# 4. Acid Base Equilibria

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1 4. Acid Base Equilibria BronstedLowry Definition of acid Base behaviour A BronstedLowry acid is defined as a substance that can donate a proton. A BronstedLowry base is defined as a substance that can accept a proton. HCl (g) + H 2 O (l) H 3 O + + Cl acid base acid base Each acid is linked to a conjugate base on the other side of the equation. HNO 3 + HNO 2 NO + H NO + Acid Base Base Acid 2 HCOOH + CH (CH ) COOH HCOO + CH (CH ) COOH Acid Base 2 Base Acid 2 In these reactions the substance with bigger Ka will act as the acid Calculating = log [H + Where [H + is the concentration of hydrogen ions in the solution Calculating 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 HNO 3 the [H + will be the same as the original concentration of the acid. For 0.M HCl the will be log[0. =.00 Always give values to 2d.p. In the exam Finding [H + from On most calculators this is [H + done by pressing = x 0 Inv (or 2 nd function)log number() Example What is the concentration of HCl with a of.35? [H + = x 0.35 = 0.045M Ionic Product for water In all aqueous solutions and pure water the following equilibrium occurs: H O (l) H + +OH This equilibrium has the following equilibrium expression Kc= [H+ [OH [H 2 O(l) Rearrange to Kc x [H O (l) =[H + [OH 2 2 Because [H 2 O (l) is much bigger than the concentrations of the ions, we assume its value is constant and make a new constant Kw Kw = [H + [OH Learn this expression At 25 o C the value of Kw for all aqueous solutions is x0 4 mol 2 dm 6 The Kw expression can be used to calculate [H + ions if we know the [OH ions and vice versa.

2 Finding of pure water Pure water/ neutral solutions are neutral because the [H + = [OH Using Kw = [H + [OH then when neutral Kw = [H + 2 and [H + = Kw At 25 o C [H + = x0 4 = x0 7 so = 7 At different temperatures to 25 o C the of Example 2 : Calculate the of water at 50ºC given that K w = pure water changes. Le Chatelier s x 0 4 mol 2 dm 6 at 50ºC principle can predict the change.the [H + = Kw = x 0 4 =2.34 x 0 7 mol dm 3 = log 2.34 x 0 7 = 6.6 dissociation of water is endothermic so It is still neutral though as [H + = [OH increasing the temperature would push the equilibrium to the right giving a bigger Calculating of Strong Base concentration of H + ions and a lower. For bases we are normally given the concentration of the hydroxide ion. To work out the we need to work out [H + using the kw expression. Strong bases completely dissociate into their ions. NaOH Na + + OH Example 3: What is the of the strong base 0.M NaOH Assume complete dissociation. Kw = [H + [OH = x0 4 [H + = kw/ [OH = x0 4 / 0. = x0 3 M = log[x0 3 =3.00 Weak acids Weak acids only slightly dissociate when dissolved in water, giving an equilibrium mixture HA +H O (l) H O + +A We can simplify this to HA H + +A 2 3 Weak acids dissociation expression Ka= [H + [A [HA The K a for ethanoic acid is.7 x 0 5 mol dm 3. The larger ka the stronger the acid Example 4 Write an equation for dissociation of propanoic acid and its ka expression CH 3 CH 2 CO 2 H H + + CH 3 CH 2 CO 2 Ka= [H + [CH CH CO 322 [CH 3 CH 2 CO 2 H pka Sometimes Ka values are quoted as pka values pka = log Ka so Ka = 0 pka Calculating of a weak acid To make the calculation easier two assumptions are made to simplify the Ka expression: )[H + =[A because they have dissociated eqm eqm according to a : ratio. 2) As the amount of dissociation is small we assume that the initial concentration of the undissociated acid has remained constant. So [HA eqm = [HA initial Ka= Simplifies to Ka= [H + [A [HA [H + 2 [HA initial

