DP Chemistry Rob Slider

DP Chemistry Rob Slider
18 Acids and Bases (AHL) DP Chemistry Rob Slider

Acid-Base Calculations

Product constant of water Kw
Note that water slightly dissociates in equilibrium with hydrogen and hydroxide ions: H2O  H+ + OH- So, the equilibrium constant is: Kw = [H+][OH-] (product constant of water) Experimentally, it has been determined in a neutral solution at 250C: [H+] = [OH-] = 10-7 (a very small amount!) Note: water is not included in the Kw equilibrium expression because it virtually remains unchanged due to its large value compared to the hydrogen ion and hydroxide ion concentrations. So, Kw = (10-7)(10-7) = 10-14 This shows how far to the left this equilibrium is

Temperature effects on Kw
Kw and temperature Temperature effects on Kw T (°C) Kw (mol2 dm-6) pH 0.114 x 10-14 7.47 10 0.293 x 10-14 7.27 20 0.681 x 10-14 7.08 25 1.008 x 10-14 7.00 30 1.471 x 10-14 6.92 40 2.916 x 10-14 6.77 50 5.476 x 10-14 6.63 100 51.3 x 10-14 6.14 In the previous slide we saw: Kw = (10-7)(10-7) = 10-14 This value applies to 250C Kw changes with temperature variations and so does the pH! Note: this does not mean that water is more acidic at higher temperatures and more alkaline at lower temperatures. Water is still neutral. It simply means that neutral is a different pH value at different temperatures.

Temperature effects on Kw
Questions Temperature effects on Kw T (°C) Kw (mol2 dm-6) pH 0.114 x 10-14 7.47 10 0.293 x 10-14 7.27 20 0.681 x 10-14 7.08 25 1.008 x 10-14 7.00 30 1.471 x 10-14 6.92 40 2.916 x 10-14 6.77 50 5.476 x 10-14 6.63 100 51.3 x 10-14 6.14 Given the values in the table, what are the [H+] and [OH-] at 00C and 1000C? What do the values of Kw tell you about whether the ionisation of water is endothermic or exothermic? Explain. Answers: At 0C : 3.38 x10-7; at 100C: 7.16x10-7 (square root of Kw) An increase of temperature increases the value of Kw. This suggests the products are favoured. According to LC Principle, this means the forward reaction is endothermic (absorbs heat)

Calculations using Kw Since Kw = (10-7)(10-7) = 10-14
We can use this constant value to calculate [H+] or [OH-] Calculating [H+] An alkali with a hydroxide ion concentration of 0.01M (10-2) Kw = (10-2)[H+] = 10-14 [H+] = 10-12 Calculating [OH-] An acid with a hydrogen ion concentration of 0.001M (10-3) Kw = (10-3)[OH-] = 10-14 [OH-] = 10-11

pH, pOH and pKw pH We have seen previously that pH = -log[H30+] pOH
This is similar to pH pOH = -log[OH-] The ‘p’ is a mathematical operation that means ‘-log’ pKw Since Kw = [H+][OH-] pKw = pH + pOH = 14 Calculating pH example 1 An alkali with a hydroxide ion concentration of 0.01M (10-2) Kw = (10-2)[H+] = 10-14 [H+] = 10-12 pH = 12 Calculating pH example 2 A substance has been found to have a hydroxide ion concentration of 10-11 [OH-] = 10-11 pOH = 11 pH = = 3

Calculating concentration from pH
Recall the inverse of the pH calculation: pH = -log[H30+] The inverse: [H30+] = 10-pH or [H+] = 10-pH On your calculator, this may be ‘2nd’ LOG or ‘10x’ Example The pH of a solution is found to be 4.6 Find[H+] [H+] = 10-pH = = 2.5x10-5 mol dm-3

