# This value, called the ionic product of water, Kw, is related to the equilibrium constant of water

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1 HYDROGEN ION CONCENTRATION - ph VALUES AND BUFFER SOLUTIONS 1. INTRODUCTION Water has a small but definite tendency to ionise. H 2 0 H + + OH - If there is nothing but water (pure water) then the concentration of H + must equal that of OH -. The actual value is 10-7 M. Thus indicating concentration by square bracket we have: [H + ] [OH - ] = M This value, called the ionic product of water, Kw, is related to the equilibrium constant of water Keq = [H + ] [OH - ] [H 2 O] As the concentration of undissociated H 2 O is large (55 M) and essentially constant, then [H + ] [OH - ] is also a constant. That is, even when impurities are present to give a preponderence of H + or OH - the relationship must hold. [H + ] [OH - ] = M H + concentration can vary between about 10 M in solution of strong acids to M in solutions of strong bases. To compact this huge range we represent hydrogen ion concentration as an exponential, designated ph. ph = -log 10 [H + ] Thus to represent a molarity of H + as a ph value we take logs and change the sign. When molarity is represented in the exponential notation its conversion to ph is easily done as we just take the value of the index and change its sign. Examples What is the ph of 0.01 M HCl? HCl is practically completely dissociated in water into H + and Cl - [H + ] = 10-2 M ph = 2 What is the ph of 0.01 M NaOH? NaOH dissociates in water to Na + and OH - Thus [OH - ] = 10-2 M But [H + ] [OH - ] = M [H + ] = M ph = 12 What is the ph of a solution where [H + ] = 2 x 10-5 M ph = -log 10 [H + ] ph = -log (2 x 10-5 ) ph = -(log 2 + log 10-5 ) ph = -(0.3-5) ph = ph = 4.7 What is the molarity of H + when ph = 7.4? ph = -log 10 [H + ] = 7.4 log [H + ] = -7.4 [H + ] = 4 x 10-8 M When [H + ] = [OH - ] the solution of course has a ph of 7 and is said to be neutral. An acid solution has a ph less than 7. An alkaline solution has a ph greater than 7.

2 2. BUFFER SOLUTIONS and SOLUTIONS of KNOWN ph It is simple to prepare: a solution of ph 2, because [H + ] = 10-2 M e.g M HCl = ph 2 or a ph 10 solution, because [H + ] = M e.g M NaOH = ph 10 But for most occasions where biological material is being studied, solutions of a ph nearer neutrality will be required. To attempt to produce a ph of 6 with 10-6 M HCl will be extremely difficult as any slight atmospheric or container contamination will alter this very small concentration. Similarly, to prepare a solution of ph 8 with 10-6 M NaOH is practically impossible just due to traces of dissolved CO 2. However, solutions can be prepared which are able to maintain a more stable H + concentration in the mid ranges of the ph scale. These solutions are called buffer solutions. They usually consist of a mixture of a weak acid or base and a definite proportion of a suitable salt. (A strong acid is one that largely dissociates in aqueous solution, i.e. it liberates many H + ions; e.g. HCl. A weak acid is one that only partially dissociates in water. Conversely, a strong base dissociates to provide many OH - ions and a weak base dissociates little giving few OH - ions). Take for example a weak acid: HA A - + H + + [ H ] = Keq Where Keq is usually referred to as Ka the dissociation constant of the acid + ] [ H ] = [ ] Ka HA log[ ] log log [ ] + HA H = Ka + + log [ H ] = log Ka + log + Let log[ H ] = log Ka = ph pka ph A = pka + log [ ] ] This equation is called the Henderson-Hasselbach equation. Suppose we now add a salt, MA, to a solution containing HA. The MA completely dissociates. The dissociation of the acid, HA, is suppressed due to the extra common ion, A -. Therefore [A - ] becomes almost entirely that of the MA, i.e. [A - ] = [SALT]. Similarly, as the dissociation of the acid is suppressed then all the acid tends to exist in the undissociated form, i.e. [HA] = [ACID].

