Applications of Aqueous Equilibria - General Chemistry | CHEM 142, Study notes of Chemistry

Download Applications of Aqueous Equilibria - General Chemistry | CHEM 142 and more Study notes Chemistry in PDF only on Docsity! Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids 8.8 Solubility Equilibria and The Solubility Product 8.9 Precipitation and Qualitative Analysis 8.10 Complex Ion Equilibria Le Châtelier’s principle for the dissociation equilibrium for HF HF(aq) H+(aq) + F-(aq) Adding any salt with F-: “Common ion effect” Ka = [H+][F-] [HF] Example 8.2 (P279-82) - I A buffered solution contains 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution, and the pH after 0.010 M of solid NaOH is added to 1.0 L of this buffer and to pure water. HC2H3O2 (aq) H+(aq) + C2H3O2 (aq) Ka = 1.8 x 10-5 = [H+] [C2H3O2-] [HC2H3O2] Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HC2H3O2]0 = 0.50 [HC2H3O2] = 0.50 – x [C2H3O2-]0 = 0.50 [C2H3O2-] = 0.50 +x [H+]0 = ~ 0 [H+] = x X mol/L of HC2H3O2 dissociates to reach equilibrium Example 8.2 (P279-82) - II Ka = 1.8 x 10-5 = = = [H+][C2H3O2-] [HC2H3O2] ( x ) ( 0.50 + x) 0.50 - x (x) (0.50) 0.50 x = 1.8 x 10-5 The approximation by the 5% rule is fine: [H+] = x = 1.8 x 10-5 M and pH = 4.74 To calculate the pH and concentrations after adding the base: OH- + HC2H3O2 H2O + C2H3O2- Before reaction: 0.010 mol 0.50 mol - 0.50 mol After rxn. (if 100%): 0.010 – 0.010 0.50 – 0.010 - 0.50 + 0.010 = 0 mol = 0.49 mol = 0.51 mol Note that 0.01 mol of acetic acid has been converted to acetate ion by the addition of the base. ~ Example 8.2 (P279-82) - III Initial Concentration (mol/L) Equilibrium concentration (mol/L) [HC2H3O2]0 = 0.49 [HC2H3O2] = 0.49 – x [C2H3O2-]0 = 0.51 [C2H3O2-] = 0.51 + x [H+]0 = 0 [H+] = x x mol/L of HC2H3O2 dissociates to reach equilibrium Ka = 1.8 x 10-5 = = = [H+][C2H3O2-] [HC2H3O2] (x)(0.51+ x) 0.49 - x (x)(0.51) 0.49 x = 1.7 x 10-5 and pH = 4.76 If the base is added to pure water without the buffer being present we get an entirely different solution: If the 0.01 mol of NaOH is added to 1.0 L of pure water the Concentration of hydroxide ion is 0.01 M. [H+] = = = 1.0 x 1012 and the pH = 12.00Kw[OH-] 1.0 x 10-14 1.0 x 10-2 When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the [HA]/[A-] ratio and thus the [H+] stay virtually constant. The Effect of Added [Acetate-] on pH of an Acetic Acid Buffer [CH3COOH] [CH3COO-]added % Dissociation* pH 0.10 0.00 1.3 2.89 0.10 0.050 0.036 4.44 0.10 0.10 0.018 4.74 0.10 0.15 0.012 4.92 * % Dissociation = x 100 [CH3COOH]dissoc [CH3COOH]init How a Buffer Works–I A buffer consists of a solution that contains “high” concentrations of the acidic and basic components. This is normally a weak acid and the anion of that weak acid, or a weak base and the corresponding cation of the weak base. When small quantities of H3O+ or OH- are added to the buffer, they cause a small amount of one buffer component to convert into the other. As long as the amounts of H3O+ and OH- are small as compared to the concentrations of the acid and base in the buffer, the added ions will have little effect on the pH since they are consumed by the buffer components. Consider a buffer made from acetic acid and sodium acetate: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Ka = or [H3O+] = Ka x [CH3COO-] [H3O+] [CH3COOH] [CH3COOH] [CH3COO-] How Does a Buffer Work? Add a strong acid to a