of many acids and bases in the presence of water, but not all.
For example:Many species act as acids/bases in solvents otherthan waterAnd even in the absence of ANY SOLVENTTheir reactivity is dependent on the solvent and/orother species present
Problems with Arrhenius’s Definition
Q8 in the 2008 paper asks about Arrhenius definition of acids
Brønsted–Lowry Definition of Acids and Bases
An acid is a proton (H+) donor molecule or ion
A base is a proton acceptor molecule or ion
H
acid base
Hproton transfer
Acid–base reactions occur simultaneously
e.g. HCl(g) + NH3(g) NH4Cl(s)
NH4Cl => made up of NH4+ and Cl–
Hydrogen chloride is a Brønsted–Lowry acid as it donates aproton
Ammonia is a Brønsted–Lowry base because it accepts a proton
H
acid base
Hproton transfer
H2O(l) + NH3(g)→ NH4+(aq) + OH-(aq)
H2O(l) + NH4+(aq) → H3O+(aq) + NH3(aq)
charge is always conserved – the sum of charges on each side is the same
0 + 0 = 1 + (-1)
0 + 1 = 1 + 0
mass is always conserved – the number of atoms of each species on each side is the same
2H + 3H, O, N = N, O, 4H + H
2H + 4H, O, N = N, O, 3H + 3H
There was a question in 2002 HSC on Brønsted-Lowry acids
SummaryLavoisier - noticed that many oxides were acidic (egnitrous oxide, sulfur trioxide)
Davy - noticed many hydrides were acidic (hydrogen sulfide, hydrogen chloride)
Arrhenius - acids dissociate to H+ and bases to OH–
Brønsted/Lowry - acids are , bases are
2005 HSC had the question “Analyse how knowledge of the composition and properties of acids has led to changes in the definition of acids.”
H
acid base
H
proton transfer
conjugate base conjugate acid
Conjugate Pairs
An acid gives up a proton to form its conjugate baseA base accepts a proton to form its conjugate acidThis is what forms a BUFFER (Q9 2005 and Q6
2008 HSC)
Base ----------------------- Conjugate acid
e.g. NH3(l) + H2O(l) NH4+(aq) + OH–(aq)
Acid ----------------------Conjugate base
Conjugate base = acid – H+
Conjugate acid = base + H+
Remember charges must balance on both sides of the equation!
Acid ------------------------- Conjugate base
e.g. NH3(l) + –OCH3(aq) NH2–(aq) + CH3OH(l)
Base ------------------------- Conjugate acid
The conjugate base of a strong acid is an extremely weak base(e.g. HCl Cl–)
The conjugate base of a weak acid is a weak base(e.g. NH4
+ NH3)
The conjugate bases of extremely weak acids (e.g. water, methanol and ethanol) are very strong bases
Q7, 2009 asked about the conjugate base of HSO4
–
A Closer Look: Electron Movement
H2O HCl H3O Cl
OH H H Cl OH H
H
Cl
As Lewis structures, with all valence electrons shown:
How does the proton change positions? Curved arrows are used to show the movement of electron pairs
O-H bond is being made
H-Cl bond is being broken
A Closer Look: Electron Movement
As Lewis structures, with all valence electrons shown:
How does the proton change positions?
Which is generally written as:
N-H bond is being made
H-Cl bond is being broken
H2SO4, HCl, HBr, HI and HNO3 are strong acids.Most other acids (e.g. ethanoic acid, citric acid, NH4
+
and carbonic acid) are weak acids.
Metallic oxides and hydroxides are generally strongbases (remember Lavoisier?).Ammonia, ethanoate (acetate) and carbonate ions are examples of weak bases.
Strong acids and bases are completely ionised.Weak acids and bases are partially ionised.
Strong and Weak Acids and Bases
Ionisation of any acid (HA) in water:HA(aq) + H2O(l) H3O+(aq) + A–(aq)
The degree of ionisation = [H3O+] × 100%[HA]
Ionisation of base (B) in water:
B(aq) + H2O(l) HB+(aq) + –OH(aq)
The degree of ionisation = [–OH] × 100%[B]
[H3O+] = concentration of H3O+ produced, get from pH
[HA] = concentration as made up, get from moles of acid in volume of solution
How much does water ionize?
