Acids, Bases and Buffers

Bronsted-Lowry Definition of Acid – Base behaviour A Bronsted-Lowry acid is defined as a substance that can donate a proton. A Bronsted-Lowry base is defined as a substance that can accept a proton. HCl (g) + H2O (l)  H3O+ (aq) + Cl- (aq) Acid 1 base 2 Acid 2 Base 1 Each acid is linked to a conjugate base on the other side of the equation. HNO3 + HNO2 ⇌ NO3 – + H2NO2+ Acid 1 Base 2 Base 1 Acid 2 HCOOH + CH3 (CH2 )2COOH ⇌ HCOO– + CH3 (CH2 )2COOH2+ Acid 1 Base 2 Base 1 Acid 2 In these reactions the substance with bigger Ka will act as the acid The acidic role of H+ in the reactions of acids with metals, carbonates, bases and alkalis acid + metal  salt + hydrogen 2CH3CH2CO2H + Mg  (CH3CH2COO)2Mg + H2 acid + alkali (NaOH)  salt + water 2HNO2 + Ca(OH)2  Ca(NO2 )2 + 2H2O acid + carbonate (Na2CO3 )  salt + water + CO2 2CH3CO2H + Na2CO3  2CH3CO2 -Na+ + H2O + CO2 2H+ + Mg  Mg2+ + H2 Ionic Equations H+ + OH–  H2O 2H+ + CO3 2–  H2O + CO2

Calculating pH pH = – log [H+ ] Where [H+ ] is the concentration of hydrogen ions in the solution Calculating pH of strong acids Strong acids completely dissociate The concentration of hydrogen ions in a monoprotic strong acid will be the same as the concentration of the acid. For HCl and HNO3 the [H+ (aq)] will be the same as the original concentration of the acid. For 0.1M HCl the pH will be –log[0.1] =1.00 Always give pH values to 2d.p. In the exam Finding [H+ ] from pH [H+ ] = 1 x 10-pH On most calculators this is done by pressing Inv (or 2nd function) log  – number(pH) Example 1 What is the concentration of HCl with a pH of 1.35? [H+ ] = 1 x 10-1.35 = 0.045M. Ionic Product for water [H+ (aq)][OH- (aq)] [H2O(l)] Kc= This equilibrium has the following equilibrium expression Kc x [H2O (l)] = [H+ (aq) ][OH- (aq)] Rearrange to Because [H2O (l)] is much bigger than the concentrations of the ions, we assume its value is constant and make a new constant Kw Kw = [H+ (aq) ][OH- (aq) ] Learn this expression At 25oC the value of Kw for all aqueous solutions is 1×10-14 mol2dm-6 The Kw expression can be used to calculate [H+ (aq)] ions if we know the [OH- (aq)] ions and vice versa Finding pH of pure water Pure water/ neutral solutions are neutral because the [H+ (aq) ] = [OH- (aq)] Using Kw = [H+ (aq) ][OH- (aq) ] then when neutral Kw = [H+ (aq) ]2 and [H+ (aq) ] = √ Kw At 25oC [H+ (aq) ] = √ 1×10-14 = 1×10-7 so pH = 7 At different temperatures to 25oC the pH of pure water changes. Le Chatelier’s principle can predict the change. The dissociation of water is endothermic (because bonds are broken) so increasing the temperature would push the equilibrium to the right giving a bigger concentration of H+ ions and a lower pH Example 2 : Calculate the pH of water at 50ºC given that Kw = 5.476 x 10-14 mol2 dm-6 at 50ºC [H+ (aq) ] = √ Kw = √ 5.476 x 10-14 =2.34 x 10-7 mol dm-3 pH = – log 2.34 x 10-7 = 6.6 It is still neutral though as [H+ (aq) ] = [OH- (aq)] N Goalby chemrevise.org Calculating pH of Strong Base For bases we are normally given the concentration of the hydroxide ion. To work out the pH we need to work out [H+ (aq)] using the kw expression. Strong bases completely dissociate into their ions NaOH  Na+ + OHExample 3: What is the pH of the strong base 0.1M NaOH Assume complete dissociation. Kw = [H+ (aq)][OH- (aq)] = 1×10-14 [H+ (aq)] = kw/ [OH- (aq)] = 1×10-14 / 0.1 = 1×10-13 M pH = – log[1×10-13 ] =13.00. In all aqueous solutions and pure water the following equilibrium occurs: H2O (l) H+ (aq) + OH- (aq)Diluting an acid or alkali pH of diluted strong acid [H+ ] = [H+ ]old x old volume new volume pH = – log [H+ ] pH of diluted base [OH– ] = [OH– ]old x old volume new volume [H+ ] = Kw [OH– ] pH = – log [H+ ] Example 7 Calculate the new pH when 50.0 cm3 of 0.150 mol dm-3 HCl is mixed with 500 cm3 of water. H+ ] = [H+ ]old x old volume new volume 0.05 0.55 [H+ (aq)] = 0.150 x [H+ (aq)] = 0.0136 pH = – log [H+ ] = -log 0.0136 = 1.87

