8. Acids and Bases HL

Give examples of strong bases

Ba(OH)2 Group 1 hydroxides such as: LiOH NaOH KOH RbOH

Define a strong acid and a strong base in terms of their dissociation in ions in water,

"Strong" in this context means that the acid is completely dissociated when added to water. This gives high electrical conductivity since there are lots of free moving charged particles. Strong acids react faster than weak ones since the strong have much higher concentrations of ions. Strong acids tend to be more dangerous than weak ones. "Weak" means partially dissociated -- a weak acid will only release very few protons into solution.

Define a strong base and a weak base in terms of their dissociation into ions in solution,

"Strong" in this context means that the base is completely dissociated when added to water. This gives high electrical conductivity since there are lots of free moving charged particles. Strong bases react faster than weak ones since the strong have much higher concentrations of ions. Strong bases tend to be more dangerous than weak ones. "Weak" means partially dissociated -- a weak base will only release very few hydroxide ions.

Describe the observable differences between solutions of strong and weak acids of the same concentration in terms of their pH, colour of Universal indicator, electrical conductivity and reactivity with metals and carbonates.

1. A weak acid has a lower concentration of hydrogen ions and hence a higher pH than a strong acid of the same concentration. This can be established using narrow range universal indicator paper or, preferably, a pH probe and meter. 2. A weak acid, because of its lower concentration of hydrogen ions, will be a much poorer electrical conductor than a strong acid of the same concentration. "Strong" in this context means that the acid is completely dissociated when added to water. This gives high electrical conductivity since there are lots of free moving charged particles. 3. Weak acids react more slowly with reactive metals, metal oxides, metal carbonates and metal hydrogencarbonates than strong acids of the same concentration. This is again due to a lower concentration of hydrogen ions in the weak acid since it is the hydrogen ions that are responsible for the typical chemical properties of acids. 4. Strong and weak acids can also be distinguished by measuring and comparing their enthalpies of neutralization. (vet dock inte hur..)

HL Calculate the pH of solutions of strong acids and alkalis of known concentration, in the latter case using the relationship pH + pOH = 14 (at 25 oC),

1. Calculate the pH of 0.01 mol dm-3 aqueous sodium hydroxide, NaOH(aq). NaOH(aq) → Na+(aq) + OH-(aq) (remember that NaOH is a strong alkali, totally ionised) [OH-(aq)] = 0.01 mol dm-3 pOH = -log10[OH-(aq)] = -log10(0.01) = 2 using pOH + pH = 14 2 + pH =14 therefore pH = (14 - 2) = 12 2. Calculate the pH of 0.01 mol dm-3 hydrochloric acid, HCl(aq). HCl(aq) → H+(aq) + Cl-(aq) remember the definition pH = -log10[H+(aq)] now [H+(aq)] = 0.01 mol dm-3 (note [H+(aq)] is equal to the concentration of the acid, because it is a strong acid that is completely ionized) therefore pH = -log(0.01) = - log10^-2 = 2

Give 1 example each of monprotic/monobasic, diprotic/dibasic, triprotic/tribasic

1. HCl 2. H2SO4 3. H3PO4

HL Calculate the pH of weak acid and bases in solution

1. Weak Acid Calculate the pH of 0.10 mol dm-3 CH3COOH given that Ka = 1.8 x 10^-5 mol dm-3 at 298 K. Ka = [H+] [CH3COO-] / [CH3COOH] But for this acid [H+] = [CH3COO-] and initially [CH3COO-] = 0.10 therefore Ka = [H+]2 / 0.10 = 1.8 x 10^-5 mol dm-3 [H+] = √1.8 x 10^-5 = 1.34 x 10^-3 mol dm-3 pH= -log(1.34x10^-3) pH = 2.87 2. Weak base The value of Kb for ammonia at 25 C is 1.8 x 10-5 mol dm^-3.. Calculate the pH of a 1.00 x 10^-2 mol dm-3 solution of the alkali. since [NH4+] = [OH-] and initially [NH3] = 1.00 x 10-2 therefore; Kb = [OH-]^ / [NH3] Kb = [OH-]^2 / 1.00 x 10^-2 = 1.8 x 10^-5 [OH-]^2 = (1.8 x 10^-5) x (1.00 x 10^-2) = 1.8 x 10^-7 [OH-] = √1.8 x 10^-7 = 4.24 x 10^-4 mol dm-3 pOH = - log 4.24 x 10^-4 = 3.37 using pH + pOH = 14.00 pH = 14.00 - 3.37 = 10.63

HL understand that the Lewis definition of an acid and base involves the transfer and acceptance of an electron pair, an involves the formation of a coordinate covalent bond,

A Lewis acid is a species capable of accepting an electron pair to form a coordinate covalent bond. A Lewis base is a species capable of donating a pair of electrons to form a coordinate covalent bond.

HL Define a buffer solution

A buffer solution is a solution that is resistant to changes in pH on the addition of small amounts of acid or alkali.

Define the terms acid deposition and acid rain

Acid deposition (a form of secondary pollution) refers to the processes by which acidic particles leave the atmosphere. Acid rain is a rain or any other form of precipitation that is unusually acidic, meaning that it possesses elevated levels of hydrogen ions (low pH). It can have harmful effects on plants, aquatic animals and infrastructure. The most well-known example is acid rain, but acidic substances may also be removed by snow and fog, as well as by dry processes involving gases and solid particles.

Describe the effects that acid deposition has upon the environment in terms of lakes and fish life, forests and plant life, the erosion of the landscape and buildings, and on human health, and

Acid deposition impacts the environment in five ways: 1. It affects the pH of lakes and rivers, which impacts the organisms living in them. -Impact on lakes and rivers: Acid deposition can enter aquatic ecosystems either directly (e.g. precipitation as rain) or indirectly as run-off. A low pH has a direct impact on aquatic organisms. Below pH 5.5 some species of fish, such as salmon, are killed, and algae and zooplankton, which are food for larger organisms, are depleted. Loss of some species can cause a knock on effect through the food chain adversely affecting other organisms (e.g. due to lack of prey). In addition, low pH prevents hatching of fish eggs. 2. It affects the availability of metal ions in soil, which goes on to affect nearby plant life or surface water. -The pH of soils is a key factor determining whether certain species of plants will grow. Aluminium ions are present in soil in a number of forms. At high pH aluminium forms an insoluble hydroxide, Al(OH)3. As the pH falls due to acid deposition the aluminium ions are released into solution. For example, if the acid deposition contains sulfuric acid, aluminium is released into nearby lakes and streams as soluble aluminium sulfate: 2Al(OH)3(s)+ 3H2SO4 (aq) → Al2(SO4)3 (aq) + 3H2O(l) Other ions such as magnesium and calcium, which are essential for plant growth, are washed out of the soil in a similar way. They are therefore unavailable for absorption by the roots of plants. Low pH also favours the release of excessive concentrations of metal ions such as iron and manganese, which may be toxic to plants. 3. It directly affects plants. - As well as damaging the soil, and lowering the availability of nutrients, acid deposition can damage plants directly. For example, it can damage leaf chlorophyll, turning leaves brown and reducing the photosynthetic ability of the plant. 4. It directly affects buildings and other materials. - Limestone and marble are forms of calcium carbonate. Many historical buildings are made from these materials, so they can be eroded by acid rain: Metallic structures, especially those made of iron (or steel) or aluminium, are readily attacked by acid deposition. Sulfur dioxide gas may directly attack iron as follows (an example of damage by dry deposition): Fe + SO2 + O2 → FeSO4 Or sulfuric acid may attack the iron (an example of wet deposition): Fe + H2SO4 → FeSO4 + H2 (ionically: Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)) It directly affects human health. - Acidic rainwater and moisture in the air can adversely affect the mucous membranes and lungs causing irritation and possibly exacerbating the symptoms for people with asthma and other respiratory conditions.

