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Systematic Treatment of Equilibrium Systematic Treatment of Equilibrium Properties of Umass Boston Systematic Treatment of Equilibrium • Methodology to deal with chemical equilibria systematically: figure out as many independent algebraic equations as there are unknown species • Independent algebraic equations: – chemical equilibrium conditions – mass balance – charge balance Properties of Umass Boston Chemical Equilibrium Example: dissolve K3PO4 into water •Dissociation reaction + 3- K3PO4 ' 3K + PO4 •Other reactions with water 3- 2- - PO4 + H2O ' HPO4 + OH 2- - - HPO4 +H+ H2O ' H2PO4 +OH+ OH •In aqueous solutions, always self-ionization of water + - H2O ' H + OH Properties of Umass Boston Charge Balance • The statement of electroneutrality: The sum of the positive charges in solution equals the sum of the negative charges in solution – to make the electroneutrality of the solution. – Creation of one positive charge (ion) must accompany the creation of one negative charge Properties of Umass Boston Charge balance Example: dissolve K3PO4 into water •Dissociation reaction +1+ 3-3- K3PO4 ' 3K ++ PO PO44 •Other reactions with water 3- 2-2- -1- PO4 + H2O ' HPO4 + OH 2- -1- - 1- HPO4 +H+ H2O ' H2PO4 +OH++OH+ OH OH •In aqueous solutions, always self-ionization of water +1+ - 1- H2O ' H ++ OH OH Properties of Umass Boston Charge balance • Example: dissolve K3PO4 1+ 3- K3PO4 ' 3K + PO4 3- 2- 1- PO4 + H2O ' HPO4 + OH 2- 1- 1- HPO4 + H2O ' H2PO4 + OH 1+ 1- H2O ' H + OH --------------------------------------------------------------------------------- • ∑(positive charge) = ∑ (negative charge) • Total positive charge = [K+]+[H+] - - 2- 3- Total negative charge = [OH ]+[H2PO4 ]+2[HPO4 ]+3[PO4 ] + + - - 2- 3- [K ] + [H ] ≡ [OH ] + [H2PO4 ] + 2[HPO4 ] + 3[PO4 ] Net charge in the solution = 0 Properties of Umass Boston Mass Balance or material balance • Statement of the conservation of matter: the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to amount of that atom (or group) delivered to the solution – Chemical process can not produce new kinds of atom. Properties of Umass Boston Mass Balance or material balance Example: dissolve K3PO4 into water •Dissociation reaction + 3- K3PO4 ' 3K3K + PO4 •Other reactions with water 3- 2- - PO4 + H2O ' HPO4 + OH 2- - - HPO4 +H+ H2O ' H2PO4 +OH+ OH •In aqueous solutions, always self-ionization of water + - H2O ' H + OH Properties of Umass Boston Mass Balance or material balance • Example: dissolve K3PO4 + 3- K3PO4 ' 3K +PO+ PO4 3- 2- - PO4 + H2O ' HPO4 + OH 2- - - HPO4 + H2O ' H2PO4 + OH + - H2O ' H +OH+ OH --------------------------------------------------------------------------------- • The amount that K3PO4 puts into the solution: + 3- 2- - [K3PO4] = 1/3[K ] = [PO4 ] + [HPO4 ] + [H2PO4 ] + 2- All K species All PO4 species Properties of Umass Boston What about activity? • In concentrated solutions, shouldn’t activity replace concentration? –Why? • Concentration rather than activity is used in both mass balance and charge balance. • Charge Balance: – Charge contributed by 0.1M H+ is exactly 0.1M • Mass Balance: – Concentration of each species counts exactly as # of atoms Properties of Umass Boston Steps to treat the equilibrium • Step 1 : write all possible reactions • Step 2: write charge balance equations • Step 3: write mass balance equations • Step 4: write the equilibrium constant for each chemical reaction – only time activity MAY be used. • Step 5: count the numbers of equation and unknowns (concentration of species) • Step 6 : solthlve the equa titfidttion to find out unknowns Properties of Umass Boston Example: dissolve K3PO4 • Step 1: All possible reactions + 3- K3PO4 ' 3K + PO4 K1 3- 2- - PO4 + H2O ' HPO4 + OH K2 2- - - HPO4 + H2O ' H2PO4 + OH K3 + - H2O ' H + OH Kw • Step 2: Charge balance + + - - 2- 3- [K ]+[H] + [H ] ≡ [OH ]+[H] + [H2PO4 ]+2[HPO] + 2[HPO4 ]+3[PO] + 3[PO4 ] •Step 3: Mass balance + [K3PO4] = 1/3[K ] 3- 2- - [K3PO4] = [PO4 ] + [HPO4 ] + [H2PO4 ] Properties of Umass Boston Example: dissolve K3PO4 • Write the equilibrium constant for each chemical reaction – only here is when activity MAY be used. + 3 3 3- K1={[K ] ©k x [PO4 ] ©PO4}/[K3PO4] 2- 2- 3- K2={[HPO4 ] ©HPO4 x [OH] ©OH}/[PO4 ] ©PO4 - - 2- 2- K3={[H2PO4 ] ©H2PO4 x [OH] ©OH}/[HPO4 ] ©HPO4 + Kw=[H ] ©H[OH] ©OH (In dilute solutions, © can be omitted (© ؆ 1 • Count the numbers of equation and unknowns (concentration of species) + + - - 2- 3- 7 unknown: [K ],[H ],[OH ],[H2PO4 ],[HPO4 ],[PO4 ], [K3PO4] 7 equations: K1, K2, K3, Kw,,g one charge balance and 2 mass balances • Solve the equation to find out unknowns In general , th e sol ubilit y o f MA i s high , if A i s f rom a weak acid Properties of Umass Boston Coupled Equilibria • The product of one reaction is the reactant of another or others Example 1: + 3- K3PO4 ' 3K + PO4 K1 3- 2- - PO4 +H+ H2O ' HPO4 +OH+ OH K2 2- - - HPO4 + H2O ' H2PO4 + OH K3 + - H2O ' H + OH Kw Example 2: 2+ - -11 CaF2(S) ' Ca + 2F Ksp=3.9 x 10 - - -11 F + H2O ' HF + OH Kb =1.5 x 10 + - -14 H2O ' H + OH Kw=1.0 x 10 Properties of Umass Boston Acids and Bases Weak Acid/Base Equilibria Properties of Umass Boston pH Revisited: Acids and Bases • Remember pH = -log[H+]? + pH is -logAH+ = -log[H ] ©H + - Kw=[H ] ©H[OH ] ©OH • Strong acids or bases -> complete dissociation • Weak acids or bases -> exist in equilibrium + − –HA ' H+ + A- ⎣⎡H ⎦⎤⎣⎡A ⎦⎤ Ka = [HA] ⎡BH + ⎤⎡OH − ⎤ + - ⎣ ⎦⎣ ⎦ –B + H2O ' BH + OH Kb = []B • Formal concentration (F): the total number of moles of a compound dissolved in a liter. Properties of Umass Boston Determining pH: Weak acids • Systematic treatment: HA ' H+ +A+ A- – Charge balance: [H+] = [A-] + [OH-] – Mass balance: F = [A-]+[HA]] + [HA] – Equilibria: + - Ka =[H= [H ][A ]/[HA] + - Kw = [H ][OH ] – 4 unknowns [A-][HA][H], [HA], [H+][OH], [OH-] and 4 equations Properties of Umass Boston Determining pH: Weak bases + - • Systematic treatment: B + H2O ' BH + OH – Charge balance: [BH+] = [OH-] – Mass balance: F=[BH+] + [B] – Equilibria: + - Kb = [BH ][OH ]/[B] + - - Kw = [H ][OH ] (all OH comes from B) – 3 unknowns [B], [BH+][OH], [OH-] and 3 equations Properties of Umass Boston Weak acid/base problems • Weak acid problem: HA ' H+ +A+ A- + − 2 ⎣⎡H ⎦⎤⎣⎡A ⎦⎤ x = = Ka [HA] F − x + - • Weak bass problem: B + H2O ' BH + OH + − 2 ⎣⎡BH ⎦⎤⎣⎡OH ⎦⎤ x = = Kb []B F − x Properties of Umass Boston The dependence of solubility on pH Properties of Umass Boston Polyprotic Acid-Base Equilibria Properties of Umass Boston Polyprotic Acids and Bases • Can donate (acid) or accept (base) more than one