Weak acids Weak acids only slightly dissociate when dissolved in water, giving an equilibrium mixture HA + H2O (l) H3O+ (aq) + A- (aq) HA (aq) H+ (aq) + A- (aq) We can simplify this to [H+ (aq)][A- (aq)] [HA (aq)] Ka= Weak acids dissociation expression The Ka for ethanoic acid is 1.7 x 10-5 mol dm-3 . The larger ka the stronger the acid [H+ (aq)][CH3CH2CO2 – (aq)] [CH3CH2CO2H(aq)] Ka= CH3CH2CO2H(aq) H+ (aq) + CH3CH2CO2 – (aq) Example 4 Write equation for dissociation of propanoic acid and its ka expression Calculating pH of a weak acid To make the calculation easier two assumptions are made to simplify the Ka expression: 1) [H+ (aq)]eqm = [A- (aq)] eqm because they have dissociated according to a 1:1 ratio 2) As the amount of dissociation is small we assume that the initial concentration of the undissociated acid has remained constant. So [HA (aq) ] eqm = [HA(aq) ] initial [H+ (aq)][A- (aq)] [HA (aq)] Ka= Simplifies to [H+ (aq)]2 [HA (aq)] initial Ka= Example 5 What is the pH of a solution of 0.01M ethanoic acid (ka is 1.7 x 10-5 mol dm-3 )? CH3CO2H(aq) H+ (aq) + CH3CO2 – (aq) [H+ (aq)][CH3CO2 – (aq)] [CH3CO2H(aq)] Ka= [H+ (aq)]2 [CH3CO2H(aq)] initial Ka= [H+ (aq)]2 0.01 1.7x 10-5 = [H+ (aq)]2 = 1.7 x 10-5 x 0.01 [H+ (aq)] = √ 1.7 x 10-7 = 4.12 x 10-4 pH = – log [H+ ] = -log (4.12 x10-4 ) pH =3.38 Example 6 What is the concentration of propanoic acid with a pH of 3.52 (ka is 1.35 x 10-5 mol dm-3 )? CH3CH2CO2H(aq) H+ (aq) + CH3CH2CO2 – (aq) [H+ (aq)][CH3CH2CO2 – (aq)] [CH3CH2CO2H(aq)] Ka= [H+ (aq)]2 [CH3CH2CO2H(aq)] initial Ka= [0.000302]2 [CH3CH2CO2H(aq)] initial 1.35 x 10-5 = [CH3CH2CO2H(aq)] = 9.12 x 10-8 /1.35 x 10-5 [H+ ] = 1 x 10-3.52 = 0.000302M [CH3CH2CO2H(aq)] = 6.75 x 10-3 M pKa Sometimes Ka values are quoted as pKa values pKa = -log Ka so Ka = 10-pKa

A Buffer solution is one where the pH does not change significantly if small amounts of acid or alkali are added to it. An acidic buffer solution is made from a weak acid and a salt of that weak acid ( made from reacting the weak acid with a strong base) Example : ethanoic acid and sodium ethanoate CH3CO2H (aq) and CH3CO2 – Na+ A basic buffer solution is made from a weak base and a salt of that weak base ( made from reacting the weak base with a strong acid) Example :ammonia and ammonium chloride NH3 and NH4+ClHow Buffer solutions work In an ethanoic acid buffer CH3CO2H (aq) CH3CO2 – (aq) + H+ (aq) Acid conjugate base In a buffer solution there is a much higher concentration of the salt CH3CO2 – ion than in the pure acid. If small amounts of acid is added to the buffer: Then the above equilibrium will shift to the left removing nearly all the H+ ions added, CH3CO2 – (aq) + H+ (aq)  CH3CO2H (aq) As there is a large concentration of the salt ion in the buffer the ratio [CH3CO2H]/ [CH3CO2 -] stays almost constant, so the pH stays fairly constant. If small amounts of alkali is added to the buffer. The OHions will react with H+ ions to form water. The Equilibrium will then shift to the right to produce more H+ ions. Overall the concentration of H+ ions and pH remains constant (but some ethanoic acid molecules are changed to ethanoate ions) CH3CO2H (aq) CH3CO2 – (aq) + H+ (aq) Learn these explanations carefully and be able to write the equilibrium to illustrate your answer The salt content can be added in several ways: a