HL Understand the importance and methodology of the technique of acid-base titration using indicators. HL Describe how acid-base indicators are substance that change colour in response to the pH of the solution they are added to.

Acid-base indicators are substances whose colour depends on the pH of the solution (see section 8.3.2). They therefore signal a change in pH by undergoing a distinct colour change. These indicators are themselves weak acids or weak bases in which the undissociated acid (HIn) and its conjugate base (In-) have different colours. If we consider an indicator HIn that is a weak acid, it exists in equilibrium in solution as follows: HIn(aq) ⇌ H+(aq) + In-(aq) colour A colour B Applying Le Chatelier's principle to this equilibrium, we can predict how it will respond to a change in the pH. Increasing [H+] by adding acid: the equilibrium will shift to the left in favour of HIn (colour A). Decreasing [H+] by adding OH- ions: the equilibrium will shift to the right in favour of In- (colour B). In other words, at low pH colour A will dominate and at higher pH colour B will dominate. Table 1 shows the colours involved for several different indicators (a selection of indicator end-points is given in section 22 of the IB data booklet).

Describe how the pH can be determined using Universal indicator or a pH meter,

Acid-base indicators change colour reversibly according to the concentration of H+ ions in the solution. This happens because they are weak acids and bases whose conjugates have different colours (see Topic 18). The colour change means that they can be used to identify the pH of a substance. Indicators are generally used either as aqueous solutions or absorbed onto 'test paper'. The pH of aqueous solutions can be measured by using Universal Indicator, either in the form of a solution or as paper. This is actually a mixture of indicators that has different colours in solutions of different pH. The exact colours usually correspond to a 'rainbow' sequence as the pH increases. Indicators may be used to measure pH, by making use of the fact that their colour changes with pH. Visual comparison of the colour of a test solution with a standard colour chart provides a means to measure pH accurate to the nearest whole number. A more accurate method of measuring pH involves using a pH probe and meter.The electrode of the pH meter is placed in the solution to be tested an a voltage is generated that is converted into a pH meter reading, displayed on the screen. The pH meter is calibrated using buffers of known pH; usually of pH 4.0, 7.0 and 10.0.

Give examples of weak bases

All others are weak bases but specifically mentioned is: NH3 Amine - R-NH2 Weak bases are composed of molecules that react with water molecules to release hydroxide ions. In general for a weak molecular base, BOH: BOH + (aq) ⇌ OH-(aq) + B+(aq) An equilibrium is established, with the majority of the base molecules not undergoing ionization or dissociation. In other words, the equilibrium lies on the left-hand side of the equation.

HL Define the expressions for the acid dissociation constant, Ka, and base dissociation constant, Kb , for weak acids and bases; and also the associated terms pKa and pKb - Kb

Ammonia is an example of a weak base that partially ionizes in water: NH3(g) + H2O(l) ⇌ NH4+(aq) + OH-(aq) In a similar argument to that used above for acids, we can consider the ionization of a weak base using the generic weak base B (which would have a lone pair of electrons, B: ) B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq) and derive an expression Kb = [BH+] [OH-] / [B] Kb is known as the base dissociation constant. The higher the value of Kb at a particular temperature, the greater the ionization and so the stronger the base. The value Kb is constant for a specific base, at a specified temperature. Its value give a measure of the strength of the base.

Understand that amphiprotic species are part of a broader group of substances that are amphoteric - they can react with both acids and bases to produce a salt and water,

Amphiprotic specifically relates to Bronsted-Lowry acid-base theory, with its emphasis on the transfer of a proton. The term amphoteric, on the other hand, has a broader meaning as it is used to describe any substance which can act as an acid or as a base, including reactions that do not involve the transfer of a proton. For example, aluminium oxide is an amphoteric oxide as it reacts with both dilute acids and alkalis. We will see in Topic 18 that this acid-base behaviour is best described by a different theory, the Lewis theory. Note that all amphiprotic substances are also amphoteric, but the converse is not true.

HL What is an alternative approach to using indicators for pH? tror inte vi måste kunna men good att keep in mind

An alternative approach to titration avoids the use of indicators and follows the changes in conductivity as an acid reacts with an alkali (Figure 11). As the acid and base react the conductivity of the solution falls as ions are removed to form water molecules. A V-shaped graph is obtained whose trough corresponds to the volume of acid (or alkali) required to achieve neutralization. Weak bases and acids, of course, have shallower gradients than strong acids or bases, and finish at lower points on conductivity graphs since their concentrations of ions are relatively smaller.

HL Characteristics of Salts of weak acids/strong bases

An example of hydrolysis by the salt of a weak acid and a strong base is that of sodium ethanoate. The ethanoate ion (CH3COO-) is the conjugate base of the parent acid. When the acid is weak this conjugate base is strong enough to cause hydrolysis: CH3COO-(aq) + H2O(l) ⇌ CH3COOH(aq) + OH-(aq) The release of OH- ions causes the pH of the solution to increase.

What are anacids'+

Antacid tablets can ease the symptoms of excess stomach acid that causes indigestion and heartburn. The active ingredients of most of these antacids tablets are metal carbonates or hydrogen carbonates - NaHCO3, CaCO3, MgCO3, for instance - or insoluble metal hydroxides - usually Mg(OH)2 or Al(OH)3.

identify molecules and ions which are acids, bases (alkalis) and amphiprotic species,

BL Acid: To act as a Brønsted-Lowry acid, they must be able to dissociate and release H+. BL Base: To act as a Brønsted-Lowry base, they must be able to accept H+, which means they must have a lone pair of electrons. Amphiprotic features must include: - To act as a Brønsted-Lowry acid, they must be able to dissociate and release H+. -To act as a Brønsted-Lowry base, they must be able to accept H+, which means they must have a lone pair of electrons. So, in summary, substances that are amphiprotic according to Brønsted-Lowry theory must possess both a lone pair of electrons and hydrogen that can be released as H+.