3 The Henderson-Hasselbach equation can now be rewritten: ph = pka + log [ SALT] [ ACID] A prime mark is usually placed over pka (i.e. pká) to denote that the value was determined and is strictly valid only for a specified concentration. But within certain limits of dilution we can see that the ph of a solution is determined simply by the ratio of salt to acid. Thus knowing the pka of the acid it is possible to prepare a solution of a required ph simply by using different ratios of salt to acid. What happens if we add OH - or H + to such a solution? Adding H + : these will combine with acidic ions, [A - ], to form undissociated HA. Similarly on adding OH -, these will combine with H +. The equilibrium HA «H + + A - will be disturbed and more acid will dissociate to replace the H + ions removed. In either case, provided the concentration of HA and A - are initially made large in comparison with H +, their concentration will not be appreciably changed. Thus ph defined above as pka + log will not be largely affected. Example: To make 100 ml of 0.1 M orthophosphate buffer at ph Some compounds can donate more than one proton. Orthophosphoric acid is an example. Each dissociation can be treated separately. Phosphoric acid dissociates in successive steps to yield 3 protons per molecule, the equilibrium constants for the invididual steps are designated K 1, K 2 and K 3 corresponding to pk 1, pk 2 and pk 3. pka values are respectively 2.12, 7.2 (pk'a at 0.1 ionic strength = 6.86) and H 3 PO 4 H + + H 2 PO - K 1 = 7.5 x 10-3 pk 1 = 2.12 H 2 PO 4 H + + H PO 2-4 K 2 = 6.3 x 10-8 pk 2 = 7.22 H PO 2-4 H + + PO 3-4 K 3 = 4.0 x pk 3 = 12.4 We are mainly concerned with the pka' at 6.86 because at ph 6.32 practically none of the ions will be as H 3 PO 4 or PO4 3-. ph = pk + log 6.32 = log [HPO4 2- ] [H 2 PO - 4] [HPO 2-4] [H 2 PO - 4] [HPO 2-4] = = [H 2 PO - 4] 3 This ratio of A - to HA can be obtained either by: adding the salt to the acid in this ratio, or by adding the alkali to the weak acid to cause it to dissociate to this proportion. Both procedures are illustrated below:

4 Making orthophosphate buffer by the salt plus acid procedure First calculate the total number of moles of salt plus acid required. 100 Need 100 ml of 0.1M buffer = 0.1 x moles = 0.01 moles We can use the potassium salts, KH 2 PO 4 and K 2 HPO 4 and have 3 ions to consider; K + -, H 2 PO 4 HPO 2-4. If we let x equal the amount of KH 2 PO 4 required and x the amount of K 2 H PO 4 required. x Then = x and x = x 4x = 0.03 x = x = Thus weigh out moles KH 2 PO 4 or NaH 2 P moles K 2 HPO 4 or Na 2 HPO 4 and make up to 100 ml with H 2 O Making orthophosphate buffer by the acid plus alkalki procedure A - + HA both came from acid which must total 0.01 moles. Three parts must remain as HA whilst 1 part is neutralised with the alkali i.e. to 0.01 moles KH 2 PO 4 add moles KOH or to 0.01 moles NaH 2 PO 4 add moles NaOH. Practical consideration in making up a buffer: Because the use of pka involves an approximation as regards the effect of concentration on ph then it is best to add the calculated reagents in a volume slightly less than finally required. The ph is then obtained with a ph meter, minor adjustments are made with acid or base, and finally the solution is made up to volume. Note however, that the ph will vary with temperature. Volumetric titrations: Example: To a weak acid solution is added increments of an alkali and the resultant ph is plotted against the amount of alkali added. A curve is obtained (Figure 1). 5 ml of M OH - added ph Figure 1 The titration of a weak acid with alkali. When the curve flattens at the mid-point, half the HA has been neutralised, i.e. [HA] = [A - ]. Thus - equals 1 and log 1 = 0. Thus the Henderson-Hasselbach equation becomes ph = pka

