Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq) after equivalence point added 30.0 mL NaOH 5.0 x 10-4 mol NaOH xs Tro, Chemistry: A Molecular Approach

Adding NaOH to HCHO2 added 10.0 mL NaOH 0.00150 mol HCHO2 pH = 3.56 0.00100 mol NaOH xs pH = 12.22 added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 25.0 mL NaOH equivalence point 0.00250 mol CHO2− [CHO2−]init = 0.0500 M [OH−]eq = 1.7 x 10-6 pH = 8.23 added 5.0 mL NaOH 0.00200 mol HCHO2 pH = 3.14 initial HCHO2 solution 0.00250 mol HCHO2 pH = 2.37 added 12.5 mL NaOH 0.00125 mol HCHO2 pH = 3.74 = pKa half-neutralization added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 15.0 mL NaOH 0.00100 mol HCHO2 pH = 3.92 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52 added 20.0 mL NaOH 0.00050 mol HCHO2 pH = 4.34 Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Titration of 25.0 mL of 0.100 M HCHO2 with 0.100 M NaOH The 1st derivative of the curve is maximum at the equivalence point pH at equivalence = 8.23 Since the solutions are equal concentration, the equivalence point is at equal volumes Tro, Chemistry: A Molecular Approach

at pH 3.74, the [HCHO2] = [CHO2]; the acid is half neutralized half-neutralization occurs when pH = pKa Tro, Chemistry: A Molecular Approach

Titrating Weak Acid with a Strong Base the initial pH is that of the weak acid solution calculate like a weak acid equilibrium problem e.g., 15.5 and 15.6 before the equivalence point, the solution becomes a buffer calculate mol HAinit and mol A−init using reaction stoichiometry calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−init half-neutralization pH = pKa Tro, Chemistry: A Molecular Approach

Titrating Weak Acid with a Strong Base at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established mol A− = original mole HA calculate the volume of added base like Ex 4.8 [A−]init = mol A−/total liters calculate like a weak base equilibrium problem e.g., 15.14 beyond equivalence point, the OH is in excess [OH−] = mol MOH xs/total liters [H3O+][OH−]=1 x 10-14 Tro, Chemistry: A Molecular Approach

Ex 16. 7a – A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0 Ex 16.7a – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the equivalence point Write an equation for the reaction for B with HA. Use Stoichiometry to determine the volume of added B HNO2 + KOH  NO2 + H2O Tro, Chemistry: A Molecular Approach

HNO2 + KOH  NO2 + H2O HNO2 NO2- OH− mols Before mols added - Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH Write an equation for the reaction for B with HA. Determine the moles of HAbefore & moles of added B Make a stoichiometry table and determine the moles of HA in excess and moles A made HNO2 + KOH  NO2 + H2O HNO2 NO2- OH− mols Before 0.00400 ≈ 0 mols added - 0.00100 mols After 0.00300 0.00100 Tro, Chemistry: A Molecular Approach

HNO2 + H2O  NO2 + H3O+ HNO2 NO2- OH− mols Before mols added - Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. HNO2 + H2O  NO2 + H3O+ Write an equation for the reaction of HA with H2O Determine Ka and pKa for HA Use the Henderson-Hasselbalch Equation to determine the pH Table 15.5 Ka = 4.6 x 10-4 HNO2 NO2- OH− mols Before 0.00400 ≈ 0 mols added - 0.00100 mols After 0.00300 Tro, Chemistry: A Molecular Approach

HNO2 + KOH  NO2 + H2O HNO2 NO2- OH− mols Before mols added - Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point Write an equation for the reaction for B with HA. Determine the moles of HAbefore & moles of added B Make a stoichiometry table and determine the moles of HA in excess and moles A made HNO2 + KOH  NO2 + H2O at half-equivalence, moles KOH = ½ mole HNO2 HNO2 NO2- OH− mols Before 0.00400 ≈ 0 mols added - 0.00200 mols After 0.00200 0.00200 Tro, Chemistry: A Molecular Approach

Ex 16. 7b – A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0 Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. HNO2 + H2O  NO2 + H3O+ Write an equation for the reaction of HA with H2O Determine Ka and pKa for HA Use the Henderson-Hasselbalch Equation to determine the pH Table 15.5 Ka = 4.6 x 10-4 HNO2 NO2- OH− mols Before 0.00400 ≈ 0 mols added - 0.00200 mols After Tro, Chemistry: A Molecular Approach

Titration Curve of a Weak Base with a Strong Acid Tro, Chemistry: A Molecular Approach