Exercises Problem 1 Calculate the pH of solutions with the following H3O+ concentrations in mol dm-3: a)10-8 b)6.8 x c)0.035 Problem 2 Calculate the pH of solutions with the following OH- concentrations in mol dm-3: a)10-2 b)7.6 x c)0.055 Problem 3 Calculate the H3O+ concentrations in solutions with the following pH values: a)0.00 b)4.3 c)2.35 d)13.7 Problem 4 What is the pH of a 3.5M solution of H2SO4? (assume complete ionisation) Problem 5 What is the pH of a 0.001M solution of Ba(OH)2? (Hint: use balanced equation)

Weak solutions (Ka & Kb)
Unlike strong acids, weak acids only partially dissociate: HA + H2O  H3O+ + A- The same is true for weak bases: B + H2O  BH+ + OH- Dissociation Constants Because of these equilibria, we have known values for the ratio between products and reactants. These are known as dissociation constants: Ka – acid dissociation constant Kb – base dissociation constant Formulae Ka = 𝐻3𝑂+ [𝐴−] [𝐻𝐴] Kb = 𝐵𝐻+ [𝑂𝐻−] [𝐵] Notice water is not in these calculations. Its concentration has been incorporated into the dissociation constant.

Calculations using Ka CH3COOH(aq) ↔ H+(aq) + CH3COO-(aq)
Ka expression for acetic acid (ethanoic acid) above is: (constant at set temp) Ka = [H+] [CH3COO-] mol dm3 [CH3COOH] pKa indicates the relative strength of a weak acid. The larger the value, the weaker the acid. Same for pKb of bases Ka can be used to find the pKa (like pH) pKa = - log Ka The larger the value of Ka the stronger the acid (more it dissociates – breaks apart) What about pKa?

Calculations using Ka CH3COOH(aq) ↔ H+(aq) + CH3COO-(aq)
Ka = [H+] [CH3COO-] mol dm3 [CH3COOH] Using the equilibrium constant expression (above) we are able to calculate [H+] and thus the pH of the acid CH3COOH(aq) ↔ H+(aq) + CH3COO-(aq) Initial conc x Final conc x-y y y 𝐾𝑎= 𝑦 [𝑦] [𝑥−𝑦] = 𝑦 2 𝑥−𝑦 (𝑦≈0) ≈ 𝑦 2 [𝑥] Assumption made based on relative concentrations of ‘x’ vs. ‘y’ to simplify calculations

Calculations using Kb NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Kb = [NH4+] [OH-] [NH3] This works the same as in the acid example NH3(aq) ↔ NH4+(aq) + OH-(aq) Initial conc x Final conc x-y y y 𝐾𝑏= 𝑦 [𝑦] [𝑥−𝑦] ≈ 𝑦 2 [𝑥]

Ka, Kb and Kw Consider the following generic acid + water reaction:
HA + H2O  A- + H3O+ ACID Conjugate BASE Now, if the conjugate base reacts with water in this reaction: A- + H2O  HA + OH- So, Ka x Kb 𝐴− [𝐻3𝑂+] [𝐻𝐴] 𝑋 𝐻𝐴 [𝑂𝐻−] [𝐴−] = [𝐻3𝑂+][𝑂𝐻−] = Kw Ka x Kb = Kw 𝐾𝑎= 𝐴− [𝐻3𝑂+] [𝐻𝐴] 𝐾𝑏= 𝐻𝐴 [𝑂𝐻−] [𝐴−]

pKa, pKb and pKw pKa + pKb = pKw = 14 Ka x Kb = Kw
From the previous slide, Ka x Kb = Kw If we take the ‘p’ of each of these values (-log) we find, pKa + pKb = pKw = 14

Summary Ka = 𝐻3𝑂+ [𝐴−] [𝐻𝐴] (acid dissociation constant; larger the number, stronger the acid) Kb = 𝐵𝐻+ [𝑂𝐻−] [𝐵] (base dissociation constant; larger the number stronger the base) pKa = - log Ka (smaller the value, stronger the acid) (pKb is the same) Ka x Kb = Kw= 10-14 pKa + pKb = pKw = 14 = pH + pOH