salt solution could be added to the acid or some solid salt added. A buffer can also be made by partially neutralising a weak acid with alkali and therefore producing a mixture of salt and acid. Calculating the pH of buffer solutions [H+ (aq)][A- (aq)] [HA (aq)] Ka= We still use the weak acids dissociation expression But here we assume the [A-] concentration is due to the added salt only [HA(aq)] [A- (aq) ] [H+ (aq) Normally we ] = Ka rearrange to We also assume the Initial concentration of the acid has remained constant, because amount that has dissociated or reacted is small. Example 8: making a buffer by adding a salt solution What would be the pH of a buffer made from 45cm3 of 0.1M ethanoic acid and 50cm3 of 0.15 M sodium ethanoate (Ka = 1.7 x 10-5 ) ? Work out the moles of both solutions Moles ethanoic = conc x vol = 0.1 x 0.045 = 0.0045mol Moles sodium ethanoate = conc x vol = 0.15 x 0.050 = 0.0075 [HA(aq)] [A- (aq) ] [H+ (aq)] = Ka We can enter moles of acid and salt straight into the equation as they both have the same new final volume 0.0045 0.0075 [H+ (aq)] = 1.7 x 10-5 x [H+ (aq)] = 1.02x 10-5 pH = – log [H+ ] = -log 1.02x 10-5 = 4.99 Example 9 : making a buffer by adding a solid salt A buffer solution is made by adding 1.1g of sodium ethanoate into 100 cm3 of 0.4M ethanoic acid. What is its pH? Ka =1.7 x10-5 Work out the moles of both solutions Moles ethanoic = conc x vol = 0.4 x 0.1 = 0.04mol Moles sodium ethanoate = mass/Mr= 1.1/82 = 0.0134 [HA(aq)] [A- (aq) ] [H+ (aq)] = Ka We can enter moles of acid and salt straight into the equation as they both have the same new final volume 0.04 0.0134 [H+ (aq)] = 1.7 x 10-5 x [H+ (aq)] = 5.07x 10-5 pH = – log [H+ ] = -log 5.07x 10-5 = 4.29 If a buffer is made by adding sodium hydroxide to partially neutralise a weak acid then follow the method below Example 10 55cm3 of 0.5M CH3CO2H is reacted with 25cm3 of 0.35M NaOH. What will be the pH of the resulting buffer solution? Moles CH3CO2H = conc x vol =0.5x 0.055 = 0.0275mol Moles NaOH = conc x vol = 0.35 x 0.025 = 0.00875 Moles of CH3CO2H in excess = 0.0275-0.00875 = 0.01875 (as 1:1 ratio) [CH3CO2H ] = moles excess CH3CO2H total volume (dm3 ) = 0.01875/ 0.08 = 0.234M CH3CO2H+ NaOH  CH3CO2Na + H2O pH = – log [H+ ] = -log 3.64 x 10-5 = 4.44 [CH3CO2 – ] = moles OH- added total volume (dm3 ) = 0.00875/ 0.08 = 0.109M ka = [H+ ] [CH3CO2 – ] [ CH3CO2H ] ka is 1.7 x 10-5 mol dm-3 [H+ ] = ka x[ CH3CO2H ] / [CH3CO2 – ] = 1.7 x 10-5 x 0.234 / 0.109 = 3.64 x 10-5. Calculating change in pH of buffer on addition of small amount of acid or alkali If a small amount of alkali is added to a buffer then the moles of the buffer acid would reduce by the number of moles of alkali added and the moles of salt would increase by the same amount so a new calculation of pH can be done with the new values CH3CO2H (aq) +OH-  CH3CO2 – (aq) + H2O (l) If a small amount of acid is added to a buffer then the moles of the buffer salt would reduce by the number of moles of acid added and the moles of buffer acid would increase by the same amount so a new calculation of pH can be done with the new values. CH3CO2 – (aq) + H +  CH3CO2H (aq) Example 11: 0.005 mol of NaOH is added to 500cm3 of a buffer where the concentration of ethanoic acid is 0.200 mol dm-3 and the concentration of sodium ethanoate is 0.250 mol dm-3 . (Ka = 1.7 x 10-5 ) Calculate the pH of the buffer solution after the NaOH has been added. Work out the moles of acid and salt in the initial buffer solution Moles ethanoic acid= conc x vol = 0.200 x 0.500 = 0.100mol Moles sodium ethanoate = conc x vol = 0.25 x 0.500 = 0.125mol Work out the moles of