Treatment for Multiple Chem Equil

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Treatment for Multiple Chemical EquilibriumMass balance and charge balance equation18 Acid-Base Equilibria3Mass and Charge Balance EquationsIn a solution of electrolytes, the stoichiometry leads to mass or material balance (mbe) and charge balance equations (cbe). The merit is to consider the various ion species and equilibria in the solution when we write these equations.In a 0.10 M NaHCO3 solution, mbe: 0.10 = [Na+] = [H2CO3] + [HCO3 ] + [CO32 ] cbe: [Na+] + [H+] = [HCO3 ] + 2 [CO32 ] + [OH]Consider these equilibrium eqn for mbe and cbe:NaHCO3 = Na+ + HCO3-HCO3- + H2O = H2CO3 + OH-HCO3- + H2O = CO32- + H3O+Charge Balance EquationsThe charge balance equation is an algebraic statement that the sum of all positive charges in the solution must equal the sum of all of the negative charges. In writing charge balance equations, the coefficient in front of each species is always the magnitude of the charge on that ion; example, for the phosphate ion, 3[PO4].

Charge balance in a solution that contains 0.0250 M KH2PO4 and 0.0300 M KOH. The sum of positive charges = sum of negative charges.Charge Balance EquationsCharge Balance Equations - An algebraic statement of electroneutrality. The concentration of the sum of the positive charges = the concentration of the sum of the negative charges.General Form of the Equation

n1[C1] + n2[C2] + = m1[A1] + m2[A2] +

n1 and m1 represent the magnitude of the charge of the ion[C1] and [A1] represent the concentration of each cation and anion respectively

Charge Balance ExamplesGiven [H+] = 5.1 x 10-12 M [K+] = 0.0550 M[OH-] = 0.0020 M[H2PO4-] = 1.3 x 10-6 M[HPO42-] = 0.0220 M[PO43-] = 0.0030 MFind the solution for the Charge Balance equation.Charge Balance ExamplesWrite the charge balance equation for a solution containing H2O, H+, OH-, ClO4-, Fe(CN)63-, CN-, Fe3+, Mg2+, CH3OH, HCN, NH3, and NH4+.Charge Balance ExamplesWrite a charge balance equation for aqueous solution of glycine, which reacts as follows:

+H3NCH2CO2- H2NCH2CO2- + H++H3NCH2CO2- + H2O +H3NCH2CO2H + OH-Charge Balance ExamplesWrite a charge balance equation for a solution of Al(OH)3 dissolved in 1 M KOH. Possible species are Al3+, AlOH2+, Al(OH)3, Al(OH)2+, and Al(OH)4-Mass Balance EquationMass Balance (Material Balance) -Statement of Conservation of Matter - The quantity of all species in a solution containing a particular atom (or group of atoms) must be equal to the amount of that atom (or group of atoms) delivered to the solution.Mass Balance Example (known concentrations)Write the mass balance equation for the acetate group of atoms in a 0.05 M solution of acetic acid.

Write the mass balance equation for the phosphate group of atoms in a 0.025 M solution of phosphoric acid.Mass Balance Example (known concentrations)Write the mass balances for K+ and for phosphate in a solution prepared by mixing 0.0250 mol KH2PO4 plus 0.0300 mol KOH and diluting to 1.0 L.

Write a mass balance equation for a 0.05 M solution of glycine in water.Mass Balance Example (known concentrations)Suppose that 0.30 g of AlOOH (FM = 59.99) plus 150 mL of 3.0 M KOH are diluted to 1.0 L to give the species Al3+, AlOH2+, Al(OH)3, Al(OH)2+, and Al(OH)4- Mass Balance Example (unknown concentrations)Write the mass balance equation for La(IO3)3.

Write the mass balance equation for a saturated solution of slightly soluble Ag3PO4, which produces PO43- and 3Ag+ when it dissolves.Mass Balance (or material-balance) EquationsThe mass-balance equation is statement of the conversation of matter. For example in considering a 0.050 M solution of acetic acidCH3CO2H which dissociates CH3CO2H(aq) < == > CH3CO2 (aq) + H+(aq)0.050 M = [CH3CO2H] + [CH3CO2]

If there are several products as in the ionization of apolyprotic acid, the mass balance must include allpossible species.

Mass balance for the solubility of magnesium hydroxide (see page 153)

Approach 1: using magnesium as our reference Step 1: because the starting material is Mg(OH)2,the amount of magnesium ions (denoted by X) we have in the solution should be half of the hydroxide released by Mg(OH)2 (denoted by Y). i.e. X = Y Note that the total amount of hydroxide (denoted by Z) in solution include the hydroxide (Y) from Mg(OH)2 and from the hydroxide water that is equal to the concentration of [H+] from water, i.e. Z = Y + [H+]; thus Y = Z [H+] Step 2: X exists in water in the forms: [Mg2+], [MgOH+], therefore X = [Mg2+] + [MgOH+], hydroxide Z exists in the forms [OH-] and [MgOH+], therefore Z = [OH-] + [MgOH+], thus Y = [OH-] + [MgOH+] - [H+]; Step 3: given that X = Y [Mg2+] + [MgOH+] = ([OH-] + [MgOH+] - [H+]); 2( [Mg2+] + [MgOH+]) = [OH-] + [MgOH+] - [H+]; 2( [Mg2+] + [MgOH+]) + [H+] = [OH-] + [MgOH+]; (this is the same as textbook)