Exercises Calculate the H3O+ concentrations, and the pH value for the following weak acids: a) HCO2H conc. = mol dm-3 Ka = 2.04 x 10-4 b) CH3CO2H conc. = 0.1 mol dm-3 Ka = 1.77 x 10-5 c) HCN conc. = 1.00 mol dm-3 Ka = 3.96 x 10-10 Calculate Ka for the following acids: d) HF conc. = mol dm-3 pH = 2.23 e) HClO2 conc. = mol dm-3 pH = 1.85 Calculate the OH- and H3O+ concentrations, and the pH value for the following weak bases: f) NH3 conc. = mol dm-3 Kb = 1.80 x 10-5 g) C5H5N conc. = 0.1 mol dm-3 Kb = 1.40 x 10-9 h) CH3NH2 conc. = 1.00 mol dm-3 Kb = 4.38 x 10-4

Buffer Solutions

What is a buffer? A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it There are 2 types: Acidic Alkaline

Acidic buffers An acidic buffer has a pH less than 7
It is often made from equal molar concentrations of a weak acid and it’s conjugate base Example: ethanoic acid and ethanoate ion CH3COOH  CH3COO H+ (weak acid) (conj. base)

Acidic buffers Take a 0.1M solution of ethanoic acid:
CH3COOH  CH3COO H+ At equil: ( M) ( M) ( M) What’s the pH of this solution? Now, say you add HCl to this equilibrium. The acetate ion is the only species available to reduce the added H+ and these are very low in concentration, so the pH will drop dramatically. What can you do to reduce this effect? -log [H+] = 2.38 Add acetate ions (sodium acetate) in equimolar amounts

Acidic buffers – adding acid
CH3COOH  CH3COO H+ If we buffer the solution by adding 1M sodium acetate, what will happen? This will have the following effect: Increase the amount of acetate ions, shifting the equilibrium to the left Increase the pH to 4.76 When adding H+ ions, the extra acetate ions will react to reduce the [H+] moving the equilibrium to the left – producing more weak acid. The solution resists changes to pH, meaning it is buffered

Acidic buffers – adding base
Add OH- CH3COOH  CH3COO H+ Adding base results in more OH- being added to the equilibrium. This extra amount of ions is removed by: CH3COOH + OH-  CH3COO- + H2O hydroxide ions are removed by reaction with undissociated ethanoic acid, shifting equilibrium right making a weak base (CH3COO-)

Acidic buffers – made with a base
An alternative way to make an acidic buffer is to add a strong base to the weak acid to produce high proportions of the weak acid and the conjugate base. MOH + HA   H2O + MA This drives the equilibrium to the right producing more salt (MA) which dissociates. This leads to: M+ + OH- + HA  H2O + M+ + A- Adding acid – conjugate base A- reacts with the added H+ to minimise the effect Adding base – weak acid HA dissociates further and the H+ reacts with the OH- to minimise the effect A 2:1 molar ratio of a weak acid and a strong base of the same concentration is used to make this buffer solution.

pH of buffers depends on concentration of conjugate pair
vol. of 0.1M acetic acid (ml) vol. of 0.1M sodium acetate (ml) 3 982.3 17.7 4 847.0 153.0 5 357.0 653.0 6 52.2 947.8 Note: you can also get various pH buffers by changing the acid/base pair.

Alkaline buffers An alkaline buffer has a pH greater than 7
It is often made from a equal molar concentrations of a weak base and it’s conjugate acid Example: ammonia and ammonium ion NH3 + H2O   NH OH- (weak base) (conj. acid)

Alkaline buffers NH3 + H2O   NH4+ + OH- (weak base) (conj. acid)
Due to the weak nature of ammonia, this equilibrium will be well to the left. We can create a buffered solution by adding ammonium chloride or a strong acid (like the acid buffer). What will this do to the equilibrium? Addition of ammonium ions (adding ammonium chloride) will shift the equilibrium even further to the left. The pH of this solution would be 9.25 You can also add ~half the moles of a strong acid (e.g. HCl) which will react with ammonia to form more ammonium ions (NH3 + H+  NH4+)