• Kw = 10–14
• [H3O+].[OH–] = 10–14
• At neutrality [H3O+] = [OH–]• Therefore [H3O+] = 10–7 M• pH = –log10[H3O+] = 7
KW [H3O ] [OH ]
Autoionisation of water:H2O(l) + H2O(l) ↔ H3O+(aq) + OH–(aq)
pH = –log10[H3O+]
A neutral solution has a pH of 7 (exactly)An acidic solution has a pH <7 ([H3O+] > [–OH])A basic solution has a pH >7 ([–OH] > [H3O+])
At 25 °C:[HO–] = 1.00 × 10–14 OR pOH = 14 – pH
[H3O+]
The pH scale and pH calculations
pH of strong acid/base calculated from molarityThe pH of a weak acid/base requires the degree of ionisation.Example: Calculate the pH of a 0.1 molar solution of ascorbic acid. Note the solution is 2.8% ionised.
i) Degree of ionisation = [H3O+] × 100% = 2.8%[ascorbic acid]
ii) 2.8 = [H3O+] × 100% 0.1
iii) [H3O+] = 2.8 × 10–3 molar
iv) pH = –log10[2.8 × 10–3] = 2.6
Exams generally have a question on calculation of pH.e.g. Q8 2005; Q10 2007; Q21 2008; Q21 2010 HSC exam
pH Calculations
A solution of a strong acid (100% ionised) has a lower pH value (i.e. higher [H3O+]) than the pH of a similar concentrated solution of a weak acid (not fully ionised).
See hydrochloric acid verses ascorbic acid in 2001 exam (Q20), 2002 (Q22) and hydrochloric verses ethanoic (acetic) acid in 2010 (Q21).
Be practical….
Generally the reaction of an acid and base produce a salt andwater.
e.g. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
The reaction of an acid and base to give salt and water is often referred to as a neutralisation reaction.
The reaction is exothermic, i.e. releases heat.
Buffers: Le Chatelier’s Principle
• Buffers are mixtures of a weak acid and weak base -usually conjugate pairs
• Buffer solutions “buffer” against changes in pH• Both components are weak and readily accept protons
added to the solution if they are a base or have protons todonate to added bases if they are an acid - Le Chatelier’sPrinciple
• Acetic acid & sodium acetateCH3COOH + B CH3COO- + BH+
CH3COO- + HA CH3COOH + A-
Le Chatelier’s Principle in action
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf
Natural / Biological Buffer Systems
• Important biological buffer systems– dihydrogen phosphate system – internal fluid of all cells
H2PO4–(aq) H+(aq) + HPO4
2–(aq)
– carbonic acid system – sea water and blood
CO2(g) + H2O(l) H2CO3(aq) H+(aq) + HCO3–(aq)
– pH of ocean has dropped from 8.1 to 7.9 in the last 20 years– Both bicarbonate and hydrogenphosphate are amphiprotic (a
substance that can donate or accept a proton, H+)
This was asked in 2004 and 2006 HSC
pH of Salt Solutions
• A salt solution may be acidic, basic or neutral• Depends on strength of parent acid and base• Strong Acid + Strong Base = Neutral Salt• Strong Acid + Weak Base = Acidic Salt• Weak Acid + Strong Base = Basic Salt
• HCl + NaOH → NaCl + H2O Neutral• HCl + NH3 → NH4Cl Acidic• NaOH + CH3COOH → CH3COONa + H2O Basic
Equivalence Point:• Where amounts of acid & base
sufficient to cause completeconsumption of both.
• Measure equivalence point withindicator, pH meter or by electricalconductivity.
Volumetric Analysis of Acids and Bases
Used to determine concentration of unknown acid/base
For example: When 20 mL of 0.1 M hydrochloric acid is titrated with 0.1 M sodium hydroxide (from a burette) the following conductance change is observed.
0 10 20 30mL NaOH (aq) added
Con
duct
ivity
As NaOH is added, the H+ gets used upand is effectively replaced by Na+ ions,which are much larger than H+ ions andhave a smaller conductivity.Conductance falls as the equivalence pointis reached. After the equivalence point, weare essentially adding NaOH to the NaClsolution. This adds extra ions and thereforeleads to an increase in conductance.
Reaction: H+ + Cl– + Na+ + HO– H2O + Na+ + Cl–
*Understanding of a conductivity graph was assessed in the 2001 exam.
Example of a Volumetric Acid–Base AnalysesFrom 2008 HSC exam, Q28:
• A standard solution was prepared by dissolving 1.314 g of sodiumcarbonate in water. The solution was made up to a final volume of250.0 mL
(a) Calculate the concentration of the sodium carbonate solutionSodium carbonate is Na2CO3, MW = 2 × 22.99 + 12.01 + 3 × 16.00= 106.0 g mol-1 (4 sig. fig.)