Approach 2: Using hydroxide as our referenceHydroxide (denoted by X) in solution comes from two sources: (1) Y amount from Mg(OH)2, which is equal to TWICE amount of magnesium ions (denoted by M here) Y = 2 M; and source (2) Z amount from water, that equals the amount of [H+] from water, i.e. Z = [H+]We now know X = Y + Z, i.e. X = Y + [H+], i.e. X = 2 M + [H+];M in solution exists in the forms [Mg2+], [MgOH+], M = [Mg2+] + [MgOH+], Therefore X = 2([Mg2+] + [MgOH+]) + [H+];Hydroxide X in solution exits in the forms of [OH-] and [MgOH+], X = [OH-] + [MgOH+], Combined the above steps: [OH-] + [MgOH+] = 2([Mg2+] + [MgOH+]) + [H+];

First: the calcium oxalate should not bear no charge, as pointed out by a student in class. If so, the charge balance equation will be 2[Ca2+] = 2[C2O42-] + [HC2O4-] + [OH-]Assuming the original question is correct, the charge balance equationwould be 2[Ca2+] = 2[C2O42-] + [HC2O4-] + [OH-] + 2[CaC2O42-]

If we also consider the autoprotolysis of H2O: H2O H+ + OH- (R1)The charge balance equation will become 2[Ca2+] + [ H+] = 2[C2O42-] + [HC2O4-] + [OH-]

Mass balance equation for 9-11Using calcium as our referenceThe amount of calcium (denoted by X) should be equal to the amount of oxalate (denoted by Y), given that they come from the same source with a 1:1 stoichoimetric relationship.X in the system exists as [Ca2+], i.e. X = [Ca2+]Y in the system exists as [C2O42-], [HC2O4-], and [H2C2O4], i.e. Y = [C2O42-] + [HC2O4-] + [H2C2O4],Since X = Y, we have [Ca2+] = [C2O42-] + [HC2O4-] + [H2C2O4] (Final answer) Using oxalate as our reference yeilds the same result as above How about using hydroxide ion as the reference of mass balance? Hydroxide ion (denoted by X) in solution comes from water, which leads to the formation of [HC2O4-] + [H2C2O4], therefore [OH-] = [HC2O4-] + [H2C2O4], Multiple chemical equilibriumEquilibrium CalculationsPreviousBaC2O4 Ba2+ + C2O42-

[Ba2+] = [C2O42-]

But what if oxalate then reacted with H2OC2O42- + H2O HC2O4- + OH-HC2O4- + H2O H2C2O4 + OH-

[Ba2+] [C2O42-]because [C2O42-] = [C2O42-] + [HC2O4-] + [H2C2O4]

Systematic Treatment of EquilibriumStep 1. Write all pertinent reactions

Step 2. Write the charge balance equation

Step 3. Write the mass balance equation

Step 4. Write the equilibrium constant for each chemical reaction (should use activities but we will ignore them)

Step 5. Count the equations and unknowns (# equations = # unknowns)

Step 6. Solve the unknownsSystematic Treatment of EquilibriumDetermining the equations used to calculate unknown concentrations is the heart of this type of problem

The hard part, however, is solving for multiple unknowns using these equationsSystematic Treatment of EquilibriumCalculate the concentration of the hydronium and hydroxide ions in water.

Calculate the concentration of Hg22+ in a saturated solution of Hg2Cl2.

Find the solubility of CaF2 in water if the pH is somehow fixed at 3.0.

Systematic Treatment of EquilibriumFind the solubility of Ba(C2O4) in water assuming a pH of 2.0. Calculate the concentration of all species in solution.Systematic Treatment of EquilibriumWe can look for two more pieces of info:Charge Balance: Electroneutrality of the solution; the sum of the positive charge in solution equals the sum of negative charges in the solution.

Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution.

27[H+] + 2[Ca2+] = [OH-] + [F-](4)2[Ca2+] = [F-] + [HF](5)Systematic Treatment of EquilibriumHow does the solubility of CaF2 depend on pH?28CaF2(s) Ca2+ + 2F-Ksp = [Ca2+][F-]2 = 3.910-11(1)F-+H2O HF + OH-

(2)H2O H+ + OH-Kw = [H+][OH-] = 1.010-14(3)Five unknowns: [Ca2+], [F-], [HF], [H+], and [OH-]Three equations mean we need more equationsSystematic Treatment of EquilibriumWe can look for two more pieces of info:Charge Balance: Electroneutrality of the solution; the sum of the positive charge in solution equals the sum of negative charges in the solution.

Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution.

