Acid base equilibria

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 base acid base Each acid is linked to a conjugate base on the other side of the equation. 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 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)]. In all aqueous solutions and pure water the following equilibrium occurs: H2O (l) H+ (aq) + OH- (aq). 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

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 an 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

pH Calculations involving Neutralisation Reactions These can be quite complex calculations working out the pH of a partially neutralised acid or the pH of the solution if too much alkali has been added and has gone past neutralisation. The method differs if the acid is strong or weak for the partially neutralised case. Strong Acid and Strong Base Neutralisations Work out moles of original acid Work out moles of base added Work out which one is in excess Work out new concentration of excess H+ ions [H+ ] = moles excess H+ total volume (dm3 ) pH = – log [H+ ] Total volume = vol of acid + vol of base added If excess acid Work out new concentration of excess OHions [OH-] = moles excess OH- total volume (dm3 ) [H+ ] = Kw /[OH– ] pH = – log [H+ ] Total volume = vol of acid + vol of base added If excess alkali Example 8 45cm3 of 1M HCl is reacted with 30cm3 of 0.65M NaOH. What will be the pH of the resulting mixture? Moles HCl = conc x vol = 1 x 0.045 = 0.045mol Moles NaOH = conc x vol = 0.65 x 0.030 = 0.0195 HCl + NaOH  NaCl + H2O Moles of HCl in excess = 0.045-0.0195 = 0.0255 (as 1:1 ratio) [H+ ] = moles excess H+ total volume (dm3 ) = 0.0255/ 0.075 = 0.34M pH = – log [H+ ] = -log 0.34 = 0.47 Example 7 15cm3 of 0.5M HCl is reacted with 35cm3 of 0.55M NaOH. What will be the pH of the resulting mixture? = 0.01175/ 0.05 = 0.235M [H+ ] = Kw /[OH– ] = 1×10-14 / 0.235 = 4.25×10-14 pH = – log [H+ ] = -log 4.25×10-14 = 13.37 Moles HCl = conc x vol = 0.5 x 0.015 = 0.0075mol Moles NaOH = conc x vol = 0.55 x 0.035 = 0.01925 HCl + NaOH  NaCl + H2O Moles of NaOH in excess = 0.01925 – 0.0075= 0.01175 (as 1:1 ratio) [OH-] = moles excess OH- total volume (dm3 ) Weak Acid and Strong Base Neutralisations Work out moles of original acid Work out moles of base added Work out which one is in excess Work out new concentration of excess HA [HA] = initial moles HA – moles OH- total volume (dm3 ) Work out concentration of salt formed [A-] [A-] = moles OH- added total volume (dm3 ) Rearrange ka = [H+ ] [A-] to get [H+ ] [HA] pH = – log [H+ ] If excess acid If excess alkali use the same method with excess alkali and strong acid above Example 9 55cm3 of 0.5M CH3CO2H is reacted with 25cm3 of 0.35M NaOH. What will be the pH of the resulting mixture? 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. Working out pH of a weak acid at half equivalence When a weak acid has been reacted with exactly half the neutralisation volume of alkali, the above calculation can be simplified considerably. At half neutralisation we can make the assumption that [HA] = [A-] ka = [H+ ] [CH3CO2 – ] [ CH3CO2H ] So [H+ (aq)] = ka And pH = pka Example 10 What is the pH of the resulting solution when 25cm3 of 0.1M NaOH is added to 50cm3 of 0.1M CH3COOH (ka 1.7 x 10-5 ) From the volumes and concentrations spot it is half neutralisation (or calculate) pH = pka = -log (1.7 x 10-5 ) = 4.77 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 11 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

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 steep 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 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 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. If a pH curve is plotted then the pH of a weak acid at half neutralisation (½ V) will equal the pKa Constructing a PH curve Calibrate meter first by measuring known pH of a buffer solution. This is necessary because pH meters can lose accuracy on storage Can improve accuracy by maintaining constant temperature •Measure initial pH of the acid • Add alkali in small amounts noting the volume added • Stir mixture to equalise the pH • Measure and record the pH to 1 dp • When approaching endpoint add in smaller volumes of alkali •Add until alkali in excess. Strong acid – Weak base Equivalence point < 7 Steep 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 Steep 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 steep 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