weak acid / salt buffer and see what happens: H+ + A- HA Original buffer pH Final pH of buffer close to original- Added H+ ions Replaced by HA Ka = [H+] [A-] [HA] [H +] = Ka [HA] [A-] How a Buffer Works–IV Suppose we add 1.0 mL of a concentrated base instead of an acid. Add 1.0 mL of 12.0 M NaOH to 1.00 L of our buffer, and let’s see what the impact is: 1.00 mL x 12.0 mol OH-/1000mL = 0.012 mol OH- This will reduce the quantity of acid present and force the equilibrium to produce more hydronium ion to replace that neutralized by the addition of the base! Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+ Initial 0.250 ---- 0.250 Change w/ base - 0.012 ---- +0.012 (if 100% completion) Equilibrium 0.238 ---- 0.262 Assuming: Again, using x as the quantity of acid dissociated we get: our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238 [H3O+] = 1.8 x 10-5 x = 1.635 x 10-5 0.238 0.262 pH = -log(1.635 x 10-5) = 5.000 - 0.214 = 4.79 After base is added! How a Buffer Works–V By adding the 1.00mL base to 1.00 L of pure water we would get a hydroxide ion concentration of: 1.00 L 0.012 mol OH-[OH-] = = 1.2 x 10-2 M OH- The hydrogen ion concentration is: [H3O+] = = = 8.3 x 10-13 Kw [OH-] 1 x 10-14 1.2 x 10-2 M pH = -log(8.3 x 10-13) = 12.1 With 1.0 mL of the base in pure water! In summary: Buffer alone: pH = 4.74 1.00 L Buffer plus 1.0 mL base: pH = 4.79 Base alone: pH = 12.1 100 mL Buffer plus 1.0 mL acid: pH = 4.3 Acid alone: pH = 0.92 The Relation Between Buffer Capacity and pH Change Te —_h, oO 0.30 0.10 Concentration (M) of buffer components 0.030 (initial pH = 4.74) 4.8 pH TT | T oT T | | | | T | T 4.9 5.0 5.1 pH and Capacity of Buffered Solutions The pH of a buffered solution is determined: mainly by the pKa of the acid, and much more weakly by the ratio [A-]/[HA]: pH = pKa + log ([A-]/[HA]) The capacity of a buffered solution is determined: by the lower magnitude of [HA] or [A-]. Buffers well when: [added base or acid] << [HA] and [A-] and [HA] ≈ [A-] Preparing a Basic Buffer Problem: The ammonia-ammonium ion buffer has a pH of about 9.2 and can be used to keep solutions in the basic pH range. What mass of ammonium chloride must be added to 400.00 mL of a 3.00 M ammonia solution to prepare a buffer ? Plan: The conjugate pair is the ammonia-ammonium ion pair which has an equilibrium constant Kb = 1.8 x 10 -5. The reaction equation with water can be written along with the Kb expression, since we want to add sufficient ammonium ion to equal the aqueous ammonia concentration. Solution: The reaction for the ammonia-ammonium ion buffer is: NH3 (aq) + H2O(l) NH4+(aq) + OH-(aq) Want: [NH4+] = [NH3] [NH3] = 3.00 mol x 0.400 L = 1.20 mol NH4Cl = 53.49 g/mol Therefore mass =NH4Cl = 1.20 mol x 53.49g/mol mass = 64.2 g NH4Cl(s) L pH = pKa + log ([base]/[acid]) For basic buffers this becomes: pH = (14.00-pKb) + log ([base]/[acid]) pH = (14.00 – log pKb) + log (0.060/0.040) = 14.00 - log (1.8x10-5) + 0.18 = 14.00 - 4.74 + 0.18 = 9.44 Thus, this named equation can be used for calculating the pH of buffered solutions of acids or bases. Or, use Henderson-Hasselbalch equation: pH Box pH [H3O+] pOH [OH-] pOH = -log[OH-] [OH-] = 10-pOH pH = -log[H3O+] [H3O+] = 10 -pH Kw = 1 x 10-14 @ 25oC [H3O+][OH-]= = 1 x 10-14 pH + pOH = = 14 @ 25oC Exact Treatment of Buffer Solutions Only important: when pKa near 7 and [HA] + [A-] <10-5 M So, read this section so you know the