[H+] + 2[Ca2+] = [OH-] + [F-](4)2[Ca2+] = [F-] + [HF](5)Ksp = [Ca2+][F-]2 = 3.910-11(1)

(2)Kw = [H+][OH-] = 1.010-14(3)(4)(5)We need to substitute to get things in terms of Ca2+ or H+[H+] + 2[Ca2+] = [OH-] + [F-]2[Ca2+] = [F-] + [HF]Solving the Equations31

Ksp = [Ca2+][F-]2 = 3.910-11Kw = [H+][OH-] = 1.010-142[Ca2+] = [F-] + [HF]Combine eq. 2 and eq. 5, we have

(B)Solving the EquationsCombine (B) and eq. 1, we have

(B)

(C)Its much simpler if we can consider the pH fixed (how would we do that?)If we can fix pHpH = 3.00[H+] = 1.010-3 MKw[HF-] =1.5[F-]Kb[F-] =0.80[Ca2+]Mass[Ca2+] = 3.910-4 MKsp

[OH-] = 1.010-11 MCalculate the Concentration of Ca2+ in a solution that is saturated in CaF(s) and buffered to pH 3.0.Write relevant solubility and acid dissociation reactions (how many different ions?)Write the Ksp (4 x 10-11) and Ka (7 x 10-4) expressionsWrite the mass balance equation for CalciumCaF2(s) Ca2+ + 2F-

HF H+ + F-Ksp = [Ca2+] [F-]2

Ka = ([H+] [F-]) / [HF][Ca2+] = ([F-] + [HF])Calculate the Concentration of Ca2+ in a solution that is saturated in CaF(s) and buffered to pH 3.0.Use the Ka expression to substitute for HF and then the Ksp expression to substitute for F-Solve for [Ca2+]; how does this differ from value if HF were strong acid?[Ca2+]1.5 = (4x10-11 / 4)1/2(1+10-3/7x10-4) = 7.7x10-6

[Ca2+] = 4x10-4 M (This is the solubility s)

If no Ka 4x10-11 = s (2s)2 s = 2x10-4[Ca2+] = ([F-] + ([H+][F-])/ Ka) = [F-](1 + [H+]/Ka)

[Ca2+] = (Ksp/[Ca2+])1/2 (1 + [H+]/Ka)

[Ca2+]1.5 = (Ksp/4)1/2 (1 + [H+]/Ka)[Ca2+] = ([F-] + [HF])Lecture 10 Proton Balance, Graphical Solutions & Ionization Fractions for Multiprotic SystemsProton balance equation (mass balance on H+)Graphical solutions for multiprotic systemsPlot species concentration (pC) versus Master Variable (pH)Diprotic [H+]: pH = pH[OH-]: pOH = pKW - pH[H2B] = BT*a0[HB-] = BT*a1[B2-] = BT*a2 36Start with System DefinitionSystem: ClosedComponents: HA & H2OSpecies: HA, A-, H2O, OH-, H+Equilibria:H2O = H+ + OH-Kw = [H+][OH-] = 10-14HA = H+ + A-KA = [H+][A-]/[HA] = 10-6Mass Balance: AT = [HA] + [A-] = 10-3 MCharge Balance: [A-] + [OH-] = [H+] 37Summary of steps Plot line for [H+]: pH = pHPlot line for [OH-]: pOH = pKW -pHPlot AT: horizontal line at pC = pAT

Plot system point: pH = pKA on pAT line38Plot AT

ATSystempoint39Lines for the acid speciesAcid species [HA]: Combine mass balance and mass action to get

In log form

At low pH (pH > pKA)

40pH dependence of pHA

HA41Lines for the acid speciesConjugate base [A-]: Combine mass balance and mass action to get

In log form

At low pH (pH > pKA)

42Including dependence of [A-] with pH

43Complete system

Intercept ~ 0.3 log below AT44

System with 10-3 M Benzoic Acid45

System with 10-3 M Benzoic Acid46What is the equilibrium pH for a solution comprised of AT M of the salt of the conjugate base, NaA?System: ClosedComponents: Na+, HA & H2OSpecies: HA, A-, H2O, OH-, H+Equilibria:H2O = H+ + OH-Kw = [H+][OH-] = 10-14HA = H+ + A-KA = [H+][A-]/[HA] = 10-pKaMass Balance: AT = [HA] + [A-]NaT = [Na+] = ATCharge Balance: [A-] + [OH-] = [H+] + [Na+]Proton Balance equation: [HA] + [H+] = [OH-]47Monoprotic Weak Acid - NaAcetate (salt of the conjugate base)

48What is the equilibrium pH for a solution comprised of AT M of the salt of the conjugate base, NaA?System: ClosedComponents: Na+, HA & H2OSpecies: HA, A-, H2O, OH-, H+Equilibria:H2O = H+ + OH-Kw = [H+][OH-] = 10-14HA = H+ + A-KA = [H+][A-]/[HA] = 10-pKaMass Balance: AT = [HA] + [A-]NaT = [Na+] = ATCharge Balance: [A-] + [OH-] = [H+] + [Na+]Proton Balance equation: [HA] + [H+] = [OH-]49Monoprotic Weak Acid - NaAcetate (salt of the conjugate base)

50Monoprotic Weak Acid - NaCN (salt of the conjugate base)

51What is the equilibrium pH for a solution comprised of BT M of a generic acid diprotic acid H2B?System: ClosedComponents: H2B & H2OSpecies: H2B, HB-,B2- H2O, OH-, H+Equilibria:H2O = H+ + OH-Kw = [H+][OH-] = 10-14H2B = H+ + HB-KA1 = [H+][HB-]/[H2B] = 10-pKa1HB- = H+ + B2-KA2 = [H+][B2-]/[HB-] = 10-pKa2Mass Balance: BT = [H2B] + [HB-] +[B2-]Charge Balance ( = PBC): 2[B2-] + [HB-] + [OH-] = [H+] 52Diprotic Acid - Carbonate system

