Chapter 9. Aqueous Solutions and Chemical Equilibria. Classifying Solutions of Electrolytes

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1 Chapter 9 Aqueous Solutions and Chemical Equilibria Classifying Solutions of Electrolytes Electrolytes solutes form ions when dissolved in water (or certain other solvents, e.g. acetonitrile) Strong (weak, non-) electrolytes HNO + H O NO + H O (completely ionized) CH COOH+ H O CH COO + H O (partially ionized) sugar (s) + water = sugar aquesous soln (completely non-ionized) HF HF is a weak electrolyte 1

2 Acids and Bases An acid donates protons, a base accepts protons An acid donates protons only in the presence of a proton acceptor (a base). A base accepts protons only in the presence of a proton donor (an acid) Bronsted-Lowry Acids and Bases These two chemists pointed out that acids and bases can be seen as proton transfer reactions. According to the Bronsted-Lowry concept: An acid donates a proton and the base accepts it. Newly formed ecies are called conjugate base and conjugate acid, reectively. acid + base conjugate base + conjugate acid HCl + NH Cl + NH Acid + base base + acid 1 1 A conjugate base is formed when an acid loses a proton (HClCl - ); A conjugate acid is formed when a base accepts a proton (NH 3 NH4 + )

3 Consider the reaction of NH 3 with H O: NH 3 (aq) + H O (aq) NH 4+ (aq) + OH - (aq) (base1) (acid) (acid1) (base) Note: NH 3 & NH 4+ are a conjugate acid-base pair. H O & OH - : also a conjugate acid-base pair Example Identify the acid and base ecies in the following equations: CO 3 - (aq) + H O(l) HCO 3- (aq) + OH - CH COOH + H O CH COO + H O (Acetic acid) (base) (hydronium) Amphiprotic ecies ( amphi- : both): a ecies that can act as either an acid or a base, depending on the other reactant. Consider water: H O + CH 3 O - OH - + CH 3 OH acid base H O + HBr H 3 O + + Br - base acid 3

HPO + HO HPO + OH - - 4 3 4 base1 acid acid1 base ( amphiprotic) H PO + H O HPO + H O - - + 4 4 3 acid1 base base1 acid NH CH COOH NH CH COO 3 glycine zwitterion sweeterian A

H O + H O H 3 O + + OH - The extent of autoprotolysis is very small.

4 HPO + HO HPO + OH base1 acid acid1 base ( amphiprotic) H PO + H O HPO + H O acid1 base base1 acid NH CH COOH NH CH COO 3 glycine zwitterion sweeterian A zwitterion is an ion that bears both a positive and a negative charge AUTOPROTOLYSIS Water undergoes self-ionization autoprotolysis, since H O acts as an acid and a base. H O + H O H 3 O + + OH - The extent of autoprotolysis is very small. The equilibrium constant expression for this reaction is: K c = [H 3O + ] [OH - ] [H O] The concentration of water is essentially constant. Therefore: [H O] K c = [H 3 O + ]. [OH - ] constant = K w Example: Calculate the hydronium and hydroxide ion concentration of pure water at 5 C and 100 C. 4

5 We call the equilibrium value of the ion product [H 3 O + ][OH - ] the ion-product constant of water. K w = [H 3 O + ] [OH - ] = 1.0 x at 5 C Using K w you can calculate concentrations of H 3 O + and OH - in pure water. [H 3 O + ] [OH - ] = 1.0 x But [H 3 O + ] = [OH - ] in pure water [H 3 O + ] = [OH - ] = 1.0 x 10-7 M If you add an acid or a base to water the concentrations of H 3 O + and OH - will no longer be equal. But K w will still hold. Examples of autoprotolysis (autoionization) H O + H O 3 H O + OH CH OH + CH OH CH OH + CH O HCOOH + HCOOH HCOOH + HCOO NH + NH NH + NH base1 + acid acid1 + base Acid / Base Strengths The strength of an acid or base is determined by the extent it dissociates in water to form H + or OH - reectively Strong acids: HA H + A (100% dissociation) + [H ] = [HA] initial Strong bases: MOH M + OH (100% dissociation) [OH ] = [MOH] initial 5

Weak Acid Weak acids only partially dissociate to give H + in H O Equilibrium Constant HA + H O H 3 O + + A - K a = [H + ][A - ] / [HA] K a acid dissociation constant The smaller the K a value, the