Alkaline buffers – adding acid
NH3 + H2O   NH OH- What will happen if you add acid to this solution? Reaction of the acid with the weak base NH3 + H+   NH4+ removal by reaction with ammonia to produce more ammonium ion – a weak acid)

Alkaline buffers – adding base
NH3 + H2O   NH OH- What will happen to the equilibrium if you add base? Adding base effectively adds OH-. This means: The ammonium ion reacts with the OH- to shift the equilibrium to the left, consuming most of the OH- ions. NH OH-   NH3 + H2O

Summary Buffer solutions resist changes in pH when acids and alkalis are added Buffers generally contain: Sufficient concentrations of a weak acid and it’s conjugate base OR weak base and it’s conjugate acid The pH of buffer solutions depend on the concentrations and type of conjugate acid/base pairs that are used. Go here  for good animations of this (chapter 8)

pH of a buffer solution Consider the dissociation of a weak acid
Commercially available buffer solutions are used to calibrate pH meters Consider the dissociation of a weak acid HA(aq) + H2O(l)  A-(aq) + H3O+ The acid dissociation constant expression is 𝐾𝑎= 𝐴− [𝐻3𝑂+] [𝐻𝐴] This can be rearranged to find [𝐻3𝑂+] 𝐻3𝑂+ =𝐾𝑎 [𝐻𝐴] 𝐴−

HA(aq) + H2O(l)  A-(aq) + H3O+
pH of a buffer solution HA(aq) + H2O(l)  A-(aq) + H3O+ Assumptions: As a large quantity of conjugate base (A-) has been added, the equilibrium shifts far left, so that equilibrium concentration of the acid is approximately equal to the initial concentration of the weak acid : [HA]eq ≈ [HA]i = [acid] The equilibrium concentration of the conjugate base ion (A- ) is approximately equal to the concentration of the salt that was added to the equilibrium. [A-]eq ≈ [A-]i = [salt]

Recall from a previous slide that
pH of a buffer solution Recall from a previous slide that 𝐻3𝑂+ =𝐾𝑎 [𝐻𝐴] 𝐴− So, considering our 2 assumptions: [HA]eq ≈ [HA]I = [acid] [A-]eq ≈ [A-]i = [salt] Note: some texts will show “ + log [salt]/[acid]”. This is the same value as “-log [acid]/[salt]” 𝐻3𝑂+ =𝐾𝑎 x [𝑎𝑐𝑖𝑑] 𝑠𝑎𝑙𝑡 Alternatively, you may solve for [H+] first and then solve for pH using pH=-log[H+] 𝑝𝐻=𝑝𝐾𝑎−𝑙𝑜𝑔 [𝑎𝑐𝑖𝑑] 𝑠𝑎𝑙𝑡 When [acid] = [salt] pH= pKa

pOH of a buffer solution
Similarly for a base equilibrium B(aq) + H2O(l)  HB+(aq) + OH-(aq) 𝑂𝐻− =𝐾𝑏 [𝐵] 𝐻𝐵+ =𝐾𝑏 [𝑏𝑎𝑠𝑒] 𝑠𝑎𝑙𝑡 Alternatively, you may solve for [OH-] first and then solve for pOH using pOH=-log[OH-] 𝑝𝑂𝐻=𝑝𝐾𝑏−𝑙𝑜𝑔 [𝑏𝑎𝑠𝑒] 𝑠𝑎𝑙𝑡 When [base] = [salt] pOH= pKb