SystempointsCT53What is the equilibrium pH for a solution comprised of BT M of the salt of the conjugate base of H2B?System: ClosedComponents: Na+, H2B & H2OSpecies: H2B, HB-,B2- H2O, OH-, H+Equilibria:H2O = H+ + OH-Kw = [H+][OH-] = 10-14H2B = H+ + HB-KA1 = [H+][HB-]/[H2B] = 10-pKa1HB- = H+ + B2-KA2 = [H+][B2-]/[HB-] = 10-pKa2Mass Balance: BT = [H2B] + [HB-] +[B2-]= 10-3 MNaT = [Na+] = BTCharge Balance: 2[B2-] + [HB-] + [OH-] = [H+] + [Na+]Proton Balance: [H2B] + [H+] = [B2-] + [OH-]54Diprotic Acid - NaHCO3 (salt of the conjugate base of H2B)

55What is the equilibrium pH for a solution comprised of BT M of the salt of the conjugate base of HB-?System: ClosedComponents: Na+, HB- & H2OSpecies: H2B, HB-,B2- H2O, OH-, H+Equilibria:H2O = H+ + OH-Kw = [H+][OH-] = 10-14H2B = H+ + HB-KA1 = [H+][HB-]/[H2B] = 10-pKa1HB- = H+ + B2-KA2 = [H+][B2-]/[HB-] = 10-pKa2Mass Balance: BT = [H2B] + [HB-] +[B2-]= 10-3 MNaT = [Na+] = 2BTCharge Balance: 2[B2-] + [HB-] + [OH-] = [H+] + [Na+]Proton Balance: 2[H2B] + [HB-] + [H+] = [OH-]56Diprotic Acid - NaHCO3 (salt of the conjugate base of HB-)

5758NUMERICAL EQUILIBRIUM CALCULATIONSMonoprotic acidWhat are the pH and the concentrations of all aqueous species in a 5 x 10-4 M solution of aqueous boric acid (B(OH)3)?Steps to solution1) Write down all species likely to be present in solution: H+, OH-, B(OH)30, B(OH)4-.592) Write the reactions and find the equilibrium constants relating concentrations of all species:H2O H+ + OH-

(i)(ii)B(OH)30 + H2O B(OH)4- + H+603) Write down all mass balance relationships:5 x 10-4 M = B = [B(OH)4-] + [B(OH)30](iii)

4) Write down a single charge-balance (electroneutrality) expressions:[H+] = [B(OH)4-] + [OH-](iv)

5) Solve n equations in n unknowns.

61EXACT NUMERICAL SOLUTIONEliminate [OH-] in (i) and (iv)[H+][OH-] = Kw[OH-] = Kw/[H+][H+] = [B(OH)4-] + Kw/[H+][H+] - [B(OH)4-] = Kw/[H+]

(v)62Solve (iii) for [B(OH)30][B(OH)30] = B - [B(OH)4-]

[H+][B(OH)4-] = KA(B - [B(OH)4-])(vi)

Now solve (v) for [B(OH)4-]- [B(OH)4-] = Kw/[H+] - [H+][B(OH)4-] = [H+] - Kw/[H+]Substitute this into (vi)

63[H+]([H+] - Kw/[H+]) = KA(B - [H+] + Kw/[H+])[H+]2 - Kw = KAB - KA[H+] + KAKw/[H+][H+]3 - Kw[H+] = KAB[H+] - KA[H+]2 + KAKw[H+]3 + KA[H+]2 - (KAB + Kw)[H+] - KAKw = 0[H+]3 + (7x10-10)[H+]2 - (3.6x10-13)[H+] - (7x10-24) = 0

We can solve this by trial and error, computer or graphical methods. From trial and error we obtain [H+] = 6.1x10-7 M or pH = 6.2164[OH-] = Kw/[H+][OH-] = 10-14/10-6.21[OH-] = 10-7.79 M

[B(OH)4-] = [H+] - Kw/[H+][B(OH)4-] = 6.1x10-7 - 1.62x10-8[B(OH)4-] = 5.94x10-7 M

[B(OH)30] = B - [B(OH)4-][B(OH)30] = 5x10-4 - 5.94x10-7 M = 4.99x10-4 M

65APPROXIMATE SOLUTIONLook for terms in additive equations that are negligibly small (multiplicative terms, even if very small, cannot be neglected.Because we are dealing with an acid, we can assume that [H+] >> [OH-] so that the mass balance becomes:[H+] = [B(OH)4-]and then[B(OH)30] = B - [H+]66

(ii)[H+]2 = KAB-KA[H+][H+]2 + KA[H+] - KAB = 0This is a quadratic equation of the form:ax2 + bx + c = 0and can be solved using the quadratic equation

67In our case this becomes:

Only the positive root has any physical meaning.[H+] = 5.92 x 10-7

We could have made this problem even simpler. Because boric as is a quite weak acid (i.e., veryKA value, very little of it will be ionized, thus[B(OH)30] >> [B(OH)4-]B [B(OH)30] = 5 x 10-4 M68

[H+]2 = 3.5 x 10-15[H+] = 5.92 x 10-7 M

It is wise to check your assumptions by back substitutinginto original equations. If the error is 5%, the approxi-mation is probably justified because KA values are at leastthis uncertain!69CALULATE THE pH OF A STRONG ACIDCompute the pH and equilibrium concentrations of all species in a 2 x 10-4 M solution of HCl.1) Species: H+, Cl-, HCl0, OH-2) Mass action laws:

