 # Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis.

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Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

I. Common ion effect Common ion: an ion that is produced from more than one solute Common ion effect: shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. Example: Consider CH 3 COOH (weak acid, partial ionizes) Add CH 3 COONa (strong electrolyte, completely ionizes) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) (*) CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) common ion Le Chatelier's principle: equilibrium (*) shift This shift to the left will also result in a decrease of the [H + ] The ionization of a weak electrolyte is decreased by adding a strong electrolyte (i.e. a salt) that produces an common ion

Ex: What is the the pH of a solution containing 0.20 M CH 3 COOH and 0.30 M CH 3 COONa at 25 o C? (K a =1.8 x10 -5 ). CH 3 COOH (aq) H + (aq) + CH 3 COO – (aq) Initial (M) Change (M) Equilibrium (M) 0.200.00 -x-x+x+x 0.20 - x 0.30 +x+x x0.30+ x K a = x(0.30+x) 0.20 - x = 1.8 x 10 -5 Ka  Ka  0.30x 0.20 = 1.8 x 10 -5 Assume 0.20 – x  0.20 0.30 + x  0.30 x = 1.2 x 10 -5 M [H + ] = [CH 3 COO – ] = 1.2 x 10 -5 M pH = -log [H + ] = 4.92

II. Buffer solution A. A solution that resist pH change when acids or base are added. B. In general, a buffer is made of –A weak acid and its salt –A weak base and its salt HF+NaF NH 3 +NH 4 Cl Add strong acid H + (aq) + HCOO - (aq) HCOOH (aq) Add strong base OH - (aq) + HCOOH (aq) HCOO - (aq) + H 2 O (l) Consider an equal molar mixture of HCOOH + HCOONa ? How does buffer work ?

Ex. Which of the following can be used to prepare for buffer solutions? (a) NaF/HF (b) NaBr/HBr, (c)Na 2 SO 3 /NaHSO 3, and (d) CH 3 COOH and NaOH. (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) SO 3 2- is a weak base and HSO 3 - is it conjugate acid buffer solution (d) CH 3 COOH is a weak base and NaOH is a strong base buffer solution CH 3 COOH + NaOH --> CH 3 COONa + H 2 O

C. pH of a Buffer solution Consider a mixture of a weak acid (HA) and its salt (XA) HA (aq) H + (aq) + A - (aq) XA (s) X + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] = - log K a + log [A - ] [HA] pH = pK a + log [conjugate base] [acid] Henderson-Hasselbalch equation pK a = -log K a As pK a, the acid strength______.

Ex: What is the the pH of a solution containing 0.20 M CH 3 COOH and 0.30 M CH 3 COONa at 25 o C? (K a =1.8 x10 -5 )? What is the pH after the addition of (a) 20.0 mL of 0.050 M NaOH or (b) 20.0 mL of 0.050 M HCl to 80.0 mL of this buffer solution? Henderson-Hasselbalch eq. pH = pK a + log [conjugate base] [acid] pH = -log(1.8 x10 -5 )+ log 0.30 0.20 = 4.74+ 0.18 = 4.92 (a) Addition of 20.0 mL of 0.050 M NaOH (0.001 mol NaOH) NaOH will react with the acid in the buffer CH 3 COOH (aq) + OH - (aq) CH 3 COO – (aq) + H 2 O (l) start (moles) end (moles) 0.0160.0010.024 final volume = 80.0 mL+ 20.0 mL=100.0 mL = 4.96 pH = 4.74 + log [0.25] [0.15] [CH 3 COOH] = =0.15 M 0.015 0.10 [CH 3 COO – ] = =0.25 M 0.025 0.10 0.0150.000.025

Continued Henderson-Hasselbalch eq. pH = pK a + log [conjugate base] [acid] pK a = 4.74 (b) Addition of 20.0 mL of 0.050 M HCl (0.001 mol HCl) HCl will react with the base in the buffer CH 3 COO – (aq) + H + (aq) CH 3 COOH (aq) start (moles) end (moles) 0.0240.0010.016 0.0230.00.017 final volume = 80.0 mL+ 20.0 mL=100.0 mL = 4.87 pH = 4.74 + log [0.23] [0.17] [CH 3 COOH] = =0.17 M 0.017 0.10 [CH 3 COO – ] = =0.23 M 0.023 0.10

D. Preparation of buffer Select acid/base pair with pK a that is closest to the desired pH. Adjust ratio of base/acid to have the desired pH. pH = pK a + log [conjugate base] [acid] Ex. How to prepare a buffer with pH = 3.85? [H + ] =10 -3.85 =1.4 x10 -4 HCOOH seems to be a good choice, K a =1.7 x10 -4 pH = 3.85 = -log(1.7 x10 -4 ) + log [HCOO – ] [HCOOH] [HCOO – ] [HCOOH] log =0.080 [HCOO – ] [HCOOH] =1.2 The buffer can be prepared by mixing 1.0 M HCOOH and 1.2 M HCOONa