Exercise An aqueous solution of 0.1M ammonia and 0.1M ammonium chloride has a pH of 9.3. The reaction is: NH3 + H2O  NH4+ + OH- Calculate the Kb for ammonia If a pH of 9.0 is needed, what should be added? Explain Calculate the new concentration of the substance added in b) Answers: 𝐾𝑏= 𝑂𝐻− [𝑁𝐻4+] [𝑁𝐻3] = 10−4.7 [0.1] [0.1] = = 2.00 x 10-5 Ammonium chloride should be added as this will shift the equil to the left(towards ammonia), reducing the [OH-] and decreasing the pH pH = 9.0 means pOH = 5, so [OH-] = 10-5 [𝑁𝐻4+]= 𝐾𝑏[𝑁𝐻3] [𝑂𝐻−] = (2𝑥10−5 )(0.1) [10−5 ] = 0.2 mol dm-3 Notice the additional volume has a negligible effect on [NH3] and has been ignored

Exercise If you wanted to make a buffer solution of pH=4.46 using ethanoic acid and sodium ethanoate, what concentrations could you use? pKa (ethanoic acid) = The reaction is: CH3COOH+ H2O  CH3COO- + H3O+ Answer: pH = pKa – log [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡] 4.5 = 4.76 – log [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡] log [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡] = 4.76 – 4.46 = 0.30 [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡] = [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡] = 2.0 So, we need a solution with twice as concentrated acid as ethanoate salt Note: it should be expected that the acid concentration will be higher than the salt. At equal concentrations, the pH = pKa = 4.76 At pH 4.46 (a lower pH), we need to add acid to shift equil right, increasing [H3O+], decreasing pH

Exercise An aqueous solution of 0.025M ethanoic acid and 0.050M sodium ethanoate is prepared. Calculate the pH of this solution given Ka = 1.74x10-5 mol dm-3 If 1.0cm-3 of 1.0M NaOH is added to 250cm3 of buffer, what will happen to the pH? Answers: 𝑝𝐻=𝑝𝐾𝑎−𝑙𝑜𝑔 [𝑎𝑐𝑖𝑑] 𝑠𝑎𝑙𝑡 = -log (1.74x10-5) – 𝑙𝑜𝑔 [.025] = =5.06 CH3COOH + OH-  CH3COO- + H2O ni (mol) 0.001 0.0125 - nc(mol) -0.001 +0.001 ne(mol) 0.0135 [ ] (mol dm-3) 0.021 0.054 So, pH = pKa – log [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡] pH = 4.76 – log [0.021] [0.054] pH = pH = 5.17 Notice the additional volume has again been ignored as it is assumed to have little effect on the overall pH

Acid-Base Titration

Titration Titration is an analytical technique that is used to determine the end point of a reaction (often Acid + Base) using an indicator or pH meter. Acid-Base Titration (basic steps): Unknown: Accurately measure a volume of base of unknown concentration using a pipette and place into a flask Indicator: Add an appropriate indicator which will change colour at the end point of the reaction Titrate: Carefully add an acid of known concentration until the end point is reached as indicated by the colour change Indicator addition Pipette Titration End point

Titration – end/equivalence
Equivalence Point The equivalence point in an acid-base reaction is the point where neutralisation occurs. The molar ratios have been reached End point The end point is where an indicator solution just changes colour permanently. Indicators change colours at specific pH ranges and are chosen to be as close to the equivalence point as possible. More later…

Titration Graphs

Strong Acid - Strong Base
Investigating the titration between: 1M HCl and 1M NaOH KeMsoft06

HCl Start with 10ml of alkali and slowly add acid measuring the pH 10 ml NaOH KeMsoft06

HCl 10 ml NaOH + almost 10 ml HCl KeMsoft06

HCl 10 ml NaOH + 10 ml HCl Now we have equivalent amounts of strong acid and strong base – notice the pH changes dramatically KeMsoft06

HCl 10 ml NaOH Now as we continue to add acid past pH = 7, the pH drops quickly at first and then more slowly KeMsoft06

1NaOH + 1HCl  NaCl + H2O NaOH + HCl = NaCl + H2O 1M M 10ml 10ml Equivalence point at pH7 Solutions mixed in the right proportions according to the equation. KeMsoft06