3) Mass balance: [HCl0] + [Cl-] = 2 x 10-4 M4) Charge balance: [H+] = [Cl-] + [OH-]70Assumptions: HCl is a very strong acid so[H+] >> [OH-] and [Cl-] >> [HCl0]Now the only source of H+ and Cl- are the dissociation of HCl, so[H+] = [Cl-](this is also apparent from the charge balance)Thus, pH = - log (2 x 10-4) = 3.70, and [Cl-] = 2 x 10-4 M.[OH-] = Kw/[H+] = 10-14/2 x 10-4 = 5 x 10-11 M

71CALCULATE THE pH OF A WEAK MONOPROTIC BASECompute the pH and equilibrium concentrations of all species in a 10-4.5 M solution of sodium acetate.1) Species: H+, Na+, Ac-, HAc0, OH-2) Mass action laws:

3) Mass balances: [HAc0] + [Ac-] = 10-4.5 M = C[Na+] = 10-4.5 M = C4) Electroneutrality: [Na+] + [H+] = [Ac-] + [OH-]Combine 3) and 4) to get proton condition: [HAc0] + [H+] = [OH-]72We cannot make any approximations relative to the concentrations of [H+] and [OH-] because acetate is a weak base and total acetate concentration is low. However, because base is weak: [Ac-] >> [HAc0] so[Ac-] 10-4.5 M = CSubstitute for [OH-] in proton condition[HAc0] + [H+] = Kw/[H+][HAc0] = Kw/[H+] - [H+] Now substitute into ionization constant expression

73[H+]C = KAKw/[H+] - KA[H+][H+]2C = KAKw - KA[H+]2[H+]2C + [H+]2KA = KAKw[H+]2(C + KA) = KAKw[H+]2 = KAKw/(C + KA) [H+] = (KAKw/(C + KA))0.5[H+] = (10-4.710-14/(10-4.5 + 10-4.70)[H+] = 6.2 x 10-8 MpH = 7.2pOH = pKw - pH = 14.0 - 7.2 = 6.8[OH-] = 1.61 x 10-7

74Rearranging the proton condition we get:[HAc0] = [OH-] - [H+] = 1.6 x 10-7 - 6.2 x 10-8 = 9.8 x 10-8

Check of assumption:[Ac-] = C - [HAc0] = 10-4.50 - 9.8 x 10-8so [Ac-] 10-4.50and the assumption made is valid.75CALCULATE THE pH OF AN AMPHOLYTECalculate the pH of a 10-3.7 M solution of sodium hydrogen phthalate (NaHP).1) Species: H2P0, HP-, P2-, H+, OH-, Na+2) Mass action expressions:COOHCOONa

3) Mass balance expressions:PT = 10-3.7 M = [H2P0] + [HP-] + [P2-]76PT = 10-3.7 M = [Na+]4) Charge balance: [H+] + [Na+] = [OH-] + [HP-] + 2[P2-]Now, substitute 3) into 4) to get proton condition:[H+] + [H2P0] + [HP-] + [P2-] = [OH-] + [HP-] + 2[P2-][H+] + [H2P0] = [OH-] + [P2-]Because both pK values are less than 7, assume:[OH-] > [OH-]. Also, because KA,2 and KA,3 are quite small, then [HPO42-] and [PO43-] are negligible compared to [H3PO40] and [H2PO4-]. The mass balance then becomes:PT = [H3PO40] + [H2PO4-]82And the charge balance expression becomes:[H+] = [H2PO4-]which can be substituted into the expression for KA,1.

KA,1PT-KA,1[H+] = [H+]2[H+]2 + KA,1[H+] - KA,1PT = 0

83[H+] = 8.986 x 10-4 MpH = 3.05pOH = pKw - pH = 14 - 3.05 = 10.95[OH-] = 1.122 x 10-11 M[H3PO40] = PT - [H2PO4-] = 10-3 - 8.986 x 10-4[H3PO40] = 1.014 x 10-4 M

84

[PO43-] = 10-16.15 M = 7.079 x 10-17 M

A check of all the assumptions shows that they are allvalid.85CALCULATION OF pH OF A VOLATILE BASECompute the pH and concentrations of all species of a solution exposed to an atmosphere of pNH3 = 10-4 atm.1) Species: NH30, NH4+, OH-, H+2) Mass action expressions:

863) Charge balance: [NH4+] + [H+] = [OH-]Assumptions: NH3 is a moderate base, so we can assume that [OH-] >> [H+] so the charge balance becomes[NH4+] = [OH-]also[NH30] = pNH3KH = 10-4(101.75) = 10-2.25 M

[OH-]2 = 10-6.75[OH-] = 10-3.375 M = 4.22 x 10-4 M87pH = pKw - pOH = 14 - 3.375 = 10.625so the assumption that [OH-] >> [H+] is valid. The concentrations are then:[OH-] = 4.22 x 10-4 M[NH4+] = 4.22 x 10-4 M[H+] = 2.37 x 10-11 M[NH30] = 5.62 x 10-3 M

88GRAPHICAL APPROACH TO EQUILIBRIUM CALCULATIONSConsider the monoprotic acid HA:

