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# 05 aqueous solutions

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### 05 aqueous solutions

1. 1. Many analytical methods rely on equilibrium systems in aqueous solution.This unit will review General concepts of aqueous solutions Chemical equilibrium Equilibrium calculations Deviations from ideal behavior
2. 2. acid base conjugate conjugate base acid conjugate conjugate base of HF acid of H2O acid base base strength acid strengthFor the general chemical reaction: kinetic equilibrium region region aA + bB cC + dDIf A and B are brought together: A There is an initial reduction in the B concentrations of A and B. C Both C and D will increase in concentration. D We reach a point where the concentrations no longer change. time
3. 3. These are dynamic equilibria Equilibrium concentrations are based on: The specific equilibrium At equilibrium, the forward and reverse rates of The starting concentrations reaction are equal. Other factors such as: Temperature Any given species is constantly changing from Pressure one form to another. Reaction specific conditions Changes in the system will alter the rates and a Altering conditions will stress a system, new equilibrium will be achieved. resulting in an equilibrium shift. aCc aDd Keq = a = activity aAa aBb [ C ]c [ D ]d Keq = [ A ] a [ B ]b
4. 4. We’ll be using molar concentrations when working with chemical equilibrium. This introduces errors that you should be aware of. While we will not work with activities, we need to know what they are. ( )Except for dilute systems, the effective concentration of ions is usually less than the actual concentration.The term activity is used to denote this effective concentration.activity ai = fi [ i ] where fi is the activity coefficient for i [ i ] is the molar concentration
5. 5. For very dilute solutions As fi -> 1, ai -> [ i ] For µ up to 0.1 0.5 Zi2 µ ! fi < 1 and ai < [ i ]1 + 0.33 "i µ! When u is > 0.1 Results in complicated behavior. In general, if µ < 0.01, we can safely use molar concentrations. [ H3O+ ] [ OH- ] Keq = [ H2O ]2 In dilute solutions, [ H3O+] and [ OH- ] is much smaller than [ H2O ]. [ H2O ] is essentially a constant of ~55.5 M.
6. 6. KW = 10-14 = [H3O+] [OH-] [H3O+] = [H3O+]water + [H3O+]HCl [OH-] = [H3O+]water Lets set x = [H3O+]water then [H3O+] = x + 1.0 x 10-8 [OH-] = x 10-14 = (x + 1.0 x 10-8) ( x)Our equation can be rearranged as x = -10-8 + [ (10-8)2 + 4x10-14]1/2 / 2 x2 + 10-8 x - 10-14 = 0 = 1.9 x 10-7 MThis quadratic expression can be solved by: pH = 6.72 -b + b2 - 4ac x= So adding a small amount of HCl to water 2a DOES make it acidic. Only the positive root is meaningful in While this approach is more time equilibrium problems. consuming, you’ll find it very useful as our problems get more complex.
7. 7. KSP expressions are used for ionic materials that At equilibrium, our system is a saturated solution are not completely soluble in water. of silver and chloride ions. Their only means of dissolving is by dissociation. The only way to know that it is saturated it to observe some AgCl at the bottom of the solution. AgCl(s) Ag+ (aq) + Cl- (aq) [ Ag+ (aq) ] [ Cl- (aq) ] As such, AgCl is a constant and KSP expressions Keq = do not include the solid form in the equilibrium [ AgCl(s) ] expressionDetermine the solubility of AgCl in water at 20oC in grams / 100 ml. KSP = [Ag+] [Cl-] = 1.0 x 10-10 At equilibrium, [Ag+] = [Cl-] so 1.0 x 10-10 = [Ag+]2 [Ag+] = 10-5 M g AgCl = 10-5 mol/l * 0.1 l * 143.32 g/mol Solubility = 1.43x10-4 g / 100 mlKSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ] KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ][CrO42-] = [CrO42-]Ag2CrO4 + [CrO42-]Na2CrO4 [CrO42-] = 0.010 M [ Ag+ ] = ( KSP / [ CrO42- ] ) 1/2With such a small value for KSP, we can assume that is [CrO42-]Ag2CrO4 negligible. = 1.1 x 10-12 / 0.010 M If we’re wrong, our silver concentration will be = 1.1 x 10-5 M significant (>1% of the chromate concentration.) [ Ag+ ] << [ CrO42- ] Then you’d use the quadratic approach. so our assumption was valid.