idea and where to find it, but it will not be on the final. Skim next 4 slides. pH and Capacity of Buffered Solutions The pH of a buffered solution is determined: mainly by the pKa of the acid, and much more weakly by the ratio [A-]/[HA]: pH = pKa + log ([base]/[acid]) The capacity of a buffered solution is determined: by the lower magnitude of [HA] or [A-]. Buffers well when: [added base or acid] << [HA] and [A-] and [HA] ≈ [A-] Summary: Characteristics of Buffered Solutions Buffered solutions contain relatively large concentrations of a weak acid and its corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+. When H+ is added to a buffered solution, it reacts essentially to completion with the weak base present: H+ + A- HA or H+ + B BH+ When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present. OH- + HA A- + H2O or OH- + BH+ B + H2O The pH of the buffered solution is determined by the ratio of the concentrations of the weak base and weak acid. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A- or G and BH+) are large compared with the amounts of H+ or OH- added. BUFFERS and The Henderson-Hasselbalch equation: pH = pKa + log ([base]/[acid]) For basic buffers this becomes: pH = (14.00-pKb) + log ([base]/[acid]) This named equation can be used for calculating the pH of buffered solutions of acids or bases, and even after adding strong acid or strong base, provided the conc. is small compared to buffer ingredients. Assume they react 100%. Titration of a strong acid (HCl) solution by a strong base (NaOH). Equivalence point: mol acid = mol base mol H+ = mol OH- Quantitative Analysis How concentrated is the acid sample? Quantify! pH=7.0 for strong acid / strong base titration Eq. Pt. Why does pH change so rapidly near equivalence point? Titrate 50.0mL 0.200M HNO3 (10.0 mmol H+) with 0.100M NaOH 1) Add 10.0 mL NaOH (1.00 mmol) 1.0 mmol OH- + 10.0 mmol H+ = 9.0 mmol H+ [H+] = 9.0 mmol/(50.0mL + 10.0 mL)=0.15M pH = 0.82 2) Add 2 mL more NaOH 1.2 mmol OH- + 10.0 mmol H+ = 8.8 mmol H+ [H+] = 8.8 mmol/(50.0mL + 12.0 mL)= 0.14M pH = 0.85 Change in pH = 0.03 3) After 99.0 mL NaOH added (9.90 mmol) 9.9 mmol OH- + 10.0 mmol H+ = 0.1 mmol H+ [H+] = 0.1 mmol/(149.0 mL) = 7 x 10-4M pH = 3.2 4) Add 2 mL more NaOH 10.1 mmol OH- + 10.0 mmol H+ = 0.1 mmol OH- [OH-] = 0.1 mmol/(151 mL) = 7 x 10-4M pOH = 3.2 pH = 14 - 3.2 = 10.8 Change in pH = +7.6 TABLE 8.3 Selected pH Values Near the Equivalence Point in the Titration of 100.0 mL of 0.100 M HCl with 0.100 M NaOH NaOH Added (mL) pH oo) 5.3 100.00 7.0 100.01 8.7 Figure 8.2: The pH curve for the titration of 100.0 ml of 0.50 M NaOH with 0.10 M HCl. Weak acid Weak acid Strong acid Vol NaOH Treat the stoichiometry Calculating the pH During a Weak Acid-Strong Base Titration–I Problem: Calculate the pH during the titration of 20.00 mL of 0.250 M nitrous acid (HNO2; Ka = 4.5 x 10-4) after adding different volumes of 0.150 M NaOH : (a) 0.00 mL (b) 15.00 mL (c) 20.00 mL (d) 35.00 mL. Plan: (a) We just calculate the pH of a weak acid. (b)-(d) We calculate the amounts of acid remaining after the reaction with the base, and the anion concentration, and plug these into the Henderson-Hasselbalch eq. Solution: HNO2 (aq) + NaOH(aq) H2O(l) + NaNO2 (aq) HNO2 (aq) + H2O(l) H3O+(aq) + NO2-(aq) (a) Ka = = = 4.5 x 10-4 [H3O+] [NO2-] [HNO2] x (x) 0.250 M x2 = 1.125 x 10-4 x = 1.061 x 10-2 pH = -log(1.061 x 10-2) = 2.000 - 0.0257 = 1.97 no base added pH = pKa + log = 3.35 + log(0.00225/0.00275) ( )[NO2-][HNO2] Calculating the pH During a Weak Acid-Strong Base Titration–II (b) 15.0 mL of 0.150 M NaOH (15.0 mL x 0.150 mmol/mL = 2.25 mmol of OH-) is added to the 20.0 mL of 0.250 M HNO2 (5.00 mmol HNO2) which will neutralize 2.25 mmol of HNO2, leaving 5.00-2.25 = 2.75 mmol HNO2, and generate 2.25 mmol of nitrite anion. Concentration (M) HNO2 (aq) + H2O(l) H3O+(aq) + NO2-(aq) Initial 0.00275/V ---- 0 0.00225/V Change -x ---- +x +x Equilibrium 0.00275/V - x ---- x 0.00225/V + x pH = 3.35 -0.0872 = 3.26 after 15.0 mL of NaOH Figure 8.4: The pH curves for the titrations of 50.0 mL of 0.10 M acids Summary: Titration Curve Calculations A Stoichiometry problem. The reaction of hydroxide ion with the weak acid is assumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined. An equilibrium problem. The position of the weak acid equilibrium is determined, and the pH is calculated. TABLE 8.2 Summary of Results for the Titration of 100.0 mL 0.050 M NH; with 0.10 M HCl Volume of 0.10 M HCl Added (mL) [NH3]o INH," Jo ([H*] pH 0 0.050 M 0 11x10°''M 10.96 4.0 mmol 1.0 mmol TO i 4X ‘ ane (Won Comim i" “~ 7 : 2.5 mmol 2.5 mmol m0 .Q* 6% : 2s) (100 + 25) mL (100 + 25) mL St i a= 5.0 mmol = i SEE EESEREe x 6 50.0 0 (100 + 50) mL 43x10 °M 5.36 5.0 mmol 1.0 mmol + nae 60,0 ° (100 + 60) mL 160 mL 2. =62x103°M *Halfway point. tEquivalence point. '[H*] determined by the 1.0 mmol of excess H*. Figure 8.6: The indicator phenolphthalein is pink in basic solution and colorless in acidic solution. Figure 8.9: pH curve of 0.10 M HCI being titrated with 0.10 M NaOH Weak Polyprotic Acid Titration Curve “eon of 40.00 mL of 0.1000 H,SO, with 0.1000 M NaOH pH = 9.86 12, at second equivalence 10. point Z HSO,-] = [SO,7-] pK,.=7.19~ 8 [HSO, 3 2 ~~ [een septa eareee ree Melee pene alee o 6-4 4 4 equivalence 23 = ; oint pKa, = 1.85 a Butter reo" ' 0 qT qT qT q 1 20 40 60 80 100 Volume of NaOH added (mL) DISSOLUTION OF SALTS AND THEIR PRECIPTATION: THE EQUILIBRIUM CONSTANT DESCRIBING SOLUBILITY Equilibria of Slightly Soluble Ionic Compounds When a solution becomes saturated and a precipitate forms, or only part of a salt dissolves as ions, we move into the area of partially soluble salts in solution. We often wish to calculate the quantity of material that remains in solution in equilibrium with the solid. This is facilitated by a special type of equilibrium constant (K) called the: “Solubility Product” or “Solubility Constant” or “Solubility Product Constant”: K = Ksp Example : Lead chromate PbCrO4 (s) Pb2+(aq) + CrO42-(aq) Ksp = [Pb2+][CrO42-] [PbCrO4 (s)] Since the “concentration” of a solid is constant, we treat it with “activity” = 1 instead, yielding the simpler expression for the solubility product constant, Ksp: Ksp = [Pb2+][CrO42-] Writing Solubility Product Expressions for Slightly Soluble Ionic Compounds Problem: Write the solubility product expression for (a) silver bromide; (b) strontium phosphate; (c) aluminum carbonate; (d) nickel(III) sulfide. Plan: Write the equation for a saturated solution, then write the expression for the solubility product. Solution: (a) Silver bromide: AgBr(s) Ag+(aq) + Br -(aq) Ksp = [Ag+] [Br -] (b) Strontium phosphate: Sr3(PO4)2 (s) 3 Sr2+(aq) + 2 PO43-(aq) Ksp = [Sr2+]3[PO43-]2 (c) Aluminum carbonate: Al2(CO3)3 (s) 2 Al3+(aq) + 3 CO32-(aq) Ksp = [Al3+]2[CO32-]3 (d) Nickel(III) sulfide: Ni2S3 (s) + 3 H2O(l) 2 Ni3+(aq) + 3 HS -(aq) + 3 OH-(aq) Ksp =[Ni3+]2[HS-]3[OH-]3 Relationship Between Ksp and Solubility at 25oC No. of Ions Formula Cation:Anion Ksp Solubility (M) 2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4 2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4 2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH)2 1:2 6.5 x 10-6 1.2 x 10-2 3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5 Determining Ksp from Solubility Problem: Lead chromate is an insoluble compound that at one time was used as the pigment in the yellow stripes on highways. It’s solubility is 4.33 x 10 -6g/100mL water. What is the Ksp? Plan: We write an equation for the dissolution of the compound to see the number of ions formed, then write the ion-product expression. Solution: PbCrO4 (s) Pb2+(aq) + CrO42-(aq) Molar solubility of PbCrO4 = x x 4.33 x 10 -6g 100 mL 1000 ml 1 L 1mol PbCrO4 323.2 g = 1.34 x 10 -8 M PbCrO4 1 Mole PbCrO4 = 1 mole Pb2+ and 1 mole CrO42- Therefore [Pb2+] = [CrO42-] = 1.34 x 10-8 M Ksp = [Pb2+] [CrO42-] = (1.34 x 10 -8 M)2 = 1.54 x 10-16 Determining Solubility from Ksp Problem: Lead chromate used to be used as the pigment for the yellow lines on roads, and is a very insoluble compound. Calculate the solubility of PbCrO4 in water if the Ksp is equal to 2.00 x 10-16. Plan: We write the dissolution equation, and the ion-product expression. Solution: Writing the dissolution equation, and the ion-product expression: PbCrO4 (s) Pb2+(aq) + CrO42-(aq) Ksp = 2.00 x 10-16 =[Pb2+][CrO42] Concentration (M) PbCrO4 Pb2+ CrO42- Initial ---------- 0 0 Change ---------- +x +x Equilibrium ---------- x x Ksp = [Pb2+] [CrO42-] = (x)(x ) = 2.00 x 10-16 x = 1.41 x 10-8 Therefore the solubility of PbCrO4 in water is 1.41 x 10-8 M Predicting the Formation of a Precipitate: Qsp vs. Ksp Qsp = Ksp : When a solution becomes saturated, no more solute will dissolve, and the solution is called “saturated.” There will be no changes that will occur. Qsp > Ksp : Precipitates will form until the solution becomes saturated. Qsp< Ksp : Solution is unsaturated, and no precipitate will form. The solubility produce constant, Ksp, can be compared to the ion-product constant, Qsp to understand the characteristics of a solution with respect to forming a precipitate. TABLE 8.6 Calculated Solubilities for CuS, Ag2S, and BizSz at 25°C Calculated Solubility Salt i (mol/L) @uSss 10) 9210 AgS 1.6x10°7 3.4x 1071’ BioS3; 1.110 7% 1.0x 10° Predicting Whether a Precipitate Will Form–I Problem: Will a precipitate form when 0.100 L of a solution containing 0.055 M barium nitrate is added to 200.00 mL of a 0.100 M solution of sodium chromate? Plan: We first see if the solutions will yield soluble ions, then we calculate the concentrations, adding the two volumes together to get the total volume of the solution, then we calculate the product constant (Qsp), and compare it to the solubility product constant to see if a precipitate will form. Solution: Both Na2CrO4 and Ba(NO3) are soluble, so we will have Na+, CrO42-, Ba2+ and NO3- ions present in 0.300 L of solution. We change partners, look up solubilities, and we find that BaCrO4 would be insoluble, so we calculate it’s ion-product constant and compare it to the solubility product constant of 2.1 x 10-10: [Ba2+] = = 0.183 M in Ba2+ For Ba2+: [0.100 L Ba(NO3)2] [0.55 M] = 0.055mol Ba2+ 0.055 mol Ba2+ 0.300 L