CT = 10-3 = [HA0] + [A-]; so [A-] = CT - [HA0]KA[HA0] = [H+][A-]KA[HA0] = [H+](CT - [HA0])KA[HA0] = [H+]CT - [H+][HA0]KA[HA0] + [H+][HA0] = [H+]CT

89CTKA - KA[A-] = [H+][A-]CTKA = [A-]([H+] + KA)

1) At pH < pKA, [H+] >> KA so [H+] + KA [H+][HA0] = CT([H+]/[H+]) = CTlog [HA0] = log CT [A-] = CTKA/[H+]log [A-] = log CT - pKA + pH

902) pH = pKA; [H+] = KA so [H+] + KA = 2[H+][HA0] = CT[H+]/(2[H+]) = CT/2log [HA0] = log CT - log 2 = log CT - 0.301[A-] = CT [H+]/(2[H+]) = CT/2log [A-] = log CT - log 2 = log CT - 0.3013) pH > pKA; [H+] > [OH-][H+] [A-]To compute composition of 10-3 M NaA solution, start with proton condition:[HA0] + [H+] = [OH-][OH-] >> [H+][HA0] [OH-]93SPECIATION DIAGRAM FOR A DIPROTIC SYSTEMConsider H2S with pK1 = 7.0, pK2 = 13.0ST = 10-3 M = [H2S0] + [HS-] + [S2-]

941) pH < pK1 < pK2; [H+] > K1 > K2

log [H2S0] = log ST

log [HS-] = log (STK1) + pH

log [S2-] = log (STK2K1) + 2pH952) pH = pK1 < pK2; [H+] = K1 > K2

log [H2S0] = log ST - 0.301

log [HS-] = log ST - 0.301

log [S2-] = log (STK2/2) + pH963) pK1 < pH < pK2; K1 > [H+] > K2

log [H2S0] = log (STK1) - pH

log [HS-] = log ST

log [S2-] = log (STK2) + pH974) pK1 < pK2 = pH; K1 > [H+] = K2

log [H2S0] = log (STK1/2) - pH

log [HS-] = log ST - 0.301

log [S2-] = log ST - 0.301985) pK1 < pK2 < pH; K1 > K2 > [H+]

log [H2S0] = log (STK1K2) - 2pH

log [HS-] = log (STK2) - pH

log [S2-] = log ST99

Speciation diagram for H2S with CT = 10-3 at 25C100IONIZATION FRACTIONSMonoprotic acid: HBB = 1 [B]/C = KA/(KA + [H+])HB = 0 [HB]/C = [H+]/(KA + [H+])1 + 0 = 1Diprotic acid: H2A

101

102

Figure 3.10a-c from Stumm and MorganThe Systematic ApproachWrite all chemical equations for systemIdentify the unknown speciesNeed as many independent mathematical equations as unknownsEquilibrium constant expressionsMass balance equation(s)Charge balance equationMake simplifying assumptions where appropriateDo the math!Check assumptionsStrong Monoprotic Acid HClChemical equationsHCl --- H+ + Cl- and H2O --- H+ + OH-Unknown species [H+], [OH-], [Cl-]We need three independent equations[H+][OH-] = Kw[Cl-] = CHCl(mass balance)[H+] = [OH-] + [Cl-](charge balance)Combine equations[H+] = Kw/[H+] + CHClRearrange to [H+]2 CHCl[H+] Kw = 0Solve either one using methods of Lecture 9 or make simplifying assumptionsSolve for 0.10 M HCl and 1.0 x 10-8 M HClWe get [H+] = 0.10 M (pH 1.00) and 1.05 x 10-7 M (pH 6.98)

Making Simplifying AssumptionsIf CHCl is large enough that [OH-] [Cl-]Then charge balance becomes [H+] = [Cl-]It works for 0.10 M HClIt doesnt work for 1.0 x 10-8 M HClMaking assumption leads to pH 8.00 solutionA basic solution of HCl? Impossible!So how do you know when to make assumptions and when not to?Thats a good questionTry them and see if answers make sense

Weak Monoprotic Acid HAChemical equationsHA --- H+ + A- and H2O --- H+ + OH-Unknown species HA, H+, A-, OH-We need four independent equationsKw and Ka expressions[HA] + [A-] = CHA(mass balance)[H+] = [OH-] + [A-] (charge balance)

Solving It ExactlyCombine Ka and mass balance, solve for [A-][H+][A-]/Ka + [A-] = CHA[A-] = CHAKa/([H+] + Ka)Put this and [OH-] = Kw/ [H+] into charge balance[H+] = Kw/ [H+] + CHAKa/([H+] + Ka)Solve directly by Lecture 9 Excel methodsOr rearrange to cubic equation and use poly on calculator

Exact Cubic Equation[H+]3 + Ka[H+]2 (Kw + CHAKa)[H+] KwKa = 0Solve for 0.10 M HA and 1.0 x 10-8 M HAGiven Ka = 1.8 x 10-5Compare with strong acidFor 0.10 M HAx3 + 1.8 x 10-5x2 1.8 x 10-6x 1.8 x 10-19 = 0x = 1.33 x 10-3 (pH 2.88)For 1.0 x 10-8 M HAx3 + 1.8 x 10-5x2 1.9 x 10-13x 1.8 x 10-19 = 0x = 1.05 x 10-7 (pH 6.98)