6 Weak Acid Weak acids only partially dissociate to give H + in H O Equilibrium Constant HA + H O H 3 O + + A - K a = [H + ][A - ] / [HA] K a acid dissociation constant The smaller the K a value, the weaker the acid Weak Base Weak bases react with water by abstracting a proton Equilibrium Constant B + H O BH + + OH - K b = [BH + ][OH - ] / [B] K b base hydrolysis constant Hydrolysis reaction with water The smaller the K b value, the weaker the base The strongest acids have the weakest conjugate bases and the strongest bases have the weakest conjugate acids. A acid (base) can be cationic, anionic, neutral. 6

7 Differentiating/Leveling Solvents In water, perchloric and hydrochloric acids are strong acids and have the same strength. In acetic acid, perchloric acid is considerably stronger than HCl; its dissociation being ~5000 times greater. Acetic acid acts as differentiating solven toward the two acids by revealing the inherent differences in their activities, and water is called leveling solvent for perchloric, hydrochloric (and nitric) acids because all three are 100% ionized in water and show no difference in strength. Equilibrium Constant aa bb... cc dd... c d [ C] [ D]... K a b [ A] [ B]... Standard State 1. Solution concentration ([X]) unit: M (mol/l).. Gas concentrations (partial pressure): atm. 3. The concentrations of pure solids, pure liquids, and solvents are omitted because they are (constant) unity. 4. K = f (T). 5. Only thermodynamic (tendency) rather than kinetic (reaction rate) conditions are considered. 7

= [H ][OH ] - w K K K [HA] [A ] a Manipulating Equilibrium Constants (1) () AB CD K EF GH K

8 Acid: HA H + A Conjugate base of HA acid, A A + H O Calculation of K a or K b from its Conjugate K b or K a K [H ][A ] [HA] - - [HA][OH ] HA + OH Kb - [A ] - [H ][A ] [HA][OH ] a b = [H ][OH ] - w K K K [HA] [A ] a Manipulating Equilibrium Constants (1) () AB CD K EF GH K (3) (1) () K K K 3 1 (3) ABEF CDGH K? (4) (1) () K K / K 4 1 (4) ABGH CDEF K? [Class practice] 8

9 Solubility Product Constant (K ) In General: A x B y (s) xa +y + yb -x [A +y ] x [B -x ] y K = [A x B y ] [A x B y ]K=K =[A +y ] x [B -x ] y {[A x B y ](solid)isconstant} Examples of K Expression AgCl() s Ag Cl K [ Ag ][ Cl ] CaF s Ca F K Ca F () [ ][ ] BaSO () s Ba SO K [ Ba ][ SO ] CaCO () s Ca CO K [ Ca ][ CO ] PbI() s Pb I K? Calculating K from the Solubility If the solubility of AgCl is 1.9x10-3 g/l, what would be the K? AgCl( s) Ag Cl, K = [Ag ][Cl ] ( s) s s K = [Ag ][Cl ] s (1.310 ) [ s g / L? M g / L g / L 5 = M ] FW.. AgCl g/ mol 9

10 The solubility of lead(ii) arsenate, Pb 3 (AsO 4 ), is 3.0x10-5 g/l, calculate its K. Pb ( AsO ) ( s) 3Pb AsO K = [ Pb ] [ AsO ] ( s) 3 s s K = [ Pb ] [ AsO ] (3 s) ( s) 108s s g L M 5 [ /? g / L g / L 8 = M ] F. W. Pb ( AsO ) g / mol 3 4 Calculating Solubility from K AgI() s Ag I ( s) s s K [ Ag ][ I ] ss s s 3 K PbI s Pb I ( s) s s K [ Pb ][ I ] = s( s) = 4s s () 3 K K ( ) 4 4 1/3 A general expression for a precipitate dissociation ( x y) A B (s) x y xa + yb ( s) xs ys K [ A] [ B] [ xs] [ ys] x y s s x y x y x y ( x y) K K ; s ( ) x y x y x y x y 1 x y 10