Examiner tip

Be careful of the technically-correct names for nitrous acid (HNO2) and sulfurous acid (H2SO3). They are nitric(III) acid and sulfuric(IV) acid, respectively. Look carefully at the oxidation number of N and S in each case.

Define an acid as a proton (H+) donor and a base as a proton acceptor (Brønsted-Lowry theory),

Brønsted-Lowry acid: Proton (H+) donor Brønsted-Lowry base: Proton (H+) acceptor

Outline and be familiar with the major acids, bases and alkalis Give examples of bases and alkalils

CaO CuO NaOH Ca(OH)2 NH3

Give examples of weak acids

Carbonic acid - H2CO3 Ethanoic acid methanoic acid R-COOH - carboxylic acid In writing the equations for the ionization reactions of weak acids and bases, it is essential to use the equilibrium sign (⇌).

HL Understand that a buffer solution consists of a solution of a weak acid or base and a solution of a salt containing its conjugate base or acid. -Composition of an acidic buffer HL Describe how an acidic or basic buffer is able to resist a change in pH due to the addition of small amounts of extra H+ or OH- ions.

Composition of an acidic buffer - Such a buffer is made by mixing an aqueous solution of a weak acid with a solution of its salt of a strong alkali. For example: mix equimolar solutions of ethanoic acid and sodium ethanoate. - ethanoic acid: CH3COOH ⇌ CH3COO- + H+ equilibrium to the left sodium ethanoate: Na+ and CH3COO- fully ionised So the mixture contains relatively high concentrations of both CH3COOH and CH3COO- , that is the acid and its conjugate base. These can be considered as 'reservoirs', ready to react with any added OH- and H+ respectively. added acid (H+ ions): H+ will combine with the base CH3COO- to form CH3COOH, therefore removing most of the added H+ and resisting any change of pH. added alkali (OH- ions): OH- will combine with the hydrogen ions to form water molecules, removing H+ ions from the equilibrium; the equilibrium responds by shifting to the right, ethanoic acid molecules dissociate to restore the H+ ion concentration (Le Chatelier) resisting any change in pH.

HL Describe how different indicators are appropriate for particular titrations of acid-base while others are not suitable for use with certain titrations.

Figures 5 and 7 show that both phenolphthalein and bromothymol blue would be appropriate indicators for titrations between strong acid and alkali. In fact the size of pH change around the equivalence point in such titrations is sufficiently great that several different indicators could be used here. Figure 8 emphasises that methyl orange would also be appropriate even though it changes colour over a very different range to phenophthalein. Looking at the titration curves for other acid-base combinations it's possible to make a judgement as to which are suitable indicators to use for a given titration. Figure 9 shows clearly that while methyl orange would be appropriate for a titration involving a strong with a weak alkali, phenolphthalein would not. Phenolphthalein would not give a sharp end point in this case as the pH range over which it changes colour does not fall in the vertical region of the curve.

Understand that some species can both donate and accept a proton depending on the circumstances of the reaction - they can be amphiprotic,

For example, water can act as both a Brønsted-Lowry acid and base depending on the other species it was reacting with. You'll remember that we commented that, in this, water was behaving as an amphiprotic species. You are probably not used to thinking of water as an acid, or indeed as a base. In the context of pH we emphasise water as a neutral substance. The point here is that Brønsted-Lowry theory describes acids and bases in terms of how they react together, so it all depends on what water is reacting with. Consider the following : CH3COOH(l) + H2O(l) → CH3COO-(aq) + H3O+(aq) CH3NH2(l) + H2O(l) → CH3NH3+(aq) + OH-(aq) In the first reaction with ethanoic acid, water is accepting a proton: that is behaving as a base. In the second reaction with methylamine, water is donating a proton: acting as an acid.

HL Describe how the pH of the solution changes during the titration of a strong acid and a strong base, noting the key regions of the curve obtained and the significance of the equivalence point.

From the values shown (de finns under "titration curves and indicators") you can see that adding a substantial volume of alkali (25.00 cm3) to the acid produces only a small change in pH. In this case the equivalence point is reached when 50.00 cm3 of alkali has been added (the acid and alkali have the same molarity). Importantly, do note that the addition of a relatively small amount of alkali past the equivalence point produces a dramatic increase in pH. The curve shows that the initial pH is low as this is a strong acid. As base is added, the increase in pH is at first very gradual, so that even when the mixture is only 1.0 cm3 away from equivalence it is still a long way below pH 7. A small addition of base around the equivalence point causes a dramatic rise in pH, with an increase of about eight units from pH 3 to 11. It is found that in most titrations involve this large jump in pH around equivalence, and this is known as the point of inflection. The equivalence point is determined as being halfway up this jump. At the equivalence point, the acid and base have exactly neutralized each other, so the solution contains salt and water only. Remember that the pH of this solution depends on hydrolysis by ions in the salt, determined by the relative strengths of the parent acid and base (section 18.3.2). The equivalence point is only at pH 7.00 in the titration of strong acids against a strong base.

HL How do you prepare a basic or acidic buffer?

Generally buffer solutions can be prepared either by mixing a weak acid or base with an equimolar solution of a salt containing its conjugate. An alternative approach to making an acidic buffer solution is to start with a weak acid and add half as many moles of strong base. A basic buffer can also be prepared by starting with a weak base and adding half as many moles of strong acid.

Give 5 examples of amphiprotic species

H2O hydrogen carbonate ion HCO3^- hydrogen sulfate ion HSO4^- dihydrogen phosphate H2PO4^- hydrogen phosphate HPO4^2-

Outline and be familiar with the major acids, bases and alkalis Give examples of acids

HCl HNO3 H2S04 CH3COOH C6H5COOH

Give examples of strong acids

HCl HNO3 H2SO4 HClO4

HL Manipulate the values of concentration, pH, pOH, Ka, pKa , Kb , and pKb to calculate various unknown parameters relating to solutions of acids and bases.