Running acid into alkali Running alkali into acid pOH = pKb The same applies to a base being neutralised by an acid This nearly horizontal section of the graph is called the buffer region. Here the pH stays relatively constant due to a buffering between the acid and the salt pH = pKa This point is called half-equivalence where half of the acid has been neutralised. Here, there are equivalent amounts of acid and salt, so 𝑝𝐻=𝑝𝐾𝑎−𝑙𝑜𝑔 𝑎𝑐𝑖𝑑 𝑠𝑎𝑙𝑡 =pKa −0 or pH = pKa

1M HCl and 1M NH3 Strong Acid - Weak Base
Investigating the titration between: 1M HCl and 1M NH3 KeMsoft06

Strong Acid - Weak Base acid alkali
Weak base so initial pH value is less than 14 1NH3 + 1HCl  NH4Cl 1M M pH starts to fall quickly as acid is added 25ml 25ml pH falls less quickly as buffer soln formed (excess NH3 and NH4Cl present) Soln at equivalence point is slightly acidic because ammonium ion is slightly acidic NH4+ + H2O  NH3 + H3O+ KeMsoft06

Strong Acid - Weak Base alkali acid
After equivalence point the soln contains NH3 and NH4Cl – a buffer soln and so resists large increase in pH so graph flattens out. alkali Very low pH indicative of a strong acid solution < pH 7 acid KeMsoft06

Strong Acid - Weak Base Running acid into alkali
Running alkali into acid KeMsoft06

1M CH3COOH and 1M NaOH Weak Acid - Strong Base
Investigating the titration between: 1M CH3COOH and 1M NaOH KeMsoft06

Weak Acid - Strong Base acid alkali 1M 1M 25ml 25ml
1CH3COOH + 1NaOH  CH3COONa acid alkali Very high pH – indicative of a strong alkali solution 1M M 25ml 25ml Soln at equivalence point is slightly alkaline because ethanoate ion is slightly alkaline pH falls less quickly as buffer soln formed (excess CH3COOH and CH3COO- present) CH3COO-+ H2O  CH3COOH + OH- KeMsoft06

Weak Acid - Strong Base pH rises less quickly as alkali
Excess of alkali present – graph same as when adding strong alkali to strong acid alkali acid pH rises less quickly as buffer soln formed (excess CH3COOH and CH3COO- present) pH starts to rise quickly as alkali is added KeMsoft06

Weak Acid - Strong Base Running acid into alkali
Running alkali into acid KeMsoft06

1M CH3COOH and 1M NH3 Weak Acid - Weak Base
Investigating the titration between: 1M CH3COOH and 1M NH3 KeMsoft06

Weak Acid - Weak Base 1CH3COOH + 1NH3 = CH3COO- + 1NH4+
alkali 1CH3COOH + 1NH3 = CH3COO- + 1NH4+ About as weak as each other - KeMsoft06

Weak Acid - Weak Base acid alkali No steep section – small addition of acid causes a large change in pH, so… Very difficult to do a titration between a weak acid and a weak base. KeMsoft06

Summary KeMsoft06

Salt Hydrolysis (pH of salts)

All salts are not neutral
The pH of a salt depends upon the relative strength of the ions that make up the salt. Very few salts are neutral Salts completely dissociate into their ions when sufficiently dilute NaCl(s)  Na+(aq) + Cl-(aq) It is possible for these ions to interact with water to produce H+ or OH- ions which results in acidic or alkaline solutions. These are known as hydrolysis reactions Remember: the stronger the acid/base, the weaker it’s conjugate

Strong Acid +Base Neutral ions
pH = 7 Neutral anions are formed from strong acids Neutral cations are formed from strong bases NaCl is a neutral salt because the ions that are formed derive from a strong acid and base HCl + NaOH  NaCl + H2O Na+ and Cl- are weak conjugates, so there is no tendency for these ions to undergo hydrolysis reactions.