Titration of a Polyprotic Acid if Ka1 >> Ka2, there will be two equivalence points in the titration the closer the Ka’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH Tro, Chemistry: A Molecular Approach

Monitoring pH During a Titration the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+] using a probe that specifically measures just H3O+ the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator Tro, Chemistry: A Molecular Approach

Monitoring pH During a Titration Tro, Chemistry: A Molecular Approach

HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq) Indicators many dyes change color depending on the pH of the solution these dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq) the color of the solution depends on the relative concentrations of Ind:HInd when Ind:HInd ≈ 1, the color will be mix of the colors of Ind and HInd when Ind:HInd > 10, the color will be mix of the colors of Ind when Ind:HInd < 0.1, the color will be mix of the colors of HInd Tro, Chemistry: A Molecular Approach

Phenolphthalein

Methyl Red Tro, Chemistry: A Molecular Approach

Monitoring a Titration with an Indicator for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH pKa of HInd ≈ pH at equivalence point Tro, Chemistry: A Molecular Approach

Acid-Base Indicators

Solubility Equilibria all ionic compounds dissolve in water to some degree however, many compounds have such low solubility in water that we classify them as insoluble we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water Tro, Chemistry: A Molecular Approach

Solubility Product the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp for an ionic solid MnXm, the dissociation reaction is: MnXm(s)  nMm+(aq) + mXn−(aq) the solubility product would be Ksp = [Mm+]n[Xn−]m for example, the dissociation reaction for PbCl2 is PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) and its equilibrium constant is Ksp = [Pb2+][Cl−]2 Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Molar Solubility solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature the molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated solution for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq) Tro, Chemistry: A Molecular Approach

PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C Write the dissociation reaction and Ksp expression Create an ICE table defining the change in terms of the solubility of the solid PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) Ksp = [Pb2+][Cl−]2 [Pb2+] [Cl−] Initial Change +S +2S Equilibrium S 2S Tro, Chemistry: A Molecular Approach

Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C Substitute into the Ksp expression Find the value of Ksp from Table 16.2, plug into the equation and solve for S Ksp = [Pb2+][Cl−]2 Ksp = (S)(2S)2 [Pb2+] [Cl−] Initial Change +S +2S Equilibrium S 2S Tro, Chemistry: A Molecular Approach

Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M Tro, Chemistry: A Molecular Approach

PbBr2(s)  Pb2+(aq) + 2 Br−(aq) Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M Write the dissociation reaction and Ksp expression Create an ICE table defining the change in terms of the solubility of the solid PbBr2(s)  Pb2+(aq) + 2 Br−(aq) Ksp = [Pb2+][Br−]2 [Pb2+] [Br−] Initial Change +(1.05 x 10-2) +2(1.05 x 10-2) Equilibrium (1.05 x 10-2) (2.10 x 10-2) Tro, Chemistry: A Molecular Approach

Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M Substitute into the Ksp expression plug into the equation and solve Ksp = [Pb2+][Br−]2 Ksp = (1.05 x 10-2)(2.10 x 10-2)2 [Pb2+] [Br−] Initial Change +(1.05 x 10-2) +2(1.05 x 10-2) Equilibrium (1.05 x 10-2) (2.10 x 10-2) Tro, Chemistry: A Molecular Approach

Ksp and Relative Solubility molar solubility is related to Ksp but you cannot always compare solubilities of compounds by comparing their Ksps in order to compare Ksps, the compounds must have the same dissociation stoichiometry Tro, Chemistry: A Molecular Approach

The Effect of Common Ion on Solubility addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt for example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2 PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) addition of Cl− shifts the equilibrium to the left Tro, Chemistry: A Molecular Approach

CaF2(s)  Ca2+(aq) + 2 F−(aq) Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C Write the dissociation reaction and Ksp expression Create an ICE table defining the change in terms of the solubility of the solid CaF2(s)  Ca2+(aq) + 2 F−(aq) Ksp = [Ca2+][F−]2 [Ca2+] [F−] Initial 0.100 Change +S +2S Equilibrium S 0.100 + 2S Tro, Chemistry: A Molecular Approach

Ex 16. 10 – Calculate the molar solubility of CaF2 in 0 Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C Substitute into the Ksp expression assume S is small Find the value of Ksp from Table 16.2, plug into the equation and solve for S Ksp = [Ca2+][F−]2 Ksp = (S)(0.100 + 2S)2 Ksp = (S)(0.100)2 [Ca2+] [F−] Initial 0.100 Change +S +2S Equilibrium S 0.100 + 2S Tro, Chemistry: A Molecular Approach