Here are some other examples based on this type of calculation for you to try so that you become practised at manipulating the different parameters involved. 1. An acid has a Ka of 4.1 x 10-6 mol dm-3 and a pH of 4.50. Calculate the concentration of the acid solution. pH= 4.50 [H+] =10^-pH [H+]= 10^-4.50 = 3.16 x10^-5 Ka = 4.1 x10^-6 Ka = [H+]^2 / [Acid solution] So, 4.1 x10^-6 = (3.16 x10^-5)^2 /[Acid solution] [Acid solution] = (3.16 x10^-5)^2 /(4.1 x10^-6) [Acid solution] ≈ 2.44 x 10^-4 2. The pH of a 0.20 mol dm-3 solution of a weak acid is 3.90. Calculate the Ka of this acid. pH= 3.90 [H+] = 10^-3.90 Ka = [H+]^2 / [weak acid] Ka= (10^-3.9)^2 / 0.20 Ka = 7.92 x10^-8 3. The pH of a solution of a weak base of concentration 3.0 x 10-2 mol dm-3 is 10.00. Calculate the pKb value of the base. pH + pOH = 14 10 + pOH = 14 pOH = 4 [OH-] = 10^-4 Kb = [OH-]^2 / 3.0 x 10^-2 Kb = (10^-4)^2 / 3.8 x10^-2 Kb = 10^-8 / 3.8 x10^-2 Kb = 3.33 x 10^-7 pKb = -log(Kb) = -log(3.33 x10^-7) pKb = 6.48 4.The pKb value of methylamine is 10.66. Calculate the concentration of an aqueous solution of methylamine that has a pH of 10.80. Kom så här långt vet ej svaret pkb = 14 - 10.66 = 3.34 pH = 10.80 pH + pOH = 14 pOH = 3.2 [OH-] = 10^-pOH [OH-] = 10^-3.2 = 6.31 x10^-4 Kb = 10^-pkb kb = 10^-3.34 Kb = [OH-]^2 /[base solution] 2.19 x10^-11= (6.31 x10^-4)^2 /[bs] [bs] = (6.31 x10^-4)^2 / ( 10^-3.34) [bs] =8.71 x 10^-4 (so you can check how you are doing the answers to these questions are: 2.44 x 10-4 mol dm-3; 7.92 x 10-7mol dm-3; 6.48; 8.71 x 10-4 mol dm-3)

HL Characteristics of Salts of strong acids/weak bases

Here the cation in the salt is a conjugate of the parent base. When the base is weak and this conjugate acid is a non-metal (e.g. NH4+), it is able to hydrolyse water: NH4+(aq) + H2O(l) ⇌ NH4OH(aq) + H+(aq) The release of H+ ions causes the pH of the solution to decrease. If the cation is a metal ion with a high charge density, Fe3+, Al3+ or Cu2+ at the centre of a hydrated complex for instance, then the solution will be acidic as the result of hydrolysis of a ligand water molecule

What is hydrolysis? google def för ingen hade.

Hydrolysis is a reaction involving the breaking of a bond in a molecule using water. The reaction mainly occurs between an ion and water molecules and often changes the pH of a solution.

Question Which of the following statements is/are true about acids and bases? I. The presence of oxide ions makes a sodium hydroxide solution alkaline II. Hydrochloric acid solution is acidic because it contains hydronium ions III. Sulfuric acid is acidic because it contains an excess of H- ions IV. Aqueous ammonia solution is alkaline because of the presence of an excess of OH- ions

II and IV

Understand that each Brønsted-Lowry acid has a conjugate base, and vice versa, with the two being linked by the exchange of an H+ ion

If a reactant is linked to a product by the transfer of a proton we call this pair a conjugate pair. The acid and base in a conjugate acid-base pair differ by just one proton. The fact that in a conjugate pair the acid always has one proton more than its conjugate base, makes it easy to predict the formula of the corresponding conjugate for any given acid or base. When writing the formula of the conjugate acid of a base, simply add one H+; when writing the conjugate base of an acid, simply remove one H+. Remember to adjust the charge by the +1 removed or added. A conjugate ACID is made by ADDING a proton (H+). So check the equation and see what product has had a proton added -- it's the conjugate acid. The conjugate base is the other product, which has had a proton removed.

What are the strengths of the strong and weak acids/bases respective conjugate.

In general, weak acids produce relatively strong conjugate bases in aqueous solutions and so the equilibrium will lie to the left (the lower line in Figure 1). Conversely, a stronger acid will ionize more and the equilibrium produced will lie to the right; the conjugate base in this case being relatively weak. Similarly, strong bases produce weak conjugate acids in aqueous solutions and weak bases produce strong conjugate acids in aqueous solution

Exmaniner tip

Indicators are weak acids or bases whose components of the conjugate acid-base pair have different colours; this conjugate acid-base pair exists in equilibrium. A list of some common acid-base indicators are given in section 22 of the data booklet.

Worked example A sample solution of sodium hydroxide contains 0.1 moles of NaOH per dm-3 . What is the pH of the solution?

It has a concentration of 0.1 mol dm-3 NaOH Sodium hydroxide is a strong alkali - it is completely ionised. So [OH-] ions = 10-1 mol dm-3 But, for an aqueous solution, the ionic product of water is constant. Kw = [H+][OH-]= 10^-14 - the equilibrium constant is products/reactants. As H2O is in great excess it can be excluded from the equation. so [H+]=10^−14/[OH−]=10^−14/10^−1=10−13 mol dm−3 pH = - log[H+] = - log 10^-13 = 13 Therefore: a sodium hydroxide solution of concentration 0.1 mol dm-3 has a pH of 13.

HL Important

It's important that you don't confuse the terms 'equivalence point' and 'end-point'. The equivalence point is where stoichiometricaliy equal amounts of acid and base have neutralised each other. The end-point is the pH at which the indicator changes colour. A titration is set up so that the pH of these two points coincides and the choice of an indicator that gives a sharp end point is important in this.

Describe the difference between the acidity of natural rainfall and the acid rain generated by anthropogenic activity - inte helt klar förklaring får återkomma och fixa

Natural rain water is acidic, with a pH of around 5.6. The acidity of rainwater comes mostly from the natural presence of CO2 found in the troposphere (the lowest layer of the atmosphere).This acidity arises from the reaction of carbon dioxide and water in the atmosphere. Carbon dioxide dissolved in water is sometimes referred to as carbonic acid, H2CO3. H2O + CO2 --> H2SO3 However, Nitric oxide (NO), also contributes to the natural acidity of rainwater, and is formed during lightning storms by the reaction of nitrogen and oxygen, the two predominant atmospheric gases. In air, NO is oxidized to nitrogen dioxide (NO2), which in turn reacts with water to give nitric acid (HNO3). Also, Sulfur dioxide, SO2, occurs naturally in the air from volcanic emissions (Figure 1) and through a series of reactions with water to produce sulfurous and sulfuric acids (H2SO3 and H2SO4). The most important sources of acid deposition are the sulfur dioxide and nitrogen oxides (primary pollutants) of anthropogenic origin (produced by human activities). Sulfur dioxide is produced in power stations and the industrial smelting of sulfide ores. When sulfur dioxide dissolves and reacts in rain water a solution of sulfurous acid is formed. Nitrogen oxides also contribute to acid deposition as primary pollutants. Nitrogen monoxide (NO), formed in vehicle and aviation jet engines, combines in the atmosphere with a hydroxyl radical to form nitrous acid (nitric(III) acid), HNO2. den här förklaringen är onödigt komplicerad eller dåligt structured kolla på denna länk om du inte förstår hehe och snälla hjälp mig fixa gav upp på denna. https://www3.epa.gov/acidrain/education/site_students/whatcauses.html

Define what is meant by a neutralization reaction and be able to state the products formed by particular combinations of acid and base, and

Neutralization reactions occur when an acid and base react together to form a salt and water only. They are exothermic reactions. Since a base is a metal oxide or hydroxide we can also usefully generalise, and represent any neutralization reaction with one or other of the following ionic equations. H+(aq) + OH-(aq) → H2O(l) 2H+(aq) + O2-(aq) → H2O(l)

Understand the usefulness of acid-base titrations as both a method of preparing a salt and an analytical procedure.