Na2CO3 = 1.314 g / 106.0 g mol-1 = 0.01240 molc = 0.01240 mol / 0.2500 L = 4.959×10-2 mol L-1
molarity =no. moles
volume (in L)
molarity is a measure of concentration
M = mV
no. moles =mass in g (that we have)
MW (molecular weight)
• This solution was used to determine the concentration of a solutionof hydrochloric acid. Four 25.00 mL samples of the acid weretitrated with the sodium carbonate solution. The average titrationvolume required to reach the end point was 23.45 mL.
(b) Write a balanced equation for the titration reaction
Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
(c) Calculate the concentration of the hydrochloric acid solution
Na2CO3 = 4.959×10-2 mol L-1 × 0.02345 L = 1.163×10-3 mol
Example of a Volumetric Acid–Base Analyses
acid + carbonate salt + carbon dioxide + water
molarity =no. moles
volume (in L)no. moles = molarity x volume (in L)
• This solution was used to determine the concentration of a solutionof hydrochloric acid. Four 25.00 mL samples of the acid weretitrated with the sodium carbonate solution. The average titrationvolume required to reach the end point was 23.45 mL.
Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
(c) Calculate the concentration of the hydrochloric acid solution
moles of HCl = 2 × 1.163×10-3 mol = 2.326×10-3 mol
[HCl] = 2.326×10-3 mol / 0.02500 L = 9.303×10-2 mol L-1
Example of a Volumetric Acid–Base Analyses
from the reaction stoichiometry, we need twice as many moles of HCl compared with Na2CO3
molarity =no. moles
volume (in L)M = m
V
(Another) Example of a Volumetric Acid–Base Analyses
From 2001 HSC exam: A household cleaning agent contains aweak base of general formula NaX. 1.00 g of this compound wasdissolved in 100.0 mL of water. A 20.0 mL sample of the solutionwas titrated with 0.1000 mol L–1 hydrochloric acid, and required24.4 mL of the acid for neutralisation. What is the molar mass ofthis base?
Step 1: write a balanced equation:
NaX(aq) + HCl(aq) NaCl(aq) + HX(aq)
• for every mole of NaX ONE mole of HCl is needed.
Step 2: calculate moles of HCl used for the titration:
= 0.1000 mol L–1 × 24.4 × 10–3 L = 2.44 × 10–3 moles (3 sig. figures)
= number of moles NaX in the 20.0 mL sample titrated.
Step 3: calculate moles of NaX in original 100 mL and therefore 1.00 g:= 100/20 (taking into account only analysed 20 mL)×2.44×10–3 = 1.22×10–2 moles
Step 4: calculate molar mass of NaX:Remember moles = 1.22 × 10–2 moles = mass (1.00 g)/molar massmolar mass = 1.00 g/1.22 × 10–2 moles = 82.0 g/mole (3 sig. figures)
Q23 in 2003, Q24 in 2005, Q21 in the 2007 and Q28 in 2010 HSC exams. For full marks, show all steps; including providing a balanced equation, volumetric analyses questions ALWAYS are assessed!
molarity =no. moles
volume (in L)no. moles = molarity x volume (in L)
we only titrated 20 mL of the original 100 mL, i.e., we only used 1/5 of the initial amount, therefore the number of moles in 1 g is 5 x our answer from above
MW x no.moles = mass in g MW =no. moles =mass in g
no. molesmass in g
MW
Indicators for Acid–Base Titrations
pH ~7 Bromothymol blue
– colour change pH 6.2–7.6
pH > 7 Phenolphthalein
– colour change pH 8.3–10
pH < 7 Methyl orange
– colour change pH 4.4–3.1
There was a question in 2007, 2009 & 2010 HSC on selection of indicators and in 2003 was to identify what indicator was used.
SO
CH3H3CHO
Br
H3C OH
BrCH3
OO
CH3
CH3
O
HO
OH
O
NN
NH3C
H3C SO
OOH
Natural indicators
• Red (Purple) Cabbage; Red cabbage containsa mixture of pigments used to indicate awide pH range. Red onion also changes frompale red in an acidic solution to green in abasic solution
• Tea; Tea is an acid-base indicator. Its colourchanges from brown in basic solution toyellow-orange in acid. This is the reasonwhy lemon juice is added to a cup of tea –not for the taste, but to reduce the intensityof the colour (called “brightening”)
• Also beetroots, cherries, many berries andgrapes contain pH sensitive pigments
Non-metal and metal oxides
CaO(s) + H2O(l) Ca(OH)2(aq) Na2O(s) + H2O(l) 2 NaOH(aq)
but …SO3(g) + H2O(l) H2SO4(aq)2 NO2(g) + H2O(l) HNO2(aq) + HNO3(aq)
Questions about the acidity/basicity of oxides was asked in 2002 and 2009 HSC