8. 8. [H3O+] [A-] [HA] Water is not included in the expressions because it is a [OH-] [BH+] constant [B] KA = [ H3O+ ] [ A- ] = 2.24 x 10-5 KA = 2.24 x 10-5 = X2 / (0.1 - X) [ HA ] Rearranging give us:Since both a H3O+ and a A- is produced for X2 + 2.24x10-5X - 2.24x10-6 = 0 each HA that dissociates: [ H3O+ ] = [ A- ] We can solve this quadratic or possibly assume that the amount of acid that [ HA ] = 0.1 M - [ H3O+ ] dissociates is insignificant (compared to the undissociated form) Lets set X = [ H3O+ ] KA = 2.24 x 10-5 [ H3O+ ] = [ A- ] [ HA ] = 0.1 M - [ H3O+ ] KA / 0.1 < 10-3 - can assume that [HA] = 0.1 M 2.24x10-5 = X2 / 0.1M X = (2.24x10-6)1/2 = 0.00150 pH = 2.82
9. 9. -2.24x10-5 + [(2.24x10-5)2 +4x2.24x10-6]1/2 Now, lets go for the exact solution X= 2 Earlier, we found that X2 + 2.24x10-5X - 2.24x10-6 = 0 X = 0.00149 pH = 2.82 -b + b2 - 4ac X= 2a No significant difference between our two answers. If you are starting with an acid, acidic conditions or the conjugate acid of a base Do your calculations using KA If starting with a base, basic conditions or the conjugate base of an acid Do your calculations using KB You can readily convert pH to pOH and KA to KB values.With complex formation, two or more species will join, forming a single, new species. If we are evaluating the decomposition of a complex, we can use a K decomposition. Mn+ + xL M(L)xn+ M(L)xn+ Mn+ + xL [ Mn+ ] [ L ]m [ M(L)x ] n+ KD = [ M(L)xn+ ] KF = [ Mn+ ] [ L ]m Since water is not a portion of these equilibria, KD = 1 / KF
10. 10. Equilibrium expressions for REDOX systems are derived from standard electrode potentials. 6Fe2+ + Cr2O72- + 14 H3O+ 6Fe3+ + 2Cr3+ + 21H2O [Fe3+]6 [Cr3+]2 KREDOX = [Fe2+]6 [Cr2O72-][H3O+]14 We’ll review how to determine and work with KREDOX when we cover the units on electrochemistry. [H3O+] [H2PO4-] K A1 = Note: [H3PO4] [H3O+] is the same for each expression. [H3O+] [HPO42-] The relative amounts of K A1 K A2 K A3 K A2 = each species can be [H2PO4-] found if the pH is known. The actual amounts can be found if pH and total [H3O+] [PO43-] H3PO4 is known. K A3 = [HPO42-] One possible ‘first step’ would be to determine the relative amounts of each species. K A1 [H2PO4-] = [H3 O+] [H3PO4] K A2 [HPO42-] = [H3 O+] [H2PO4-] K A3 [PO43-] = [H3O+] [HPO42-]
11. 11. 7.5 x 10-2 [H2PO4-] Total phosphate = 1.0 M so 10-7 = [H3PO4] = 750000 1.0 = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-] 6.2 x 10-8 [HPO42-] = = 0.62 This is referred to as a mass balance. 10-7 [H2PO4-] Based on the ratio, we know that [H3PO4] and 4.8 x 10-13 [PO43-] [PO43-] are not present at significant levels for = = 4.8x10-6 this mass balance so: 10-7 [HPO42-] These ratios show that only H2PO4- and HPO42- 1.0 = [H2PO4-] + [HPO42-] are present at significant levels at pH 7. [HPO42-] Since you now know the concentration of H2PO4-, = 0.62 you can now sequentially solve the other [H2PO4-] equilibrium expressions.[HPO42-] = 0.62 [H2PO4-] This approach is very useful when dealing with complex equilibria.1.0 = [H2PO4-] + 0.62 [H2PO4-] = 1.62 [H2PO4-] The mass balance is on means of eliminating ‘insignificant’ species based on addition or[H2PO4-] = 1.0 / 1.62 = 0.617 M subtraction