proton. • Diprotic (2), triprotic (3),… amino acid • A molecule that can both donate and accept a proton is call “amphiprotic” Properties of Umass Boston Diprotic acid and base (amino acid) • Diprotic acid: + + H2L ' HL + H Ka1 - + HL ' L + H Ka2 •Diprotic base - - L + H2O ' HL + OH Kb1 + - HL + H2O ' H2L + OH Kb2 Properties of Umass Boston Diprotic acid and base (amino acid) • Diprotic acid: + + H2L ' HL + H Ka1 - + HL ' L + H Ka2 • Diprotic base - - L + H2O ' HL + OH Kb1 + - HL + H2O ' H2L + OH Kb2 Ka1·Kb2 = Kw Ka2·Kb1 = Kw Properties of Umass Boston Dissociation of diprotic acid + • For the solution of H2L + + -3 H2L ' HL + H Ka1 = 4.69 x 10 - + -10 HL ' L + H Ka2 =1.79 x 10 Assumption: since Ka1>>Ka2 the dissociation of HL is insignificant + compared to the dissociation of H2L . + H2L can be treated as a monoprotic acid: only need Ka1 Properties of Umass Boston Dissociation of diprotic base • For the solution of L- - - -5 L + H2O ⇄ HL + OH Kb1=Kw/Ka2=5.48x10 + - -12 HL + H2O ' H2L + OH Kb2=Kw/Ka1=2.13x10 Assumption: since Kb1>>Kb2 the reaction of HL with H2O (hydrolysis) -. is insignificant compared to the reaction of L with H2O. - L can be treated as monobasic: only need Kb1 Properties of Umass Boston What happens to HL – intermediate form • Say, dissolve 0. 050 M of HL into water -10 Dissociation (acid): Ka = Ka2 = 1.79 x 10 HL ' H+ + L- Hydrolysis (base): -12 Kb=Kb2=2.13x10 + - HL + H2O ' H2L +OH+ OH • The two reactions are related and proceed to nearly equal extttent, so even th thhKough Ka >K> Kb, the hy dro lys is reac tion canno t be ignored. • Resort to systematic treatment of equilibrium Properties of Umass Boston Treatment of the intermediates • Systematic treatment: + - HL ' H + L Ka2 + - HL + H2O ' H2L + OH Kb2 + + H2L ' HL + H Ka1 + + − − + − + − Charge balance:[H ]+[H 2 L ] = [L ]+[OH ] ⇒[H 2 L ]−[L ]+[H ]−[OH ] ⎡H + ⎤ ⎡L− ⎤ HL K K = ⎣ ⎦ ⎣ ⎦ ⇒ ⎡L− ⎤ = [ ] a2 a2 ⎣ ⎦ + []HL ⎣⎡H ⎦⎤ ⎡H L+ ⎤ ⎡OH − ⎤ HL K K K = ⎣ 2 ⎦ ⎣ ⎦ ⇒ ⎡H L+ ⎤ = []b2 = HL ⎡H + ⎤ b2 b2 ⎣ 2 ⎦ − []⎣ ⎦ []HL ⎣⎡OH ⎦⎤ Kw K K K K = K ⇒ K = w ⎡OH − ⎤ = w a1 b2 w a1 ⎣ ⎦ + Kb2 ⎣⎡H ⎦⎤ HL ⎡H + ⎤ + + Kb2 + []⎣ ⎦ ⎣⎡H 2 L ⎦⎤ = [HL]⎣⎡H ⎦⎤ ⇒ ⎣⎡H 2 L ⎦⎤ = Kw Ka1 Properties of Umass Boston Treatment of the intermediates + [ HL ][ H ] [ HL ]K a 2 + K w − + + [ H ] − + = 0 K a1 [ H ] [ H ] + 2 [ HL ][ H ] + 2 − [ HL ]K a 2 + [ H ] − K w = 0 K a1 + 2 [ HL ] [ H ] ( + 1) = K a 2 [ HL ] + K w K a1 K K [ HL ] + K K [ H + ] = a1 a 2 a1 w K a1 + [ HL ] No assump(ption was made (two unknowns Properties of Umass Boston • Mass Balance: − + F = [FL]+ [L ]+[H 2 L ] + [HL]ka2 [HL][H ] = [HL]+ + + [H ] ka1 + Fka2[H ] [HL] = + + 2 ka1[H ]+ ka1ka2 +[H ] K K [HL]+ K K [H + ] = a1 a2 a1 w Ka1 +[HL] + 4 2 + 3 2 + 2 2 2 + 2 ka1[H ] +(ka1 +Fka2)[H ] +(ka1 kw −k1kw)[H ] −(ka1ka2 +ka1 kw)[H ]+ka1 ka2kw =0 Properties of Umass Boston Approximations K K [HL]+ K K [H + ] = a1 a2 a1 w Ka1 +[HL] • Since HL is weak acid/base, the [HL]≈F (added) then K K F + K K [H + ] = a1 a2 a1 w Ka1 + F •Assum e, Ka2[[]HL]>>Kw,,t the n K K F [H + ] = a1 a2 Ka1 + F • Assume, F>>Ka1 + [H ] = Ka1Ka2 pH= 1/2(pKa1+pKa2) Properties of Umass Boston Summary for diprotic acid • Starting material (Page 211) –H2A • Treat as monoprotic acid, Ka=Ka1, ignore Ka2 2- • Use Ka2 to solve for [A ].
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