11 Solubility of a Precipitate in Pure Water How many grams of AgCl (fw = 143.3) can be dissolved in ml of water at 5 o C? (K [AgCl] =1.8X10-10 ) AgCl() s Ag Cl ( s) s s K =[ Ag ][ Cl ]= ss s s ( K ) (1.810 ) 1/ 10 1/ M How many grams of AgCl (fw = 143.3) can be dissolved in ml of water at 5 o C? [from previous calculation: s =1.35X10-5 M] weight (# of moles) formula weight ( CV ) formulaweight ( C s) 5 3 = [ ( ) ( )] 143.3( / ) -4 =.0 10 g = 0.0 mg M L g mol The Common Ion Effect In AgCl( s) Ag Cl system, add a strong electrolyte containing Ag or Cl ions (the same as the dissociated ions--common ions) eg.., NaCl Na AgNO Ag Cl or NO 3 3 In the presence of common ions, the solubility of the precipitate will dramatically decrease Common Ion Effect. 11

12 Calculate the solubility of slightly soluble barium fluoride, BaF, in (a) pure water, and (b) 0.15 M NaF. (K = 1.0x10-6 ) (a) In pure water BaF s Ba F () ( s) s s K [ Ba ][ F ] s( s) 4s 3 K s ( ) ( ) M 6 1/3 1/3 (b) In 0.15 M NaF BaF ( s) Ba F ; NaF Na F (0.15 M ) ( s) s s[ F ] remaining [ F ] remaining 0.15M K [ Ba ][ F ] s(0.15) 0.05s 6 K s M Check the assumption: s + [F - ] = 8.8x = 0.15 M Assumption was valid. BaF solubility in 0.15 M NaF is ~1/140 times that in pure water. Class Practice: Calculate the molar solubility of Ag CO 3 in a solution that is M in Na CO 3. (K =8.1X10-1 ) (key: 1.0x10-5 M) 1

13 EXAMPLE: Calculate the molar solubility of Ag CO 3 that results when 00. ml of M AgNO 3 is mixed with 100. ml of M Na CO 3. K = 8.1 X 10-1 The chemical reaction : AgNO3NaCO3 AgCO3( s) NaNO3 moles 1 1 initial moles At eq. moles (:10 molar ratio, Na CO in excess) ( 0.00) ( 0.00) # of moles [CO 3 ] = = M solution volume in L Ag CO ( s) Ag CO K [ Ag ] [ CO ] ( s) s s[ CO ] [ CO ] 3 remaining 3 remaining K [ Ag ] [ CO ] ( s) (0.030) s ( ) / 6 M Check the assumption : [ CO ] s [ CO ] M remaining Assumption was valid! Precipitation Calculations xa + yb? initial conc. [ A] [ B] 0 i A B (s) As A B (s) xa + yb K [ A] x [ B] i y x y eq eq x y Solubility quotient ( Q ) (or ion product) x y Q [ A] [ B] c i i c If Qc K, precipitation occures (saturated solution) If Qc K, NO precipitation (unsaturated solution) If Q K, the reaction mixture is at equilibrium c 13

14 Suppose a solution has [Ca + ] and [SO 4 - ] of M and M, reectively. If these concentrations were doubled by evaporating half the water in the solution, would precipitation of CaSO 4 occur given that the K for calcium sulfate is.4 x 10-5? Solution: Ca SO CaSO? 4 4 Qc [ Ca ][ SO4 ] (0.005 )(0.0041) -5 = K You would expect precipitation to occur. A solution of M lead(ii)nitrate, Pb(NO 3 ), was poured into 456 ml of M sodium sulfate, Na SO 4. Would a precipitate of lead(ii)sulfate, PbSO 4, be expected to form if 55 ml of the lead nitrate solution were added? (The K for lead(ii)sulfate is 1.7 x10-8 )? Pb SO4 PbSO4 Solution: ( CV ) ( CV ) Pb SO4 Qc [ Pb ][ SO4 ] Vtotal Vtotal ( M55 ml) ( ) = (456 55) ml (456 55) ml -5-4 = M M) -9 = K You would NOT expect precipitation to occur. Completeness of Precipitation Lead chromate, PbCrO 4, is a yellow pigment used in paints. Suppose 0.50 L of a 1.0x10-5 MPb(C H 3 O ) and 0.50 L of a 1.0x10-3 M K CrO 4 solution are mixed. Calculate the equilibrium concentration of Pb + ion remaining in the solution after PbCrO 4 precipitates. What is the percentage of Pb + remaining in solution after the precipitation has occurred. (The K for PbCrO 4 is 1.8x10-14 ). Pb + CrO PbCrO (s) Ini. # moles = = Final. # moles Final. Conc. (M) ie.., solid PbCrO 4 is in contact with a solution -4 containing M CrO 4 ions