5 The pka is therefore seen to be numerically equal to the ph at which half of the acid is dissociated (neutralised). In other words, the value of the pka of a given weak acid can be obtained by mixing equal quantities of the weak acid and its salt. Notice that the half neutralisation point marks the steepest part of the graph, i.e. it is the point at which the ph changes least on adding OH - (or H + ). Thus if we require a buffering solution we use a weak acid whose pka is close to the desired ph. As long as is close to 1 then relatively large changes [ A ] in [HA] and [A - ] will result in only small changes in ph. To obtain a ph one unit away from the pka then must change tenfold in order that log changes by 1. At a ph of one unit above and below the pka' the ratio of salt to acid is 10 and 0.1 respectively. Now very small changes in [HA] and [A - ] will cause large changes in ph. If 4 mmoles H + are added to a 1 ml solution containing 5 mmoles A - and 50 mmoles HA. The ratio of then changes from to. 54 The addition of the acid has thus lead to a ph shift from pka + (-1) to pka + (-1.7). 3. THE MEASUREMENT OF ph Electrometric ph determinations: The voltage is measured between two half cells, one of known potential and the other an unknown, which contains as part of the chemical system the H + ion solution under test. The standard electrode has been set as the Hydrogen electrode. This consists of a piece of inert metal such as platinum covered with platinum black by electrolysing a solution of platinic chloride. The platinum black is saturated with hydrogen gas and maintained in this condition. The platinum black can catalyse the reaction. ½H 2 H + + e - The metal in fact becomes covered with electrons until equilibrium has been reached. This then has a certain potential. At unit activity of H + ions this potential is arbirtarily called zero. The voltage of other half cells is given in reference to this standard. The usual reference cell is the calomel electrode, i.e. metallic mercury in contact with KCl solution saturated with mercurous chloride. The other half cell is now nearly always the glass electrode (Figure 2). Silver/silver electrode HCl ph sensitive glass Figure 2 A silver electrode for ph measurement.

6 The glass behaves as a site for hydrogen ion transfer and the potential at the surface changes as H + ion are acquired or lost. Thus we have: Reference electrode Salt Bridge Electrode reversible to H + e.g. calomel : Hg saturated KCl soln. test soln:glass:hcl AgCl:Ag The ph is directly proportional to the e.m.f. and the ph is given by: 0 ph = E E RT F E = voltage measured with potentiometer. E 0 = contribution by reference half cell. F = Faraday s constant R = Universal gas constant T = Absolute temperature At 25 c voltage change = 59 mv/ph unit. Indicator solutions: The ph can be determined by the use of indicator solutions. Acid-base Indicators are weak organic acids; the colour of the undissociated acid molecule is different from that of the ionised form. There is an equilibrium between the ions and HA HA A - + H + and the acid will have its own pka. In a strongly acid solution the dissociation of the indicator will be suppressed and the colour will be predominantly that of HA. If the solution becomes strongly alkaline the dissociation will near completion and the colour will become that of A -. However from one ph unit below the pka to one ph unit above the pka the ratio of changes slowly from 0.1 to 10. Thus the indicator phenol red, pka 7.9 will contain at ph 6.9 ten times as many molecules of the yellow undissociated form as of the red dissociated form. It could not be used to differentiate between a ph of 5 and 4 as the proportion of molecules changing from red to yellow would be too small for observation. Similarly, at ph 10 it would be virtually all red and could not differentiate between two solutions of ph 10 and 12. Thus indicators are useful in differentiating ph values only within the range of about one ph unit above the pka of the indicator. Indicators are used for two main purposes: to measure the ph of a given solution. to indicate when a titration is complete.

7 For the first use we need an indicator with a pka near the ph to be measured. E.g., say we are titrating a weak acid to find how concentrated it is. At what point do we terminate the titration? At the pka of the acid we have only titrated half of it. Suppose we would like to know the amount of acid to within 0.1%. We must continue to titrate until a ph is reached where 99.9% of the HA is converted to A -. Take a weak acid pk = 5 At 99.9% Neutralisation = = 3 [ ] ph pka A = ph = 5+3 ph = 8 Thus for this degree of accuracy we need to select an indicator which will give its main colour change three ph units above the pka of the acid being titrated.

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