Neutralization reactions of acids with bases are exothermic - heat is released. The enthalpy of neutralization is defined as the enthalpy change that occurs when an acid and a base react together to form one mole of water. Neutralization reactions are often used in the laboratory to calculate the exact concentration of an acid or an alkali when the other concentration is known. The technique known as acid-base titration involves reacting together a carefully measured volume of one of the solutions (step a in Figure 3), and adding the other solution gradually until the so-called equivalence point is reached where they exactly neutralize each other (step b in Figure 3). As both solutions are colourless we need to use an indicator to determine when the equivalence point has been reached. The indicator chosen must change colour as the acid and base exactly neutralize each other.The titration is carried out until the indicator just changes colour. Then the titration is repeated several times to collect reproducible results. The salt produced in the reaction can be prepared and crystallised from the reaction mixture if the titration is repeated without the indication once the end point has been determined.

HL Understand that the range of Lewis acid-base reactions also includes organic reactions involving nucleophiles (Lewis bases) and electrophiles (Lewis acids) and that the 'curly arrow' annotation organic reaction mechanisms shows the type of transfer of electron pairs envisaged by the Lewis definitions.

Organic compounds are predominantly covalent and many organic reactions result from a process of breaking covalent bonds and forming new ones. Reaction mechanisms, such as nucleophilic (lewis base) or electrophilic (Lewis acid) substitutions, also involve electron pair transfers and can therefore described by the Lewis acid-base theory. In describing organic reaction mechanisms we often use a system of 'curly arrows' to depict the movement of electron pairs in the breaking and formation of bonds. The arrow directly represents this movement and so shows which molecular species is acting as a Lewis acid or base.

Describe how the gases involved in the production of acid rain are generated by our use of fossil fuels and their combustion - warning one million equations coming your way Describe, including summarizing equations, how these oxides produce the acids involved,

SULPHUR Coal and oil contain sulphur, meaning that they release sulphur when they are being burned. sulfur remains a contaminating presence in coal and fuel oil used in power stations. Sulfur (in coal or fuel oil) is oxidized during the combustion process in the power station and forms sulfur dioxide. Sulphurous acid S(s) + O2(g) -->SO2(g) SO2(g) + H2O (l) --> H2S03 (aq) OR Sulphuric acid S(s) + O2(g) -->SO2(g) SO2 (g)+ 1/2 O2 --> SO3 SO3(g) + H2O (l) --> H2S04 (aq) NITROGEN This reaction takes place in the both vehicle petrol and diesel engines, though the level of pollution is greater from diesel engines as their operating temperature is higher. The use of a catalytic converter (catalyseur) in vehicle exhaust systems reduces the level of these emissions. This reaction also takes place in the jet engines of aircraft. If you heat and squash air for example in an internal combustion engine, it will react. N2 (g) + O2(g) --> 2NO (g) N2 (g) 2O2(g) --> 2NO2(g) Nitric acid 2NO (g) +O2(g) --> 2NO2 (g) H2O(l) + 4NO2(g) + O2(g) --> 4HN03 (aq) Nitrous acid 2NO2 (g) + H2O(l) --> HNO3 (aq) + HNO2 (aq) These four acids (sulfuric, sulfurous, nitric and nitrous acids) are much stronger than carbonic acid, H2CO3. They dissociate to a much greater extent (almost 100% in the case of nitric and sulfuric acids) and may lead to a rainwater pH as low as 2. This represents a concentration of H3O+(aq) ions over 1000 times greater than in normal rainwater.

HL Define the expressions for the acid dissociation constant, Ka, and base dissociation constant, Kb , for weak acids and bases; and also the associated terms pKa and pKb - pKa and pKb

So, as with the concentrations of H+ and OH- ions and Kw, it is useful to convert Ka and Kb values into their negative logarithms (to the base 10) and use these values for comparisons (see Tables 1a and 1b). These values are known as pKa and pKb. pKa = - logKa pKb = - logKb - pKa and pKb numbers are usually positive and have no units - the relationship between Ka and pKa and between Kb and pKb is inverse - a change of one unit in pKa or pKb represents a 10 fold change in the value of Ka or Kb (look at the values for methanoic acid and ethanoic acid; also remember that this is a feature of the relationship between pH and [H+]), and note that pKa and pKb must be quoted at a specific temperature. Make sure that you are aware that a higher pKa value means a weaker acid and a higher pKb value means a weaker base. Remember that the values for pKa are given in section 21 of the data booklet.

Good to know..

The half-equivalence point in the graph as exactly half of the acid has been neutralized and converted into salt, while the other half of the acid in the flask has not yet reacted. This mixture, having equal quantities of a weak acid and its salt, is therefore a buffer. This explains why the pH in this region is relatively resistant to change in pH on the addition of small amounts of base. The pH at the half-equivalence point gives us an easy way to calculate pKa because at this point [acid] = [salt], we can substitute these values into the equilibrium expression of the acid: Ka = [H+][A-] / [HA] assuming that [acid] = [HA] and [salt] = [A-], then [HA] = [A-] and these terms can be cancelled, and: Ka = [H+] and therefore pKa = pH The pH at the half-equivalence point can be read directly from the graph, and so the pKa of the acid can be deduced.

HL Understand that the weaker the acid the lower the value of Ka but the higher the value of pKa ; and that a similar relationship exists between base strength and the values of Kb and pKb ,

The higher the value of Ka at a particular temperature, the greater the dissociation, and so the stronger the acid. The higher the value of Kb at a particular temperature, the greater the ionization and so the stronger the base. Make sure that you are aware that a higher pKa value means a weaker acid and a higher pKb value means a weaker base. Remember that the values for pKa are given in section 21 of the data booklet.

HL Describe how the salts of strong acid/base reactions have a pH of around 7.0 in aqueous solution but those of weak acid-base combinations have pH values above or below 7.0 due to salt hydrolysis.

