18 Acids and Bases (AHL)

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!) So, Kw = (10-7)(10-7) = 10-14 This shows how far to the left this equilibrium is

Kw and temperature In the previous slide we saw: Kw = (10-7)(10-7) = 10-14 This value applies to 250C Kwchanges 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.

Questions 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 x10-8; 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 = 14-11 = 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 = 10-4.6 = 2.5x10-5mol dm-3

Exercises Problem 1 Calculate the pH of solutions with the following H3O+ concentrations in mol dm-3: a)10-8b)6.8 x 10-3 c)0.035 Problem 2 Calculate the pH of solutions with the following OH- concentrations in mol dm-3: a)10-2b)7.6 x 10-3 c)0.055 Problem 3 Calculate the H3O+ concentrations in solutions with the following pH values: a)0.00b)4.3c)2.35d)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 = Kb= 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] 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 0 0 • Final conc x-y y y 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 0 0 • Final conc x-y y y

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 = [] = Kw Ka x Kb = Kw

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

Summary • Ka = (acid dissociation constant; larger the number, stronger the acid) • Kb = (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. = 0.0100 mol dm-3Ka= 2.04 x 10-4b) CH3CO2H conc. = 0.1 mol dm-3Ka= 1.77 x 10-5c) HCN conc. = 1.00 mol dm-3Ka= 3.96 x 10-10 Calculate Ka for the following acids: d) HF conc. = 0.200 mol dm-3 pH = 2.23e) HClO2 conc. = 0.0200 mol dm-3 pH = 1.85 Calculate the OH- and H3O+ concentrations, and the pH value for the following weak bases: f) NH3 conc. = 0.0100 mol dm-3 Kb= 1.80 x 10-5g) C5H5N conc. = 0.1 mol dm-3 Kb= 1.40 x 10-9h) 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: (0.09583M) (0.00417M) (0.00417M) 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 CH3COOH  CH3COO- + H+ Addingbase results in more OH- being added tothe equilibrium. This extra amountof ions is removedby: • CH3COOH + OH-  CH3COO- + H2O • hydroxideions are removed by reaction with undissociated ethanoic acid,shifting equilibrium right makinga weak base (CH3COO-) Add OH-

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 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   NH4+ + 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   NH4+ + OH- What will happen if youadd 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   NH4+ + 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. NH4+ + 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 baseORweak 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  http://www.pearsonhotlinks.co.uk/9780435994402.aspx for good animations of this (chapter 8)

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

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]

pH of a buffer solution Recall from a previous slide that So, considering our 2 assumptions: [HA]eq ≈ [HA]I = [acid] [A-]eq≈ [A-]i = [salt] Alternatively, you may solve for [H+] first and then solve for pH using pH=-log[H+] When [acid] = [salt] pH= pKa

pOH of a buffer solution Similarly for a base equilibrium B(aq) + H2O(l) HB+(aq) + OH-(aq) Alternatively, you may solve for [OH-] first and then solve for pOH using pOH=-log[OH-] 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: = 10-4.7 = 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 = = 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) = 4.76. The reaction is: CH3COOH+ H2O  CH3COO- + H3O+ Answer: pH = pKa – log 4.5 = 4.76 – log log = 4.76 – 4.46 = 0.30 = 100.30 = 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 = 1.74x10-5mol dm-3 If 1.0cm3of 1.0M NaOH is added to 250cm3of buffer, what will happen to the pH? • Answers: • = -log (1.74x10-5) – = 4.76+0.30=5.06 So, pH = pKa – log pH = 4.76 – log pH = 4.76 + 0.41 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

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

Strong Acid - Strong Base HCl 10 ml NaOH + almost 10 ml HCl KeMsoft06

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

Strong Acid - Strong Base 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

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

Strong Acid - Strong Base Running acid into alkali Running alkali into 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 pOH = pKb 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 The same applies to a base being neutralised by an acid

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

Weak base so initial pH value is less than 14 pH starts to fall quickly as acid is added acid alkali Strong Acid - Weak Base 1NH3 + 1HCl NH4Cl 1M 1M 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

18 Acids and Bases (AHL)