Strong Base/Weak Acid Basic anions
pH > 7 Weak acids form conjugate bases that can react with water to form hydroxide ions H2CO3 + 2NaOH  Na2CO3 + H2O Na2CO3  2Na+ + CO32- CO H2O  HCO OH- In the above reactions: NaOH is a strong base so the weak conjugate, Na+ will not react with water to form hydrogen ions Carbonic acid is weak, so the carbonate ion will react with water to a small extent to form OH- ions Basic anions are formed by weak acids

Strong Acid/Weak Base Acidic cations
pH < 7 HCl + NH3  NH Cl- A donatable hydrogen must be available on the cation. NH H2O  NH3 + H3O+ The ammonium ion comes from the weak base, ammonia, so a hydrolysis reaction can occur to a small extent producing hydronium ions.

Weak Acid + Weak Base Depends on Ka/Kb
When combining weak acids and bases, the pH of the salt will depend on their relative strengths. If Ka > Kb – acidic If Kb > Ka – basic If Ka = Kb - neutral Example: CH3COOH + NH3  NH4+ + CH3COO- Both of the salts formed in this reaction will react with water, but the effects cancel resulting in a nearly neutral solution Ammonium Acetate

Acidic complex ions  2+ Fe3+ O H Fe3+ O H + H+
Small ions with multiple charges can hydrolyse water. The high charge density pulls electrons towards the metal ion causing a proton to be released, decreasing the pH 2+ Fe3+ O H Fe3+ O H  + H+ An O-H bond may be broken releasing a proton Other metal ions that are able to hydrolyse water include Be2+ and Al3+

Salt Hydrolysis Summary
Type of reaction Example Reaction Salt produced Ion that hydrolyses water Nature of final solution Strong acid Strong Base HCl + NaOH NaCl Neither ion Neutral pH = 7 Weak acid Strong Base CH3COOH + NaOH NaCH3COO Anion Basic pH>7 Strong Acid Weak Base HCl + NH3 NH4Cl Cation Acidic pH<7 Weak Acid CH3COOH + NH3 NH4CH3COO Anion & Cation Depends on Ka/Kb Metal complex [Al(H20)6]3+ [Al(H20)5OH]2+ + H+ NA Metal complex with multiple charge

Indicators

Indicators – colour changers
Indicators can change colours depending upon the pH of the solution they are in. Indicators are themselves weak acids or weak bases and as they lose or gain H+, they form substances that have distinct colours. In general, an indicator molecule (In) Hin(aq)  H+(aq) + In-(aq) Let’s look at some examples…

Litmus (red or blue) Add base (hydroxide)
“HLit” is litmus – a weak acid – is red in solution When hydroxide (base) is added to this weak base, “Lit-” is formed and is blue When hydrogen ions (acid) is added, we shift back to the red “HLit” . Base Add acid (hydrogen ion) Source of diagrams: So, Acid turns blue litmus red Alkali turns red litmus blue Acid

Methyl Orange (yellow or red)
Methyl orange is yellow below pH 3.7 Methyl orange is red above pH 3.7 Notice the transfer of a proton between the different coloured compounds

Phenolphthalein Here is a common acid-base indicator that changes colour at pH 9.3 What colour is the acid? What colour is the base?

Phenolphthalein Here is a common acid-base indicator that changes colour at pH 9.3 What colour is the acid? colourless What colour is the base? pink

Kind (pH = pKa) Recall the generic equation for an indicator
Hin(aq)  H+(aq) + In-(aq) We can write an Equilibrium expression for this weak acid 𝐾𝑎= 𝐻+ [𝐼𝑛−] [𝐻𝑖𝑛] When the equilibrium is balanced, [𝐼𝑛− ]= [𝐻𝑖𝑛] so, 𝐾𝑎= 𝐻+ [𝐼𝑛−] [𝐻𝑖𝑛] This means Ka = [H+] or pKa = pH So the pKa of the indicator, tells us where the colour change will occur

Choosing an Indicator 3.7 9.3 KeMsoft06

KeMsoft06

Strong Acid - Weak Base KeMsoft06

Weak Acid - Strong Base KeMsoft06

Weak Acid – Weak Base KeMsoft06

18 Acids and Bases The end

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