The reaction between a strong acid and a strong base forms a salt in which both the conjugate acid and the conjugate base are weak. As a result there is virtually no hydrolysis of water by the ions of the salt, and the pH is close to neutral. However, some salts dissolve in water to form either acidic or alkaline solutions. This is because one of the ions reacts with the water to release an excess of either hydroxide or hydrogen ions. This phenomenon is called salt hydrolysis and occurs when the salts are formed from weak acids or weak bases. The pH of such salts in solution depends on whether and to what extent their ions, which are conjugate acids and bases, react with water and hydrolyse it, releasing H+ or OH- ions. strong acid + strong base - Neutral - neutral approx. 7.0 weak acid + strong base - basic - > 7 strong acid + weak base - acidic - < 7 weak acid + weak base - depends on strengths of conjugates - cannot generlize

How are the terms acid and base relative?

The terms 'acid' and 'base' are also relative terms. If two concentrated acids are reacted together, then the weaker acid of the two will be 'forced' to act as a base. For example, when concentrated nitric and sulfuric acids are reacted together in a 1 : 2 molar ratio, a so-called nitrating mixture is formed which contains a cation known as the nitronium ion, NO2+. This cation is involved in the nitration of aromatic organic compounds such as benzene

difference between weak/strong and dilute/concentrated?

The terms strong and weak refer to how an acid or alkali ionizes when it dissolves in water. The terms dilute and concentrated refer to the proportions of acid or alkali and water in the solution. Therefore, a 0.1 mol dm-3 solution of hydrochloric acid can be described as a dilute solution of a strong acid and a 0.1 mol dm-3 solution of ethanoic acid can be described as a dilute solution of a weak acid.

HL Calculate the pH of the solution at any given point during the titration of a strong acid with a strong alkali.

The values shown in the table are for the titration of 50 cm3 of 0.10 mol dm-3 HCl(aq) with a 0.10 mol dm-3 NaOH solution. In this case the acid is in the flask, and the alkali added from the burette. First find concentration and then pH moles of solute / volume (dm^3)of solution Volume of alkali added 0.00 cm3 Calculation [H+] = 0.1 mol dm3 = 1.0 x 10-1 pH = 1.00 25.00 cm3 initial mol acid = 0.10 x 0.050 = 0.0050 mol mol base added = 0.10 x 0.025 = 0.0025 mol so mol H+ = 0.0050 - 0.0025 = 0.0025 mol total volume = 0.075 dm3 so [H+] = 0.0333 pH = 1.50 50.00 cm3 all of acid has been neutralised by alkali to give NaCl solution pH = 7.00 51.00 cm3 mol base added = 0.10 x 0.051 = 0.00510 mol mol base remaining = 0.0051 - 0.0050 = 0.0001 mol mol OH- = 0.0001 volume = 0.101 dm3 [OH-] = 0.00099 so pOH = 3 pH = 11

Describe both pre-combustion and post-combustion methods that are in place to prevent and remedy the effects of acid deposition.

There are three methods by which sulfur dioxide emissions from power stations can be limited: 1. The coal or oil can be refined to remove the sulfur before combustion. This is what was done effectively in the refining of gasoline (petrol) and diesel for use in road vehicles. 2. Fluidized bed combustion (FBC) is a method by which the amount of sulfur oxides resulting from combustion is lowered. - A fluidized bed suspends the solid coal on an upward flowing jet of air during combustion. Coal dust is mixed with limestone powder (calcium carbonate, CaCO3) and blasted into the furnace with a jet of air. The jet of air suspends the solid particles so they flow like a fluid. At the high temperatures of the furnace, calcium carbonate decomposes to form calcium oxide: CaCO3 → CaO + CO2 The calcium oxide neutralizes the sulfur dioxide as it is formed. Further oxidation with oxygen produces particles of calcium sulfate: 2CaO + 2SO2 + O2 → 2CaSO4 The calcium sulfate must be removed by electrostatic precipitation. 3. Flue gas desulfurization (FGD) (Figure 2) removes sulfur dioxide from the exhaust gases before they leave the power station flue (chimney). - Sulfur dioxide emissions can be removed from the flue gases by passing the gases into a flue gas desulfurization tower (a 'scrubber': Figure 1). In the tower the gases are passed through a sprayed suspension of calcium carbonate and calcium oxide in water. The product is calcium sulphite (sulfate(IV)). CaCO3 + SO2 → CaSO3 + CO2 CaO + SO2 → CaSO3 The calcium sulfite (calcium sulfate(IV)) is then further oxidized, producing calcium sulfate (gypsum): 2CaSO3 + O2 → 2CaSO4 The first method is a pre-combustion method while the second and third methods are post-combustion methods for reducing sulphur dioxide emissions.

HL Understand the significance of the ionic product of water, Kw, as the basis of the measurement of pH and pOH in aqueous solutions,

There is an important relationship worth remembering - and that can be worked out by taking negative logs of both sides of the expression for Kw Kw = [H+(aq)] [OH-(aq)] = 1.0 x 10-14 mol2 dm-6 at 298 K - log10 Kw = -log10[H+(aq)] + (- log10[OH-(aq)]) = -log10 10-14 or pKw = pH + pOH = 14 (at 25 oC or 298 K)

Understand that acid deposition is an international problem as it can be spread by the prevailing winds to regions that were not involved in the generation of the problem

These acids - derived from sulfur dioxide and nitrogen oxides - may be deposited not only in rain water, but also in snow and fog. Fog is a particular problem for high-altitude forests. The lower temperature at high altitudes causes water vapour to condense out of the atmosphere, forming a moist 'blanket' of acidic fog which surrounds trees and results in heir lengthy exposure to adverse conditions. In dry deposition polluting particles and gases stick to the ground via dust and smoke in the absence of precipitation. This form of deposition is dangerous however because precipitation can eventually wash pollutants into streams, lakes, and rivers.

HL understand that this definition of acid-base behaviour broadens our understanding of such behaviour to include reactions that do not necessarily involve proton transfer,

This Lewis definition is an extension of the earlier one in that it takes the requirement of a base having a lone pair of electrons and stresses that the base donates that lone pair to an acid. The Lewis definitions of acids and bases are much broader and include a whole range of compounds that don't transfer hydrogen ions (protons). Some reactions are classified as acid-base reactions under the Lewis definitions that are not regarded as acid-base reactions under the Brønsted-Lowry theory. Therefore the terms Lewis acid and Lewis base are often reserved for species which are Lewis acids and bases but not Brønsted-Lowry acids and bases.

HL Understand that a buffer solution consists of a solution of a weak acid or base and a solution of a salt containing its conjugate base or acid. -Composition of an basic buffer HL Describe how an acidic or basic buffer is able to resist a change in pH due to the addition of small amounts of extra H+ or OH- ions.