acid and salt in buffer after the addition of 0.005mol NaOH Moles ethanoic acid = 0.100 – 0.005 = 0.095 mol Moles sodium ethanoate = 0.125 +0.005 = 0.130 mol [CH3COOH (aq)] [CH3COO- (aq) ] [H+ (aq)] = Ka We can enter moles of acid and salt straight into the equation as they both have the same new final volume 0.095 0.130 [H+ (aq)] = 1.7 x 10-5 x [H+ (aq)] = 1.24x 10-5 pH = – log [H+ ] = -log 1.24x 10-5 = 4.91 A carbonic acid– hydrogencarbonate equilibrium acts as a buffer in the control of blood pH Buffering action in blood Equilibrium H2CO3 ⇌ H+ + HCO3 – Adding alkali reacts with H+ so the above Equilibrium would shift right forming new H+ and more HCO3 – The H2CO3 /HCO3 – buffer is present in blood plasma, maintaining a pH between 7.35 and 7.45.

Titration curves There are 4 main types of curve 1. Strong acid and strong base 2. Weak acid and strong base 3. Strong acid and weak base 4. Weak acid and weak base The Key points to sketching a curve: Initial and final pH Volume at neutralisation General Shape (pH at neutralisation) Strong acid – Strong base pH 7 1 13 25 cm3 of base Long vertical part from around 3 to 9 pH at equivalence point = 7 You may also have to work out the neutralisation volume from titration data given in the question. These are done by standard titration calculations from module 1. The equivalence point lies at the mid point of the extrapolated vertical portion of the curve. e.g. HCl and NaOH N Goalby chemrevise.org Half neutralisation volume [H+ (aq)][A- (aq)] [HA (aq)] Ka= For weak acids At ½ the neutralisation volume the [HA] = [A-] So Ka= [H+ ] and pKa = pH If we know the Ka we can then work out the pH at ½ V or vice versa Weak acid – Strong base e.g. CH3CO2H and NaOH pH 7 1 13 V cm3 of base At the start the pH rises quickly and then levels off. The flattened part is called the buffer region and is formed because a buffer solution is made Steep part of curve >7 (around 7 to 9) Equivalence point >7 pH starts near 3 ½ V 8 pH 7 1 13 25 cm3 of base Strong acid – Weak base Equivalence point < 7 Vertical part of curve <7 (around 4 to 7) Weak acid – Weak base e.g. CH3CO2H and NH3 e.g. HCl and NH3 pH 7 1 13 25 cm3 of base No vertical part of the curve Choosing an Indicator Indicators can be considered as weak acids. The acid must have a different colour to its conjugate base HIn (aq) In- (aq) + H+ (aq) colour A colour B We can apply Le Chatelier to give us the colour. In an acid solution the H+ ions present will push this equilibrium towards the reactants. Therefore colour A is the acidic colour. In an alkaline solution the OHions will react and remove H+ ions causing the equilibrium to shift to the products. Colour B is the alkaline colour. An indicator changes colour from HIn to In- over a narrow range. Different indicators change colours over a different ranges The end-point of a titration is reached when [HIn] = [In-]. To choose a correct indicator for a titration one should pick an indicator whose end-point coincides with the equivalence point for the titration An indicator will work if the pH range of the indicator lies on the vertical part of the titration curve. In this case the indicator will change colour rapidly and the colour change will correspond to the neutralisation point. pH 7 1 13 25 cm3 of base strong base weak base strong acid weak acid pH range for phenolphthalein pH range for methyl orange Use methyl orange with titrations with strong acids but not weak acids Colour change: red acid  yellow alkali (orange end point) Only use phenolphthalein in titrations with strong bases but not weak bases- Colour change: colourless acid  pink alkali