3 2

4 Example 5 What is the of a solution of 0.0M ethanoic acid (ka is.7 x 0 5 mol dm 3 )? CH CO H H + +CH CO [H + [CH CO [H + Ka= 2.7x 0 5 [H + = Ka= [CH CO H [CH 3 CO 2 H 3 2 initial 0.0 [H + 2 =.7 x 0 5 x 0.0 = log [H + = log (4.2 x0 4 ) [H + =.7 x 0 7 = 4.2 x 0 4 =3.38 Example 6 What is the concentration ofpropanoic acid with a of 3.52 (ka is.35 x 0 5 mol dm 3 )? CH CH CO H H + +CH CH CO [H + = x = M [H + [CH CH CO Ka= Ka= [H + 2 [ CH CO H x 0 5 = [CH 3 CH 2 CO 2 H [CH initial [CH CH CO H = 9.2 x 0 8 /.35 x 0 5 [CH CH CO H = 6.75 x 0 3 M [CH CH CO H initial Working out 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. ka = [H + [CH 3 CO 2 [ CH 3 CO 2 H At half neutralisation we can make the assumption that [HA = [A Example 7 What is the of the resulting solution when 25cm 3 of 0.M NaOH is added to 50cm 3 of 0.M CH 3 COOH (ka.7 x 0 5 ) From the volumes and concentrations spot it is half neutralisation (or calculate) So [H + = ka And = pka = pka = log (.7 x 0 5 ) = 4.77 Diluting an acid or alkali of diluted strong acid [H + = [H + old x old volume new volume = log [H + of diluted base [OH = [OH old x old volume new volume [H + = K w [OH = log [H + Example 8 Calculate the new when 50.0 cm 3 of 0.50 mol dm 3 HCl is mixed with 500 cm 3 of water. H + =[H + xold volume [H [H + = old =0.50 x new volume 0.55 = log [H + = log =.87 Comparing the of a strong acid and a weak acid after dilution 0, 00 and 000 times Because is a logarithmic scale, diluting a strong acid 0 times will increase its by one unit, and diluting it 00 times would increase its by two units Weak acids would not change in the same way as when they are diluted. They increase by less than unit CH CH CO H +H O H O + +CH CH CO Diluting the weak acid pushes the equilibrium to the right so the degree of dissociation increases and more H + ions are produced meaning increases less than expected 3

7 Titration curves Constructing a PH curve Measure initial of the acid Add alkali in small amounts noting the volume added Stir mixture to equalise the Measure and record the to dp When approaching endpoint add in smaller volumes of alkali Add until alkali in excess Calibrate meter first by measuring known of a buffer solution. This is necessary because meters can lose accuracy on storage Can improve accuracy by maintaining constant temperature Strong acid Strong base e.g. HCl and NaOH There are 4 main types of curve. Strong acid and strong base 2. Weak acid and strong base 3 3. Strong acid and weak base Long vertical part from around 3 to 9 4. Weak acid and weak base 7 at equivalence point = 7 25 cm 3 of base 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. The equivalence point lies at the mid point of the extrapolated vertical portion of the curve. The Key points to sketching a curve: Initial and final Volume at neutralisation General Shape ( at neutralisation) Weak acid Strong base 3 e.g. CH 3 CO 2 H and NaOH Half neutralisation volume For weak acids [H + [A starts 7 Equivalence point >7 Vertical part of curve >7 (around 7 to 9) near 3 25 cm 3 of base Ka= [HA buffer region and is formed because a buffer solution is made At the start the rises quickly and then levels off. The flattened part is called the

8 At ½ the neutralisation volume the [HA = [A So Ka= [H + and pka = If we know the Ka we can then work out the at ½ V or vice versa 6

9 Strong acid Weak base e.g. HCl and NH 3 Weak acid Weak base e.g. CH 3 CO 2 H and NH No vertical part of the curve 7 Vertical part of curve <7 (around 4 to 7) 7 Equivalence point < 7 25 cm 3 of base 25 cm 3 of base Choosing an Indicator Indicators can be considered as weak acids. The acid must have a different colour to its conjugate base An indicator changes colour from HIn to In over a narrow range. Different indicators change colours over a different ranges The endpoint of a titration is reached when [HIn = [In. To choose a correct indicator for a titration one should pick an indicator whose endpoint coincides with the equivalence point for the titration HIn colour A In + H + 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 OH ions will react and remove H + ions causing the equilibrium to shift to the products. Colour B is the alkaline colour. An indicator will work if the 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. Only use phenolphthalein in titrations with strong bases but not weak bases 3 strong base Colour change: colourless acid pink alkali Use methyl orange with titrations with strong acids but not weak acids Colour change: red acid yellow alkali (orange end point) weak base range for phenolphthalein 7 weak acid strong acid range for methyl orange 25 cm 3 of base N

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