The Effect of pH on Solubility for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide and the lower the pH, the higher the solubility higher pH = increased [OH−] M(OH)n(s)  Mn+(aq) + nOH−(aq) for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M2(CO3)n(s)  2 Mn+(aq) + nCO32−(aq) H3O+(aq) + CO32− (aq)  HCO3− (aq) + H2O(l) Tro, Chemistry: A Molecular Approach

Precipitation precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur Q = Ksp, the solution is saturated, no precipitation Q < Ksp, the solution is unsaturated, no precipitation Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions Tro, Chemistry: A Molecular Approach

a supersaturated solution will precipitate if a seed crystal is added precipitation occurs if Q > Ksp Tro, Chemistry: A Molecular Approach

Selective Precipitation a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different Tro, Chemistry: A Molecular Approach

precipitating may just occur when Q = Ksp Ex 16.13 What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater? precipitating may just occur when Q = Ksp Tro, Chemistry: A Molecular Approach

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M Tro, Chemistry: A Molecular Approach

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10-10 M Tro, Chemistry: A Molecular Approach

Qualitative Analysis an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme wet chemistry a sample containing several ions is subjected to the addition of several precipitating agents addition of each reagent causes one of the ions present to precipitate out Tro, Chemistry: A Molecular Approach

Qualitative Analysis Tro, Chemistry: A Molecular Approach

Group 1 group one cations are Ag+, Pb2+, and Hg22+ all these cations form compounds with Cl− that are insoluble in water as long as the concentration is large enough PbCl2 may be borderline molar solubility of PbCl2 = 1.43 x 10-2 M precipitated by the addition of HCl Tro, Chemistry: A Molecular Approach

Group 2 group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+ all these cations form compounds with HS− and S2− that are insoluble in water at low pH precipitated by the addition of H2S in HCl

Group 3 group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides all these cations form compounds with S2− that are insoluble in water at high pH precipitated by the addition of H2S in NaOH

Group 4 group four cations are Mg2+, Ca2+, Ba2+ all these cations form compounds with PO43− that are insoluble in water at high pH precipitated by the addition of (NH4)2HPO4 Tro, Chemistry: A Molecular Approach

Group 5 group five cations are Na+, K+, NH4+ all these cations form compounds that are soluble in water – they do not precipitate identified by the color of their flame

Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq) Complex Ion Formation transition metals tend to be good Lewis acids they often bond to one or more H2O molecules to form a hydrated ion H2O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq) ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H2O)2+ the attached ions or molecules are called ligands e.g., H2O Tro, Chemistry: A Molecular Approach

Complex Ion Equilibria if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H2O)2+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) + 2 H2O(l) generally H2O is not included, since its complex ion is always present in aqueous solution Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) Tro, Chemistry: A Molecular Approach

Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) Formation Constant the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) the equilibrium constant for the formation reaction is called the formation constant, Kf Tro, Chemistry: A Molecular Approach

Formation Constants Tro, Chemistry: A Molecular Approach

Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Write the formation reaction and Kf expression. Look up Kf value Determine the concentration of ions in the diluted solutions Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Tro, Chemistry: A Molecular Approach

Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) [Cu2+] [NH3] [Cu(NH3)22+] Initial 6.7E-4 0.11 Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4 Equilibrium x

Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Substitute in and solve for x confirm the “x is small” approximation [Cu2+] [NH3] [Cu(NH3)22+] Initial 6.7E-4 0.11 Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4 Equilibrium x since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid Tro, Chemistry: A Molecular Approach

The Effect of Complex Ion Formation on Solubility the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands AgCl(s)  Ag+(aq) + Cl−(aq) Ksp = 1.77 x 10-10 Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) Kf = 1.7 x 107 adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+ Tro, Chemistry: A Molecular Approach

Solubility of Amphoteric Metal Hydroxides many metal hydroxides are insoluble all metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH− some metal hydroxides also become more soluble in basic solution acting as a Lewis base forming a complex ion substances that behave as both an acid and base are said to be amphoteric some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+ Tro, Chemistry: A Molecular Approach

Al3+ Al3+ is hydrated in water to form an acidic solution Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+(aq) + H3O+(aq) addition of OH− drives the equilibrium to the right and continues to remove H from the molecules Al(H2O)5(OH)2+(aq) + OH−(aq)  Al(H2O)4(OH)2+(aq) + H2O (l) Al(H2O)4(OH)2+(aq) + OH−(aq)  Al(H2O)3(OH)3(s) + H2O (l) Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach