Equilibrium in Solutions

Transcription

1 1 The law of mass action Equilibrium in Solutions Chemical reactions may be reversible or irreversible. In this latter case, the concentrations of the reactants decrease as the reaction proceeds, to reach some equilibrium values. Consider the hypothetical reaction: A + B C + D Let the initial concentrations of A and B be c AO and c BO respectively. The initial concentrations of C and D are normally 0. Fig. (6.1) shows that, as the reaction proceeds, the concentrations of the reactants decrease while those of the products increase. After enough time, equilibrium concentrations are reached. They are given the symbols [A], [B], g.mole/dm 3 Conc. c Ao or c Bo [A] or [B] [C] or [D] c Co or c Do time Fig.(6.1) Variation of conc. with time The law of mass action states that, at equilibrium, there will exist a constant ratio between the product of the concentration of the products and that of the reactants. This is written in the form: [C].[D] = K e (1) [A].[B] Where K e is called the equilibrium constant of the reaction.

2 This constant is a function of temperature and the total pressure (in reactions involving gases). The latter equation indicates that, if the concentration of one of the products, in the equilibrium mixture, decreases, then the concentrations of the reactants have to decrease in order that the الطرف اايسررفRHS remains constant. This means that the reaction will proceed in the forward direction.this is the case when one of the products can be eliminated from solution. For example, consider the following reaction in solution: NH 4 Cl + NaOH = NaCl + NH 4 OH This reaction is reversible with a low value of K e. This means that it is far from being complete. If, however, the solution is heated, then NH 4 OH will decompose to NH 3 and water vapour, causing the concentration of the NH 4 OH in solution to decrease. This will move the reaction towards the production of more NH 3. In the following sections are presented the applications of this rule concerning equilibrium in solution. 2 Dissociation of weak acids and bases Consider a weak acid HA (such as acetic acid, carbonic acid, ) of concentration c ma g.mole/dm 3. In solution, its dissociation to ions will only be partial, in contrast with strong acids (such as HCl, HNO 3, ) which are fully dissociated in solution. HA + H 2 O = A + H 3 O + The dissociation of water being negligible, equation (1) becomes : [A ].[H 3 O + ] = K a (2) [HA] Consider one g.mole of the acid in V dm 3 of solution. Hence c ma = 1/V g.mole/dm 3 If, out of this 1 mole, the number of dissociating moles = x, then there will be (1 x) mole acid left and, as x mole of the acid dissociate, we get x g. ion of A and H 3 O +. Therefore, at equilibrium [A ] = [H 3 O + ] = x and [HA] = 1 x V V Replacing in equation (2), we get : x. x V V = K a (3) 1 - x V

3 Recalling that c ma = 1/V, we get : c ma.x 2 = K a (4) 1 x A result known as Otswald's rule If the value of x is very low, which is a common case, then x can be neglected with respect to 1, and the latter equation becomes : c ma.x 2 K a or x K a /c ma (5) In this case: [H 3 O + ] = c ma.x c ma.k a (6) The value of K a, known as the acid dissociation constant, is usually very low. For example, it is 3.2x10-8 for hypochlorous acid HClO and 6.4x10-5 for benzoic acid,. Equation (5) shows that as c ma decreases, that is, as the solution becomes more diluted, the extent of dissociation increases. This rule does not apply in case of concentrated acids. The same analysis applies for the dissociation of weak bases. An equation similar to (6.4) is obtained for the dissociation reaction: MOH = M + + OH c mb.x 2 = K b (7) 1 x The value of K b, known as the base dissociation constant is also very low for weak bases. 3 Ionic product of water Water dissociates very weakly in solution: 2 H 2 O = H 3 O + + OH Applying the law of mass action, we get: [H 3 O + ].[OH ] = K w (8) The constant K w is known as the ionic product of water. At 25 C, it equals For pure water, [H 3 O + ] = [OH ] = 10-7 In order to avoid using negative powers, it was found more convenient to use the following definition for the ph value أنرب ph = - log [H 3 O + ] (9) Similarly, we define poh = - log[oh ], pk w = - log = 14 Hence, ph + poh = pk w = 14 (10) For pure water, and since [H 3 O + ] = [OH ], we get ph = 7

4 In acidic medium, the main source of H 3 O + ions is the dissociation of the acid. Even for weak acids, it is common to نتغاضر انر اignore the dissociation of water and consider that [H 3 O + ] = c o of the acid. This is normally higher than 10-7, so that [H 3 O + ] > 10-7 and hence, ph < 7 In basic medium, the main source of OH ions is the dissociation of the base.here also, it is common to ignore the dissociation of water and consider that [OH ] = c o of the base. This is normally lower than 10-7, so that [OH ] > 10-7 and hence, [H 3 O + ] < 10-7 and hence ph > 7 This is summarized in the following diagram: Acid neutral basic 4 Solubility product When a small quantity of a sparingly soluble شريح االربانا ionic salt is added to water it first dissolves, then, as the quantity is increased, saturation تشرع is reached. At that stage, there will exist an equilibrium between the concentrations of the ions in solution and the undissolved salt: MA (solid) = M + (aqueous) + A (aqueous) Since the amount of solid salt is almost constant, then, the law of mass action can be written as: [M + ].[ A ] = K s (6.11) The value K s is called the solubility product and is constant for a certain salt at constant temperature. It is always measured at 25 C. The following table shows the values of the solubility product of some common sparingly soluble salts. Salt BaSO 4 CaCO 3 AgCl PbS ZnS K s 1.0x x x x x10-24 Table 1 Solubility products of some common salts

5 Consider now the case of MA. In case of a pure salt in solution, the ionic concentrations of the two ions will be equal: [M + ] = [A ] = x g mole/ dm 3 and hence x 2 = K s For an ion like PbCl 2, we get : PbCl 2 = Pb Cl and hence 2 [Pb ++ ] = [Cl ] = 2x, and since [Pb ++ ].[Cl ] 2 = K s, we get : (x).(2x) 2 = K s, from which we get : 4x 2 = K s Similarly, in the case of Bi 2 S 3, we get : (2x) 2.(3x) 3 = 108x 5 = K s In all of the previous examples, x represents the molar concentration of the salt dissolved in the saturated solution. In a chemical reaction involving the formation of a sparingly soluble product MA, this product will precipitate when [M + ].[A ] > K s. Now, consider the presence of another source of the M + or the A ion in solution in addition to the ionization of MA. To understand what happens in this case, consider, for example, the solubility of AgCl in water: [Ag + ].[Cl ] = 1.1x10-10 (i) In case of pure water, if [Ag + ] = x, then, [Ag + ] = [Cl ] = x Hence, x 2 = 1.1x10-10 and x = 1.05x10-5 g mole/dm 3 Consider now the solubility of AgCl in a 0.1M solution of NaCl The main source of Cl ion is now NaCl and [Cl ] 0.1 g mole/dm 3 Substituting in (i), [Ag + ].(0.1) = 1.1x10-10, hence [Ag + ] = 1.1x10-9 This shows that the solubility of AgCl has decreased from 105x10-5 to 1.1x10-9 g mole/dm 3 due to the presence of the common Cl ion. This is known as the common ion effect. This principle states that: The solubility of a salt in solution is reduced in presence of a common ion in solution. 5 Titration of a strong acid with a strong base Consider the reaction of a strong acid AH with a strong base MOH. In solution, both are fully ionized. The number of moles of the base necessary to complete the reaction is n B. It is equal to the number of moles of acid n A. This is apparent from the reaction: AH + MOH = MA + H 2 O It will be more convenient to express the molar concentrations of acid and base as M A and M B resp. (not to be mistaken with mol. weights) then: M A = n A / V A and M B = n B / V B Hence n A = M A.V A and n B = M B.V B

6 So, the condition for a complete neutralization is: M A.V A = M B.V B (12) When this condition is تتيقر اfulfilled, we say that the end point of hasامعاسفةtitration been reached. In the following analysis, we are going to assume that a strong base is gradually added to a constant volume V A of a strong acid, so that the volume of the base V B is variable. We follow now the variation of the ph of the solution with increasing V B. (a)in case of pure acid, V B = 0 and the concentration of H 3 O + in solution is that of the acid M A. So that: ph = - log M A (13) (b)if the amount of base is less than that required for neutralization: In that case, M B.V B < M A.V A and we are left with an excess of acid the number of moles of which is M A.V A - M B.V B. Therefore, [H 3 O + ] = M A.V A - M B.V B The ph value is therefore: ph = - log M A.V A - M B.V B (14) (c) At the end point: To understand what happens at the end point, we need to follow the slope of the ph - V B curve. To this effect, we first calculate d[h 3 O + ] dv B This gives: - M B.( ) - 1.(M A.V A - M B.V B ) = -2 M A.V A ( ) 2 ( ) 2 Now, since ph = - log [H 3 O + ] = ln [H 3 O + ], so that the slope is d ph = d[h 3 O + ]/dv B d V B [H 3 O + ] We notice that the slope reaches as [H 3 O + ] = 0 This will happen when the end point is reached at ph =7. At that point conditon (12) is fulfilled. (d)if the amount of base is more than that required for neutralization As the amount of base added increases over that required for neutralization, we are left with an excess number of moles of base that equals M B.V B - M A.V A so that [OH ] = M B.V B - M A.V A

7 Hence poh = - log M B.V B - M A.V A From equation (10), we get : ph = 14 + log M B.V B - M A.V A (15) Fig. (2) shows the ph V B curve This curve shows a very important feature. It is clear that the presence of a point of inflection at ph = 7 at which the slope is infinite means that the variation of this relation is very rapid about this value. That is, any slight variation of V B has for effect to cause the ph to vary appreciably. This is clear from the following numerical example. If the neutralization point has been reached,then we are in presence of equal moles of acid and base. Let us have 0.1 g moles of each in a 1 dm 3 of each (that is a total volume of 2 dm 3 ).If now we add only a few drops of base, say the equivalent of 1 cm 3 then the number of moles of base in excess of the acid is equal to M B.V B = 0.1X0.001 moles = moles phا 7 Fig. (2) ph / V B titration curve ( strong acid + strong base ) Then [OH ] = /2.005 from which [OH ] , poh 3.3 Hence, ph = So we notice that 1 cm 3 of base added to 2 dm 3 of neutral solution has raised the ph from 7 to 10.7 We conclude from the previous example that it is difficult to reach exactly the end point in the neutralization of a strong acid with a strong base. V B ا

8 6 Titration of a weak acid with a strong base This is the case when a weak acid HA ( such as acetic acid ) is titrated against a strong base MOH ( such as sodium hydroxide ). (a) In case of pure acid Since the acid is weak then the value of [H 3 O + ] can be approximately calculated from equation (6) : [H 3 O + ] = c ma.x c ma.k a Hence, we get: ph = log c ma.k a ( or log M A.K a ) (16) (b) If the amount of base is less than that required for neutralization: This means that we have an excess number of moles of acid that equals M A.V A - M B.V B, the concentration of which will be : [HA] = M A.V A - M B.V B (i) The portion of the acid that has reacted with the base through the reaction: AH + OH = A + H 2 O must yield acid ions having the same concentration as that of the base : [A ] = [OH ] = M B.V B (ii) V A +V B The excess acid remaining will dissociate according to the reaction; HA + H 2 O = H 3 O + + A Applying the law of mass action on that dissociation: [H 3 O + ].[A ] = K a (iii) [HA] Substituting from (i) and (ii) in (iii), we get: [H 3 O + ]. M B.V B V A +V B = K a M A.V A - M B.V B From which we get: [H 3 O + ] = Ka. ( M A.V A - M B.V B ) M B.V B Finally, we can write: ph = -log Ka. ( M A.V A - M B.V B ) (17) M B.V B

9 (c) At the end point: At that stage, all of the acid has reacted with all of the base: M A.V A = M B.V B (12) The resulting salt MA being that of a weak acid, will partially hydrolyze according to the scheme: MA + H 2 O = MOH + HA, or A + H 2 O = OH + HA It is clear that [OH ] = [HA] (i) Now, we write the law of mass action as applied to the last reaction: [OH ].[HA] = K h (18) [A ] Where K h is an equilibrium constant for hydrolysis. From (i), we get: [OH ] 2 = K h (19) [A ] Note now that [A ] represents the concentration of the salt MA present at equilibrium. The number of moles of the salt formed by the neutralization reaction is equal to the number of moles of acid or base obtained from equation (12) = M A.V A = M B.V B The concentration of the salt is therefore: [MA] = [A ] = M A.V A Substituting in equation (19), we get: [OH ] 2 = K h. M A.V A Taking logarithms and recalling that ph =14 - poh, we get: ph = log K h. M A.V A (20) 2 When the relation ph = f (V B ) is plotted from equation (17), we get the curve shown in Fig. (3). It is clear from this curve that the point of inflection is not accompanied with an infinite slope, meaning that the الاسزدادانشكلافجائ shape of the curve at the end point does not raise abruptly Also, the ph at which the end point occurs, as calculated from equation (20) is higher than 7. It should be noted that the value of K h can be obtained by combining equations (2) and (8) since : [A ].[H 3 O + ] = K a (2) and [H 3 O + ].[OH ] = K w (8) [HA] Hence, K w = [OH ].[HA] = K h (21) K a [A ]

10 ph 10 V B Fig. (2) ph / V B titration curve ( weak acid + strong base ) (d)if the amount of base is more than that required for neutralization The remaining alkali will have a concentration equal to M B.V B - M A.V A This is equal to [OH ], since the base, being strong, will be totally ionized. Therefore the value of poh = - log M B.V B - M A.V A And we get back to equation (15): ph = 14 + log M B.V B - M A.V A (15) Similar equations can be deduced for the titration of a strong acid with a weak base. In this case, the end point will be reached at ph < 7. 7 Indicators An indicator is a solution which is used to detect theاإسرتالال end point in a neutralization reaction. It is usually a weak organic acid that changes its colour according to the ph of the medium. Let the indicator be HIn. It dissociates according to the reaction: HIn + H 2 O = In + H 3 O +, and [In ].[H 3 O + ] = K i [HIn]

11 Hence, [H 3 O + ] = K i.[hin] (22) [In ] Therefore, ph = - log [H 3 O + ] = log 1 [H 3 O + ] And, ph = log [In ] (23) K i.[hin] As a rule, the colour of [HIn] must be different from [In ]. In order that the colour of In prevails سرردد, its concentration should be at least 10 times grater than that of the non-ionized acid HIn. That is, the colour of the solution will be that of In if: [In ] > 10 (24) [HIn] Equation (23) can be re-written as: ph = log [In ] - log K i = log [In ] + pk i [HIn] [HIn] Condition (24) is therefore written in the form: ph > 1 + pk i (25) Similarly, the colour of HIn will prevail if: ph < pk i - 1 (26) Therefore, the ph will change in the range: pk i - 1 < ph < 1 + pk i (27) This means that, in order to choose a proper indicator for a certain neutralization reaction, its end point should lie in the range specified by equation (6.27). For example, phenol phtaleine has a K i value of 7x10-10, pk i = 9.15 and can therefore be used for the neutralization of a weak acid and a strong base since it will change its colour in the ph range Methyl orange, on the other hand, has K i = 2x10-4, pk i = 3.7 and can be used for the titration of a strong acid and a weak base. It will change colour in the ph range Finally, نعاداالشمسااlitmus, has K i = 3x10-7, pk i = 6.5 and will change its colour in the Ph range , making it suitable for the titration of a strong acid against a strong base.

12 8 Buffer solutions These are solutions which are not لحرتاحراسةsensitive to variations in ph when acids or bases are added. They consist of a weak acid HA and one of its salts MA, such as acetic acid and sodium acetate. The weak acid undergoes hydrolysis: [A ].[H 3 O + ] = K a (2) [HA] The salt is fully ionized: MA = M + + A So that [A ] = [MA] Equation (2) is therefore written as: [MA].[H 3 O + ] = K a [HA] The acid molar concentration of the acid is [HA] and the salt concentration is [MA], we get: [H 3 O + ] = K a. [HA] [MA] Since ph = - log [H 3 O + ] = log 1, we get : [H 3 O + ] ph = log [MA] - log K a or ph = pk a + log [MA] [HA] [HA] The ph of a buffer solution can be therefore obtained from the relation: ph = pk a + log [salt] (28) [acid] 8 Solved examples Example 1 Calculate the ph of a 0.1M solution of acetic acid, Ka = 1.8x10-5. Solution: From equation (6.16) : ph = log c ma.k a ph = log (0.1).(1.8x10-5 ) ph = 2.87

13 Example 2 What is the ph of 0.01M solution of nitrous acid, K a = Solution: The concentration of the acid is low and its equilibrium constant is relatively elevated. So it is better not to use the approximation of the last example. Otswald's rule has to be applied in its original form: c ma.x 2 = K a (4) 1 x (0.01).x 2 = x x x = 0 solving we get : x = 0.2 [H 3 O + ] = c ma.x = 0.01.(0.2) = 0.002, ph = 2.7 Note that if the approximate form (6.4) were used, we would have obtained : ph = log (0.01).(0.0005) = 2.65 Example dm 3 of a 0.001M solution of CaCl 2 is added to 0.5 dm 3 of a 0.001M solution of MgSO 4. If the solubility product of CaSO 4 = 2x10-5, will CaSO 4 precipitate? If the molar concentrations of the two salts are equal and the same volumes are used, what should be the minimum concentration of either salt for CaSO 4 to precipitate? Solution: The number of moles of CaCl 2 and of MgSO 4 = (0.001).(0.5) =.0005 The total volume being 1 dm 3, [Ca ++ ] = [SO4 ] = g.ion/dm 3 Hence [Ca ++ ]. [SO4 ] = (0.0005).(0.0005) = 2.5x10-7 < S p Let the minimum molar concentration of each salt = x, then precipitation will occur if x 2 > S p, that is : x 2 > 2x10-5 or x > i.e. the minimum concentration of either salt = g mole/dm 3

14 Example 4 Calculate the ph of a solution obtained by adding 0.35 dm 3 of 0.1M sodium hydroxide solution to 0.8 dm 3 of 0.05 HCl solution. Solution: The number of moles of base = 0.35x0.1 = moles The number of moles of acid = 0.8x0.05 = 0.04 moles The number of moles of the base is not sufficient for neutralization. There will remain = moles acid. The total volume = = 1.15 dm 3 Since all of the acid is ionized, [H 3 O + ] = 0.005/1.15 = mole/dm 3 ph = - log = 2.36 Example 5 Calculate the ph at the end point when 0.5 dm 3 of 0.1M solution of hypochlorous acid is neutralized by a 0.05 M solution of sodium hydroxide. ( K a = 3.2x10-8 ) Solution: The volume of base at the end point is obtained from : (0.5).(0.1) = V b. (0.05), hence V b = 1 dm 3 The value of K h can be obtained from : K h = /3.2x10-8 = 3.12x10-5 ph = log K h. M A.V A (6.20) 2 ph = 14 + (0.5).log (3.12x10-5 ).(0.1)(0.5) ph = 10.5 Example 6 What is the percent decrease in ph value when 1 cm 3 of HCl 0.1M is added on 1 dm 3 of pure water. Repeat the calculations when the same is added on 1 dm 3 of a buffer solution containing 0.05 moles acetic acid and 0.05 moles sodium acetate. (K a = 1.8x10-5 ) Solution: HCl is totally ionized, then the number of moles of H 3 O + = 0.001x0.1 = moles. [H 3 O + ] = / mole/dm 3, giving ph = 4 Since the original ph of water is 7, then : The percent decrease in ph = [( 7-4 ) / 7 ]x100% 43% If we consider the buffer solution :

15 The concentration of both acid and salt are equal = 0.05/1 = 0.05 Its ph = - log K a + log 0.05/0.05 = - log K a = - log 1.8x10-5 = 4.74 As HCl is added, it will react with sodium acetate giving off an equivalent number of moles of acetic acid and decreasing the number of moles of the salt by an equivalent amount : the number of moles of [H 3 O + ] = 0.05x = the number of moles of salt = 0.05x = from equation (6.28) : ph = log ( /0.0501) = The percent decrease in ph = [( )/4.74]x100% = 0.036% Questions 1- Give reasons for what follows: - A solution of iron (III) chloride turns litmus to red. - The reaction between ammonium chloride and sodium hydroxide does not go to completion unless the solution is heated. - The ph of a 10-7 M HCl solution is not If the dissociation constant of acetic acid is 1.8x10-5, find the ph of the following mixtures of acetic acid ( 0.1M ) and sodium hydroxide (0.1 M): pure acid 20 dm 3 of acid and 10 dm 3 of base 20 dm 3 of acid and 20 dm 3 of base 20 dm 3 of acid and 30 dm 3 of base pure base 3- If the dissociation of H 2 X takes place on two steps: H 2 X = HX + H + K 1 HX = X + H + K 2 And K 1 >> K 2. Prove that [X ] K 2 4- What is the solubility of barium sulphate in pure water? What would be its solubility in a solution 0.05M sodium sulphate? cm 3 of nitric acid 0.02M are mixed with 150 cm 3 of the same acid 0.01M. Calculate the ph of this mixture. How much water has to be added to raise the ph by 10%? 6- Human blood has a constant ph of 7.4 due to an equilibrium between HPO4 and H 2 PO 4 ions. If the equilibrium constant of the reaction H 2 PO 4 = HPO 4 + H + is 6.4x10-8, Find the ratio between these two ions.