15 PbCrO (s) Pb + CrO [Initial] [At e.q.] (s) s s K [Pb ][CrO 4 ] = s ( ) = s = [Pb ] = M % of Pb in solution vs precipitated: 100% = % A complete precipitation! Fractional Precipitation Fractional precipitation is the technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth. (Ba, Sr ) BaCrO ( Sr ) + + KCrO KCrO 4 SrCrO ( K K ) [BaCrO 4] [SrCrO 4] 4 Separation by Precipitation EXAMPLE: Can Fe 3+ and Mg + be separated quantitatively as hydroxides from a solution that is 0.10 M in each cation? If the separation is possible, what range of OH - concentrations is permissible. K =[Fe 3+ ][OH - ] 3 = X K =[Mg + ][OH - ] = 7.1 X

16 Assume quantitative separation requires that the concentration of the less soluble material to have decreased to < 1x10-6 M before the more soluble material begins to precipitate. Assume [Fe 3+ ] = 1.0x10-6 M What will be the [OH - ] required to reduce the [Fe 3+ ]to[fe 3+ ] = 1.0x10-6 M? K =[Fe 3+ ][OH - ] 3 = x10-39 x10-39 [OH - ] 3 = = x x10-6 [OH - ] = 1.3x10-11 M K =[Mg + ][OH - ] = 7.1 X 10-1 What [OH - ] is required to begin the precipitation of Mg(OH)? [Mg + ] = 0.10 M (0.10 M)[OH - ] = 7.1 X x10-1 [OH - ] = = 7.1x [OH - ] = 8.4x10-6 M Since 1.3x10-11 Mislessthan8.4x10-6 M, the Fe 3+ will be quantitatively removed long before the Mg(OH) begins to form. Therefore, the separation is possible between [OH - ] 1.3x10-11 M and [OH - ] = 8.4x10-6 M. 16

17 Effect of ph on Solubility () CaF s Ca F AgCl() s Ag Cl AgCN () s Ag CN What changes in their solubilities if (1) + HCl () + HNO 3 (3) H SO 4 F + H - + CN + H - + HF (weak acid) HCN (weak acid) Ca + SO CaSO (s) Which of the following does affect the solubility of a precipitate? 17

18 A saturated solution of Cu(II) hydroxide has a ph of 10. Calculate the K for the Cu(II) hydroxide. Acid-Base Dissociation Constants Example Calculating concentrations of H 3 O + and OH - in solutions of a strong acid or base. Calculate the concentrations of hydronium ion and hydroxide ion at 5 C in 0.10 M HCl. [H 3 O + ] Calculation of Weak Acids HA [H3O ][A ] 3 Ka HA + H O H O + A = c ~0 0 K [ c x] = x or x + K xk c 0 a HA a a HA [HA] [][] x x cha x x x Ka = [ c x ] + Ka Ka 4KacHA x [H3O ] If K very small and/or c large : c x c a HA HA HA x [H O ] K c + 3 a HA HA a HA (if c / K 1000, Err% 1.6% ) 18

19 [OH - ] Calculation of Weak Bases B + H O BH + OH K = c B b 0 ~0 K [ c x] = x or x + K xk c 0 b B b b B - Kb Kb 4KbcB x [OH ] If K very small and/or c large : c xc [BH ][OH ] B [B] [][] x x cb x x x Kb = [ c x ] b B B B x[oh] K c ; [HO ]= K /[OH] - b B 3 w Buffer Solutions A buffer is a solution characterized by the ability to resist changes in ph with dilution or with addition of limited amounts of acid or base. Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Thus, a buffer solution contains both an acid ecies and a base ecies in equilibrium. 19

20 Strong base neutralizes weak acid Strong acid neutralizes weak base How do we calculate the ph of a buffer? Consider mixture of salt NaA and weak acid HA. NaA (s) HA (aq) Na + (aq) + A - (aq) H + (aq) + A - (aq) [H + ] = K a [HA] [A - ] Take log -log [H + ] = -log K a -log [HA] [A - ] -log [H + ] = -log K a + log [A- ] [HA] Remember px = -logx so pk a = -log K a ph = pk a + log [A- ] [HA] K a = [H+ ][A - ] [HA] Henderson-Hasselbalch equation ph = pk a + log [conjugate base] [acid] Assumption: Changes in [A - ] & [HA] will be negligible (within 5%) if K a < 0.01 & K a < 0.01 [base] [acid] use initial conc. of acid and base to calculate ph 0

21 What is the ph of a solution containing 0.30 M HCOOH and 0.5 M HCOOK?(Potassium formate) Mixture of weak acid and conjugate base! Initial (M) Change (M) Equilibrium (M) Common ion effect 0.30 x x 0.5 HCOOH pk a = 3.77 HCOOH (aq) x +x x x H + (aq) + HCOO - (aq) 0.5 +x ph = pk a + log [HCOO- ] [HCOOH] x ph = log [0.5] [0.30] = 4.01 Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na CO 3 /NaHCO 3. (d) NaHCO 3 (a) HF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO - 3 is a weak base and HCO 3- is its conjugate acid buffer solution (d)? Amphiprotic ecies Calculate the ph of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the ph after the addition of 0.0 ml of M NaOH to 80.0 ml of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) ph = pk a + log [NH 3] [NH 4+ ] pk a = 9.5 ph = log [0.30] [0.36] = 9.17 K a = 5.6 x poh = 14 - ph = 4.73 [OH - ] = 1.8 x 10-5 M 1

Calculate moles of all components. Remember we have 80.0 ml of the buffer solution @ 0.30 M NH 3 /0.36 M NH 4 Cl and 0.0 ml of 0.050 M NaOH. start (moles) end (moles) Moles NH 4+ = (0.080 L)(0.

22 Calculate moles of all components. Remember we have 80.0 ml of the buffer 0.30 M NH 3 /0.36 M NH 4 Cl and 0.0 ml of M NaOH. start (moles) end (moles) Moles NH 4+ = (0.080 L)(0.36 M) = 0.09 moles Moles NH 3 = (0.080 L)(0.30 M) = 0.04 moles Moles NaOH = (0.00 L)(0.050 M) = moles final volume = 80.0 ml ml = 100 ml NH + 4 (aq) + OH - (aq) H O (l) + NH 3 (aq) Moles/Volume(L) [NH 4+ ] = [NH 3 ] = ph = log [0.5] [0.8] ΔpH = 0.03 = 9.0 Property of Buffer Solutions The ph remains essentially independent of dilution. Very small changes in ph when limited amounts of acid or base are added (ph <0.01) How do we build a better buffer? Add approximately equal quantities of acid and base Have relatively high concentrations of acid and base the larger the [acid] & [base] the greater the buffer capacity How do we prepare a buffer at a given ph? Choose acid/base conjugate pair from table Check to be sure that they are unreactive in the system used pk a ph typical rule of thumb ph + 1 = pk a

Create buffer with ph = 7.50. ph + 1 = pk a Look for pk a s 6.5 8.5 => K a s 3 x 10-7 3 x 10-9 HOCl / OCl - K a = 3.

3 x 10-7 Calculate quantities of acid and base. ph = pk a + log [conjugate base] [acid] H CO 3 / HCO 3 - K a1 = 4.3 x 10-7 pk a1 = 6.

23 Create buffer with ph = ph + 1 = pk a Look for pk a s => K a s 3 x x 10-9 HOCl / OCl - K a = 3.5 x 10-8 H PO - 4 / HPO - 4 K a = 6. x 10-8 H AsO - 4 / HAsO - 4 K a = 8 x 10-8 H CO 3 / HCO - 3 K a1 = 4.3 x 10-7 Calculate quantities of acid and base. ph = pk a + log [conjugate base] [acid] H CO 3 / HCO 3 - K a1 = 4.3 x 10-7 pk a1 = 6.37 ph - pk a = log [HCO 3 - ] [H CO 3 ] = 13.6 = = 1.13 [HCO 3 - ] [H CO 3 ] Set [H CO 3 ] = M (NOTE: This is a judgment call.) then [HCO 3 - ] = 13.6 [H CO 3 ] = M 3

24 Acid Rain and The Buffer Capacity of Lakes 4

Chemical Equilibrium. What is the standard state for solutes? a) 1.00 b) 1 M c) 100% What is the standard state for gases? a) 1 bar b) 1.

Chemical Equilibrium. What is the standard state for solutes? a) 1.00 b) 1 M c) 100% What is the standard state for gases? a) 1 bar b) 1. Chemical Equilibrium Equilibrium constant for the reaction: aa + bb + cc + dd + [C ] c [D ] d... equilibrium constant K = [ A] a [B ] b... [] = concentration relative to standard state molarity (M): for