This buffer can be made by mixing an aqueous solution of a weak base with a solution of its salt with a strong acid. For example: mix equimolar solutions of ammonia and ammonium chloride. ammonia solution: NH3 + H2O ⇌ NH4+ + OH- equilibrium to the left ammonium chloride: NH4+ and Cl- fully ionised The mixture contains relatively high concentrations of both NH3 and NH4+ , that is the base and its conjugate acid. These can be considered as 'reservoirs', ready to react with any added H+ and OH- respectively. added acid (H+ ions): H+ will combine with the hydroxide ions to form water molecules, removing OH- ions from the equilibrium; the equilibrium responds by shifting to the right, ammonia molecules dissociate to restore the OH- ion concentration (Le Chatelier) resisting any change in pH. added alkali (OH- ions): OH- will combine with the acid NH4+ to form NH3 and H2O, therefore removing most of the added OH- and resisting any change of pH.

HL Define the expressions for the acid dissociation constant, Ka, and base dissociation constant, Kb , for weak acids and bases; and also the associated terms pKa and pKb - Ka

We have seen the equilibrium reaction for a particular example of a weak acid; let's just consider the generic weak acid, HA, dissociating in water: HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) Kc = [H3O+] [A-] / [HA] [H2O] However, the [H2O] is very large and can be considered constant. Therefore, this expression becomes Kc [H2O] = [H3O+] [A-] / [HA] or Ka = [H3O+] [A-] / [HA] Ka is known as the acid dissociation constant (or acid ionization constant). The higher the value of Ka at a particular temperature, the greater the dissociation, and so the stronger the acid. The values of Ka is constant for a specific acid, at a specified temperature. Its value give a measure of the strength of the acid.

HL Describe the behaviour of weak acids and bases in terms of the partial dissociation of their molecules into ions in aqueous solution,

Weak acids and bases, unlike strong acids and alkalis are not completely dissociated into ions in an aqueous solution This means we cannot deduce the concentrations of ions in their solutions from the initial concentrations, as the ion concentrations will depend on the extent of dissociation that has occurred.

Understand the difference between wet and dry deposition

Wet deposition e.g. when pollutants are incorporated into the clouds or falling raindrops and result in acidified rain or snow. Wet deposition CaCO3(s) + H2SO4 (aq)--> CaSO4 (aq/s) +H2O(l) +CO2(g) CaCO3(s) + 2HNO3 (aq)--> Ca(NO3) (aq) +H2O(l) +CO2(g) Dry deposition, when atmospheric pollutants are removed by gravity or direct contact under dry conditions e.g. when emissions of ash or dry particles from power stations are absorbed directly onto plants and buildings. 2CaCO3(s) +2SO2(g) + O2(g)

HL Characteristics of Salts of weak acids/weak bases

When a weak acid reacts with a weak base, they form a salt in which both the conjugates are relatively strong and carry out hydrolysis. The pH of the solution therefore depends on the relative Ka and Kb values of the acids and bases involved. If the parent acid and base have approximately the same degree of weakness then the salt produced may well have a pH around 7.0.

HL Examiner tip

When presented with a pH curve, ensure that you are able to find the pH at the equivalence point, and also be able to explain why the pH changes so rapidly at this point.

As the temperature increases, does the pH of water increase or decrease?

When water ionises into H+ ions and hydroxide ions, a bond is being broken so energy must be put in, i.e. it is endothermic. At a higher temperature, this forward direction of the equilibrium will be favoured, so there will be more H+ ions present in solution and the pH decreases.

HL Understand that the value of Kw increases with temperature as the dissociation of water molecules into ions is endothermic,

When water ionises into H+ ions and hydroxide ions, a bond is being broken so energy must be put in, i.e. it is endothermic. At a higher temperature, this forward direction of the equilibrium will be favoured, so there will be more H+ ions present in solution and the pH decreases. That is, when the temperature increases then [H+] increases and the pH decreases. At 50 C the pH of pure water is 6.77; while at 10oC it is 7.27. So, Kw increases with temperature as the dissociation of water molecules into ions is endothermic, Kw = [H+(aq)] [OH-(aq)] = 1.0 x 10-14 mol2 dm-6 at 298 K, In other words, the pH of pure water is 7.00 only when the temperature is 298 K. Be careful though, at temperatures above and below this, despite changes in the pH value, water is still a neutral substance as its H+(aq) concentration is equal to its OH-(aq) concentration. It does not become acidic or basic as we heat it or cool it!

Define the ionic product of water, Kw = [H+] [OH-] = 1.00 x 10-14 mol dm-3 at 298 K, and be able to use it to calculate the pH of an alkali solution, and

When water is purified by repeated distillation its electrical conductivity falls to a very low, constant value. This is evidence that pure water dissociates to a very small extent to form ions: H2O(l) ⇌ H+(aq) + OH-(aq) If the equilibrium law is applied to the first equation: Kw = [H+(aq)] × [OH-(aq)] where Kw represents a constant known as the ionic product constant of water. At 298 K (25˚C) we find that the measured concentrations of H+(aq) and OH-(aq) in pure water are 1.0 × 10-7 mol dm-3, therefore: Kw = [H+(aq)] × [OH-(aq)] = 1.0 × 10-7 × 1.0 × 10-7 = 1.00 × 10-14 mol2 dm-6

HL Describe how indicators are themselves weak acids or bases that change colour over a particular pH range (usually the pKa ± 1), and that this range differs for different indicators.

You'll notice from the table that indicators do not necessarily change colour around pH 7. Most are weak acids and each have their own value of pKa and change colour over a pH range that centres on that value. The change in colour of bromothymol blue, known as its change point or end-point, happens in the range pH 6.0 - 7.6, which is within the approximate range of the pKa ± 1.0 (Figure 4). The end point lies where the equilibrium is balanced between the acid and its conjugate base, that is where [HIn] = [In-], here the indicator is exactly in the middle of its colour change and the pH = pKa. (Note that methyl orange is actually a weak base).

Calculate values for pH, [H+] and [OH-] for a solution given appropriate related data.

[H+] x [OH-] = 10^-14 M2 at 298k ( 25 degrees Celsius) pH + pOH = 14 pH = -log[H+] [H+] = 10^-pH [OH-] = 10^-pOH

HL Describe and sketch the titration curves obtained for the various different combinations of strong and weak acid and alkali; in particular noting the buffering regions and the positions of the equivalence point in each case.