Now Thats Interesting!Notice that the pH of 0.10 M HA (weak acid) is much different than the pH of 0.10 M HCl (strong acid)pH 2.88 compared to 1.00But the pH of the very dilute weak acid is the same as the very dilute strong acid!Both are pH 6.98I wonder why that is?Actually, I know, but do you?Making Simplifying AssumptionsLook what happens if [OH-] [A-] in charge balanceYou did this automatically in Chem 106 by saying [H+] = [A-] = xCharge balance becomes[H+] = CHAKa/([H+] + Ka) or[H+]2 + Ka [H+] CHAKa = 0Just a quadraticIn Chem 106, it was x2/(CHA x) = Ka

More AssumptionsNow, if [H+] CHA (5% rule)The quadratic on previous slide becomes[H+]2 + Ka [H+] CHAKa = 0[H+]2 + ([H+] CHA)Ka = 0[H+]2 CHAKa = 0[H+] = CHAKa Which is what you did in Chem 106 most of the timeWell do more of this sort of thing laterFor now, lets just look at more examples of mass balance and charge balance equations

NaA, Salt of Weak Acid HASpecies in solutionNa+, A-, HA, H+, OH-Mass balances[Na+] = CNaA[HA] + [A-] = CNaACharge balance[Na+] + [H+] = [A-] + [OH-]Plus Kw and Ka for exact solution

Buffer Solution Containing HA and NaASpecies in solutionNa+, A-, HA, H+, OH-Mass balances[Na+] = CNaA[HA] + [A-] = CHA + CNaACharge balance[Na+] + [H+] = [A-] + [OH-]Plus Kw and Ka for exact solutionWeak Diprotic Acid H2ASpecies in solutionH+, OH-, H2A, HA-, and A2-Mass balance[H2A] + [HA-] + [A2-] = CH2ACharge balance (see Example 6.8)[H+] = 2[A2-] + [HA-] + [OH-]Is the 2 in the right place? Should it be ?Plus Kw, Ka1, and Ka2 for exact solutionSolution of MgCl2Species in solutionH+, OH-, Mg2+, Cl-Mass balances[Mg2+] = CMgCl2 and [Cl-] = 2CMgCl2Charge balance[H+] + 2[Mg2+] = [OH-] + [Cl-]Neglecting H+ and OH-, we have2[Mg2+] = [Cl-]Is that right?Plus Kw for exact solutionSlightly Soluble Salt of a Weak Acid (Example 6.10)Species in solutionH+, OH-, Cd2+, S2-, HS-, H2SMass balance[Cd2+] = [S2-] + [HS-] + [H2S]Charge balance2[Cd2+] + [H+] = 2[S2-] + [HS-] + [OH-]Plus Ksp, Kw, Ka1, Ka2 for exact solutionCan you spot the approximations you made in Chem 106 to do a solubility problem like this?Examples 6.7 and 6.11[Ag(NH3)2]Cl completely dissociates into Ag(NH3)2+ and Cl-Ag(NH3)2+ dissociates in two steps, with two equilibrium constantsAlong with Kw and Kb for NH3, we have four of the equations we needThere are eight species in solutionWe need four more equationsExamples 6.7 and 6.11 ContinuedThree mass balances (Example 6.7)[Cl-] = 1.00 x 10-5 MOne based on total silver-containing speciesOne based on total nitrogen-containing speciesNote the coefficient of 2 on [Ag(NH3)2+]Charge balance (Example 6.11)The last equation in Example 6.7 is not an independent equationIt can be derived by subtracting the silver mass balance from the charge balanceExamples 6.12 and 6.13Additional illustrations of the processI dont agree with the marginal note on page 208.Its true that it is often possible to write more equations than are necessaryWhich means not all of them are independentCharge balance is usually pretty easy to write, so I like to use it rather than try to find all of the mass balance equationsPolyprotic Acids

First, notice that the pH where two species concentrations are the same is around the pKa for that equilibrium. In fact, for polyprotic acids with pKa's that differ by over 3 to 4 units, the pH is equal to the pKa.

Interesting Features of a plots

Take for example the point where [H3PO4]=[H2PO4-]. The equilibrium equation relating these two species is If we take the -log10, or "p", of this equation

Since [H3PO4]=[H2PO4-], and log10(1) = 0, pH=pKa1

pKa1pKa2pKa3Second, you might notice that the concentrations of the conjugate bases are maximum half-way between the pKa points.

For example, the point where [H2PO4-] is a maximum lies half-way between between pKa1 and pKa2. Since H2PO4- is the major species present in solution, the major equilibrium is the disproportionation reaction.