E. Buffer capacity Quantity of acid or base that can be added before pH changes significantly –Buffers work best when the ratio of [base]/[acid] is 1. –The greater the concentrations of both [base] and [acid], the ________ buffering capacity Ex. Which of the following has the greater buffering capacity? (a) 1.0 L of 1.0 M NaF + 1.0 M HF solution, or (b) 1.0 L of 0.1 M NaF + 0.1M HF solution greater

III. Titration Acid-base titration: to determine the concentration of unknown solution Standard solution (titrant): a solution of known concentration Equivalence point : the point at which the reaction is complete Indicator : a substance changes color at/near the equivalence point.

A. Strong acid-strong base titration NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l) Titration curve: plot of pH as a function of mL

Q. A 40.0 mL sample of 0.200 M HCl is titrated with 0.200 M NaOH. Calculate the pH of the soution after the following volume of NaOH have been added. (a) 35.0 mL, (b) 39.0 mL, (c) 41.0 mL, and (d) 45.0 mL. H + (aq) + OH - (aq) H 2 O (l) (a)mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0350 L x 0.200 M = 0.00700 mol excess HCl: 0.00800 mol - 0.00700 mol = 0.00100 mol Final volume = (40.0 + 35.0)mL =75.0 mL [H + ]=0.00100 mol/ 0.0750 L = 0.0133 M pH= -log[H + ]= 1.875 (b) mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0390 L x 0.200 M = 0.00780 mol excess HCl: 0.00800 mol - 0.00780 mol = 0.00020 mol Final volume = (40.0 + 39.0)mL =79.0 mL [H + ]=0.00020 mol/ 0.0790 L = 0.0025 M pH= -log[H + ]= 2.60 Initial pH = 0.699

continued H + (aq) + OH - (aq) H 2 O (l) (c)mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0410 L x 0.200 M = 0.00820 mol excess NaOH: 0.00820 mol - 0.00800 mol = 0.00020 mol Final volume = (40.0 + 41.0)mL =81.0 mL [OH – ]=0.00020 mol/ 0.0810 L = 0.0025 M pOH= -log[OH – ]= 2.61 pH=14.0 -pOH = 11.39 (d) mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0450 L x 0.200 M = 0.00900 mol excess NaOH: 0.00900 mol - 0.00800 mol = 0.00100 mol Final volume = (40.0 + 45.0)mL =85.0 mL [OH – ]=0.00100 mol/ 0.0850 L = 0.0118 M pOH= -log[OH – ]= 1.929 pH=14.0 -pOH = 12.071

B. Weak acid-strong base titration CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7): Buffering region

Q. Exactly 100 mL of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) 0.01 0.0 0.01 NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0.050.00 -x-x+x+x 0.05 - x 0.00 +x+x xx [NO 2 - ] = 0.01 0.200 = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x2 0.05-x = 2.2 x 10 -11 0.05 – x  0.05x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

B. Weak acid-strong base titration curve 1.Initial pH of a weak acid 2.Before equivalence point: buffering region 3.At the equivalence point: a weak base 4.After the equivalence point: excess strong base raises pH Ka=2 x10 -6 Ka=2 x10 -8 Ka=2 x10 -4 pH=pKa equivalence point 50 mL of 0.1 M weak acid is titrated with 0.1 M NaOH

C. Strong acid-weak base titration HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): H + (aq) + NH 3 (aq) NH 4 Cl (aq) Buffering region

IV. Acid-base indicator HIn (aq) H + (aq) + In - (aq)  10 [HIn] [In - ] Color of acid (HIn) predominates  0.1 [HIn] [In - ] Color of conjugate base (In - ) predominates

The titration curve of a strong acid with a strong base

Q: Which indicator(s) would you use for a titration of HCOOH with NaOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein

V. Solubility equilibria A. Solubility product constant, K sp AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ][F - ] 2 Ag 2 SO 4 (s) 2Ag + (aq) + SO 4 2 - (aq) K sp = [Ag + ] 2 [SO 4 2 - ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 - (aq) K sp = [Ca 2+ ] 3 [PO 3 3 - ] 2

B. Solubility and K SP a. Dissolution of an ionic solid in aqueous solution: Q = K sp Saturated solution Q < K sp Unsaturated solution No precipitate Q > K sp Supersaturated solution Precipitate will form b. Determine K sp from solubility or solubility from K sp Molar solubility (mol/L): moles of solute in 1 L of a saturated solution Solubility (g/L): grams of solute dissolved in 1 L of a saturated solution.