Buffer Solutions 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. H+ + OH –  H2O The Equilibrium will then shift to the right to produce more H+ ions. CH3CO2H (aq) CH3CO2 – (aq) + H+ (aq) Some ethanoic acid molecules are changed to ethanoate ions but 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. Learn these explanations carefully and be able to write the equilibrium to illustrate your answer. The buffer contains a reservoir of HA and A – ions [CH3CO2H (aq)] [CH3CO2 – (aq) ]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 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. 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 salt.Example 12: 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 13 : 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 N Goalby chemrevise.org If a buffer is made by adding sodium hydroxide to partially neutralise a weak acid then follow the method below Example 14 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 15: 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-level Chemistry exemplar for required practical No. 9 To investigate how pH changes when a weak acid reacts with a strong base: Investigation of how the pH of a solution of ethanoic acid changes as sodium hydroxide solution is added. Student sheet This experiment investigates how the pH of a solution of ethanoic acid changes as sodium hydroxide solution is added. The results are plotted in a graph which shows the general pattern of how the pH changes when a weak acid reacts with a strong base. It is necessary initially to calibrate a pH meter so as to give accurate pH values for each pH reading. Requirements You are provided with the following:  two 50 cm3 burettes  two funnels  100 cm3 beaker  deionised (or distilled) water in a wash bottle  stand and clamp  pH meter/probe  0.100 mol dm–3 sodium hydroxide solution  0.100 mol dm–3 ethanoic acid solution  standard pH buffer solutions at pH 4.00, 7.00 and 9.20  stirring rod  graph paper. Suggested method Part 1 Calibrate the pH meter a) Rinse the pH probe thoroughly with deionised water, and shake it gently to remove excess water. Place the probe in the standard pH 7.00 buffer solution provided, ensuring that the bulb is fully immersed. Record the pH reading in a suitable table. b) Repeat this process using the standard pH 4.00 and 9.20 buffer solutions. Rinse the pH probe thoroughly with deionised water before taking each reading. Record the pH readings in your table. c) Plot a graph of your recorded pH reading (x-axis) against the pH of the buffer solution. Your graph may be a straight line or a curve.This calibration graph will be used in the next part of the experiment to convert pH readings into more accurate pH values.Part 2 The measurement of the pH of the mixture of acid and alkali a) Rinse a burette with the 0.100 mol dm–3 solution of ethanoic acid provided and then fill the burette with this solution, ensuring that it is filled below the tap. Label this burette so that you do not confuse it with the second burette. b) Use the burette to transfer exactly 20.0 cm 3 of ethanoic acid to a clean 100 cm3 beaker. c) Rinse a second burette with the 0.100 mol dm–3 NaOH solution provided and then fill this second burette with this solution, ensuring that it is filled below the tap. d) Rinse the pH probe with distilled or deionised water and clamp it so that its bulb is fully immersed in the ethanoic acid solution in the beaker. Use a rod to stir the solution gently and record the pH reading in a suitable table. e) Using the second burette, add exactly 2.0 cm3 of the NaOH solution to the beaker containing the ethanoic acid. Stir the mixture gently with the glass rod and measure the pH of the mixture. Record the pH reading. Add the NaOH solution in 2.0 cm 3 portions from the second burette to the ethanoic acid in the beaker until 18 cm 3 of the NaOH solution have been added. Take a pH reading after each addition of NaOH solution, and in each case record the pH reading in your table. Then add the NaOH solution in 0.20 cm3 portions until 22.0 cm3 is reached. Then add the NaOH solution in 2.0 cm 3 portions again until 40 cm3 have been added. f) Rinse the pH probe with distilled or deionised water when you have taken all of your readings.Analysing the data a) Use your calibration graph from Part 1 to adjust, as appropriate, the pH readings obtained in your experiment in Part 2. These corrected pH values should be entered into a new column in your table of results. b) Plot a graph of the corrected pH values from Part 2 (y-axis) against volume of sodium hydroxide solution added. c) Join the points in the most appropriate way. d) Comment on the shape of the curve.