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What is the pH of a solution of 0.05M of sulfuric acid?

pH = 1 Sulfuric acid is a diprotic acid (H2S04); it dissociates to release two protons per molecule. So the concentration of protons is twice that of the acid itself, giving 0.1 mol dm-3. The pH is the negative log of this value, giving 1.0

Define pH as - log10[H+],

pH = − log[H+(aq)] This can be rearranged to give: [H+] = 10^−pH So we can readily see that water, where [H+] = 10-7, has a pH of 7.00 at 298 K (25oC).

Understand how the scale is based on the nature of ionization in water, which is neutral and has an equal concentration of H+ and OH- ions at 298 K; [H+] = [OH-] = 1.00 x 10-7 mol dm-3,

pH = − log[H+(aq)] This can be rearranged to give: [H+] = 10^−pH So we can readily see that water, where [H+] = 10-7, has a pH of 7.00 at 298 K (25oC).

Calculate the pH of an acid solution given [H+], and vice versa,

pH = − log[H+(aq)] [H+] = 10^−pH

Deduce changes in [H+(aq)] when the pH of a solution changes by more than one pH unit.

pH is a logarithmic scale (a change of one pH unit really means a factor of 10 change in [H+]) If a solution of pH 1 is diluted 10 fold it will now be pH = 2. If a solution of pH 1 is diluted 100 fold it will now be pH =3 If a solution of pH 1 is diluted 1000 fold it will now be pH=4 If a solution of pH 1 is diluted 10,000 fold it will now be pH=5 So what is the [H+] change when pH goes from 12 to 9? Since each pH "jump" is a factor of 10, and the pH changes by 3 "jumps" then the [H+] changes by10x10x10 = 1000 times increase in [H+]. OR pH 12, [H+] = 10^-12M, pH 9, [H+]=10^-9M, so [H+] has increased by a factor of 1000.

Understand the structure of the pH scale ranging from 0 to 14, and centred on neutrality at pH = 7.0,

pH measures the concentration of H+ ions, Strong acid: Very low pH (1 - 3) Weak acid: Low pH (4 - 6) Neutrality: 7 Weak base: High pH (8 - 10) Strong base: Very high pH (11 - 14)

Worked example Calculate the pH of 0.01 mol dm-3 hydrochloric acid, HCl(aq). HCl(aq) → H+(aq) + Cl-(aq)

remember the definition pH = -log10[H+(aq)] now [H+(aq)] = 0.01 mol dm-3 (note that [H+(aq)] is equal to the concentration of the acid, because it is a strong acid that is completely ionized.) therefore pH = -log10(0.01) = - log1010-2 = 2

Which of the acids listed below is the strongest acid thought to be present in acid rain? 1. sulfuric(VI) acid, H2SO4 2. nitric(V) acid, HNO3 3. nitric(III) acid, HNO2 4. sulfuric(IV) acid, H2SO3

sulfuric(VI) acid, H2SO4 Explanation #1 is correct; sulfuric(VI) acid is the strongest of all the four acids listed, being stronger that nitric(V) acid. The higher oxidation state acids are the stonger of each of the pairs of acids associated with each element, nitrogen and sulfur.

Understand that a base must have a lone pair of electrons in its structure to be able to accept a proton by the formation of a coordinate bond,

svaret i frågan... A base must have a lone pair of electrons in its structure to be able to accept a proton by the formation of a coordinate bond,

Recognise an acid/conjugate pairing, and deduce the conjugate species given the identity of one of the pair.

-- e.g. CH3COOH as acid (has one more H proton) CH3COO- as conjugate base

Outline the characteristic reactions of an acid with a metal, a base and a metal carbonate,

5 reactions of Acids Metal In general, reactive metal + dilute acid → salt + hydrogen 2HNO3 + 2Li --> 2LiNO3 + H2 Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g) Acid + metal hydroxide → salt + water H2SO4 + 2KOH --> K2SO4 + 2H2O Acid + metal carbonate → salt + (CO2 +water) (h2co3) 2HCl + CaCO3 --> CaCl2 + H2O + CO2 Acid + metal oxide → salt + water 2CH3CO2H + MgO --> Mg(CH3CO2)2 + H2O acid + hydrogen carbonate --> salt + water + carbondioxide H3PO4 + 3NaHCO3 --> Na3PO4 + 3H2O +3CO2

state the general equations for these characteristic reactions and write specific balanced equations for particular examples,

5 reactions of Acids Metal In general, reactive metal + dilute acid → salt + hydrogen 2HNO3 + 2Li --> 2LiNO3 + H2 Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g) Acid + metal hydroxide → salt + water H2SO4 + 2KOH --> K2SO4 + 2H2O Acid + metal carbonate → salt + (CO2 +water) (h2co3) 2HCl + CaCO3 --> CaCl2 + H2O + CO2 Acid + metal oxide → salt + water 2CH3CO2H + MgO --> Mg(CH3CO2)2 + H2O acid + hydrogen carbonate --> salt + water + carbondioxide H3PO4 + 3NaHCO3 --> Na3PO4 + 3H2O +3CO2

Understand the distinction between an alkali and a base,

A base that is soluble in water is an alkali

What are salts?

A salt is an ionic compound formed when the replaceable hydrogen of an acid is completely or partly displaced by a metal (ion). Salts formed by replacing all of the hydrogen are termed normal salts; those formed by replacing only part of the hydrogen are termed acid salts.

How does an decrease in the pH of an aqueous solution effect the concentration of H+ and OH- ?

An increase in the H+ concentration and a decrease in the OH− concentration.

HL Understand that for a conjugate acid-base pair the relationship between the values of pKa and pKb is given by the relationship pKa + pKb = 14,

For any conjugate acid-base pair: Ka x Kb = Kw pKa + pKb = pKw pKa + pKb = 14.00 at 298 K

HL Examiner tip

Make sure that you can recognize species that act as Lewis acids; these species are lone pair acceptors, and are either a positively charged ion or have an empty orbital to accept a lone pair. In contrast, Lewis bases are species that are a negatively charged ion or have a lone pair of electrons to donate.

What is an alkali?

Many metal oxides or hydroxides are bases. Some bases dissolve in water to give an excess of hydroxide ions (OH-) in solution. A base which is soluble in water is called an alkali. For example: Na+OH-(s) + aq → Na+(aq) + OH-(aq)

Understand that the scale decreases from 7 to 0 with increasing acidity, and increases from 7 to 14 with increasing alkalinity,

Strong acid: Very low pH (1 - 3) Weak acid: Low pH (4 - 6) Neutrality: 7 Weak base: High pH (8 - 10) Strong base: Very high pH (11 - 14)

The calculation of the pH of a buffer is covered in more detail in Option B (Biochemistry) and Option D (Medicinal chemistry). men slänger in en equation här ändå <3

This is formalized in the Henderson-Hasselbalch equation: pH = pKa + log (conjugate base/acid)

Identify sulfur dioxide and the oxides of nitrogen as the main pollutants causing acid deposition,

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