This equilibrium cannot be used to solve for pH because [H3O+] doesn't occur in the equilibrium equation. We solve the pH problem adding the first two equilibria equations +

Since the disproportionation reaction predicts [H3PO4]=[HPO42-]

Note that when we add chemical equilibria, we take the product of the equilibrium equations. Taking the -log10 of the last equationWhat about the Z word??Zwitterions (German for Double Ion) a molecule that both accepts and losses protons at the same time.EXAMPLES??? How about AMINO ACIDS

neutralamino-protonatedzwitterionBoth groups protonatedcarboxylic-deprotonatedwhy activity of proteins are pH dependent

Lets look at the simplest of the amino acids, glycine

H2Gly+

glyciniumHGlyGly-

glycinate

K1K2

In water the charge balance would be,

Combining the autoprotolysis of water and the K1 and K2 expressions into the charge balance yields:

HGlyGly-H2Gly+Diprotic Acids and Bases

2.)Multiple EquilibriaIllustration with amino acid leucine (HL)

Equilibrium reactions

low pHhigh pHCarboxyl groupLoses H+ammonium groupLoses H+

Diprotic acid:

Diprotic Acids and Bases

2.)Multiple EquilibriumsEquilibrium reactionsDiprotic base:

Relationship between Ka and Kb

pKa of carboxy and ammonium group vary depending on substituentsLargest variationsPolyprotic Acid-Base Equilibria Diprotic Acids and Bases

3.)General Process to Determine pHThree components to the process

Acid Form [H2L+]

Basic Form [L-]

Intermediate Form [HL]

low pHhigh pHCarboxyl groupLoses H+ammonium groupLoses H+Polyprotic Acid-Base Equilibria Diprotic Acids and Bases

3.)General Process to Determine pHAcid Form (H2L+)Illustration with amino acid leucine

H2L+ is a weak acid and HL is a very weak acid

K1=4.70x10-3K2=1.80x10-10

Assume H2L+ behaves as a monoprotic acid

Polyprotic Acid-Base Equilibria Diprotic Acids and Bases

3.)General Process to Determine pH0.050 M leucine hydrochloride

+ H+H2L+HLH+0.0500 - xxxK1=4.70x10-3

Determine [H+] from Ka:Determine pH from [H+]:Determine [H2L+]:Polyprotic Acid-Base Equilibria Diprotic Acids and Bases

3.)General Process to Determine pHAcid Form (H2L+)What is the concentration of L- in the solution?

[L-] is very small, but non-zero. Calculate from Ka2

Approximation [H+] [HL], reduces Ka2 equation to [L-]=Ka2

Validates assumptionPolyprotic Acid-Base Equilibria Diprotic Acids and Bases

3.)General Process to Determine pHFor most diprotic acids, K1 >> K2Assumption that diprotic acid behaves as monoprotic is validKa Ka1

Even if K1 is just 10x larger than K2Error in pH is only 4% or 0.01 pH units

Basic Form (L-)

L- is a weak base and HL is an extremely weak base

Assume L- behaves as a monoprotic base


3.)General Process to Determine pH0.050 M leucine salt (sodium leucinate) L-HLOH-0.0500 - xx x

Determine [OH-] from Kb:Determine pH and [H+] from Kw:Determine [L-]:


3.)General Process to Determine pHBasic Form (L-)What is the concentration of H2L+ in the solution?

[H2L+] is very small, but non-zero. Calculate from Kb2

Validates assumption [OH-] [HL], Fully basic form of a diprotic acid can be treated as a monobasic, Kb=Kb1Polyprotic Acid-Base Equilibria Diprotic Acids and Bases

3.)General Process to Determine pHIntermediate Form (HL)More complicated HL is both an acid and base

Amphiprotic can both donate and accept a proton

Since Ka > Kb, expect solution to be acidicCan not ignore base equilibrium

Need to use Systematic Treatment of Equilibrium

Polyprotic Acid-Base Equilibria Step 1: Pertinent reactions:Step 2: Charge Balance:Step 3: Mass Balance:Step 4: Equilibrium constant expression (one for each reaction):

Diprotic Acids and Bases

3.)General Process to Determine pHIntermediate Form (HL)

Step 6: Solve:


3.)General Process to Determine pHIntermediate Form (HL)Substitute Acid Equilibrium Equations into charge balance:

All Terms are related to [H+]Multiply by [H+]


3.)General Process to Determine pHIntermediate Form (HL)Step 6: Solve:

Factor out [H+]2:

Rearrange:Polyprotic Acid-Base Equilibria Diprotic Acids and Bases


Multiply by K1 and take square-root:

Assume [HL]=F, minimal dissociation:(K1 & K2 are small)



Calculate a pH:


3.)General Process to Determine pHIntermediate Form (HL)Step 7: Validate Assumptions

Assume [HL]=F=0.0500M, minimal dissociation (K1 & K2 are small).Calculate [L-] & [H2L+] from K1 & K2:

[HL]=0.0500M >> 9.36x10-6 [H2L+] & 1.02x10-5 [L-] Assumption ValidPolyprotic Acid-Base Equilibria Diprotic Acids and Bases

3.)General Process to Determine pHIntermediate Form (HL)Summary of results: [L-] [H2L+] two equilibriums proceed equally even though Ka>Kb Nearly all leucine remained as HL

Range of pHs and concentrations for three different formsSolutionpH[H+] (M)[H2L+] (M)[HL] (M)[L-] (M)Acid form0.0500 M H2A1.881.32x10-23.68x10-21.32x10-21.80x10-10Intermediate form0.0500 M HA-6.068.80x10-79.36x10-65.00x10-21.02x10-5Basic form0.0500 M HA2-11.216.08x10-122.13x10-121.64x10-34.84x10-2Polyprotic Acid-Base Equilibria Diprotic Acids and Bases

3.)General Process to Determine pHSimplified Calculation for the Intermediate Form (HL)

Assume K2F >> Kw:Assume K1

Treatment for Multiple Chem Equil

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