Q. What is the solubility of silver chloride in g/L ? K sp =1.6 x 10 -10 AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) 0.00 +s+s +s+s ss K sp = s 2 s = K sp  s = 1.3 x 10 -5 [Ag + ] = 1.3 x 10 -5 M[Cl - ] = 1.3 x 10 -5 M Solubility of AgCl = 1.3 x 10 -5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x = 1.9 x 10 -3 g/L

Q. The solubility of SrF 2 is found to be 1.1 x 10 -2 g in 100 mL of aqueous solution. Determine the value of K sp of SrF 2. SrF 2 (s) Sr 2+ (aq) + 2 F - (aq) Initial (M) Change (M) Equilibrium (M) 0.00 +s+s +2s s 2s 2s K sp = [Sr 2+ ][F – ] 2 = s(2s) 2 s: Solubility of SrF 2 in M = -s-s 1.1 x10 -2 g SrF 2 /(125.6 g/mol) 0.100 L soln = 4 s 3 = 8.74 x 10 –4 M = 2.7 x 10 -9

Q. If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? (K sp = 8.0 x10 -6 ) Ca(OH) 2 (s) Ca 2+ (aq) + 2 OH - (aq) Which ions are present in solution? Only possible precipitate is _________(solubility rules). Is Q > K sp for Ca(OH) 2 ? [Ca 2+ ] 0 = 0.100 M x 1.00 L/(1.00 L +0.00200 L)=0.100 M [OH - ] 0 = 0.00200 L x 0.200 M /(1.00L + 0.00200 L)=4.0 x 10 -4 M K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10 -6 Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form Na +, OH -, Ca 2+, Cl -. Ca(OH) 2

C. Factors affecting solubility 1.Common ion effect: common ion decreases the solubility. CaF 2 (s) Ca 2+ (aq) + 2F - (aq) If NaF is added, equilibrium shifts __________. left Ex. What is the molar solubility of CaF 2 in (a) pure water and (b) 0.010 M NaF? (K sp =3.9 x10 -11 ) K sp = 3.9 x 10 -11 =s(2s) 2 = 4s 3 s = 2.1 x 10 -4 NaF (s) Na + (aq) + F - (aq) [F - ] = 0.010 M [F - ] = 0.010 + 2s  0.010 K sp = (0.010) 2 x s s = 3.9 x 10 -7 CaF 2 (s) Ca 2+ (aq) + 2 F - (aq) (a) (b) CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Initial (M) Change (M) Equilibrium (M) 0.00 0.010 s 0.010+2s -s +s +2s

2. Effect of pH Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutions removeadd In acidic solution, solubility_________. Why? increases CaF 2 (s) +2 H + (aq) Ca 2+ (aq) + 2 HF (aq) Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) Ex. CaF 2 (s) Ca 2+ (aq) + 2 F - (aq) Ex. F – (aq) + H + (aq) HF (aq) Ex. Which salt will be more soluble in acidic solution than in pure water (a) ZnCO 3, (b) AgCN, and (c) BiI 3. (a) and (b) because CO 3 2– and CN – react with H +

Q. What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br - and Cl - at a concentration of 0.010 M? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] K sp = 1.6 x 10 -10 AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x 10 -13 K sp = [Ag + ][Br - ] [Ag + ] = K sp [Br - ] 7.7 x 10 -13 0.010 = = 7.7 x 10 -11 M [Ag + ] = K sp [Cl - ] 1.6 x 10 -10 0.010 = = 1.6 x 10 -8 M 7.7 x 10 -11 M < [Ag + ] < 1.6 x 10 -8 M

VI. Complex ion equilibria and solubility 1.Complex ion: an ion containing a central metal cation (Lewis acid) bonded to one or more molecules or ions (Lewis base). 2.Solubility of slightly soluble salts can be increased by complex ion formation AgCl (s) Ag + (aq) + Cl – (aq) Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) By adding NH 3, solubility of AgCl increases due to 3.The formation constant or stability constant (K f ) is the equilibrium constant for the complex ion formation. K f = =1.7 x10 7 [Ag(NH 3 ) 2 + ] [Ag + ][NH 3 ] 2 KfKf stability of complex

VII. Qualitative analysis

Solution contains Na +, K +, NH ions 4 + Solution containing ions of remaining groups Solution containing ions of all cation groups Solution containing ions of remaining groups Group 2 precipitates CuS, CdS, SnS, Bi 2 S 3 Group 1 precipitates AgCl, Hg 2 Cl 2, PbCl 2 Solution containing ions of remaining groups Filtration +HCl Filtration +H 2 S Group 4 precipitates BaCO 3, CaCO 3, SrCO 3 Filtration +Na 2 CO 3 Group 3 precipitates CoS, FeS, MnS, NiS ZnS, Al(OH) 3,Cr(OH) 3 Filtration +NaOH Flow chart for separation of cations

Flame test of cations lithium sodium potassiumcopper

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