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Tro, Chemistry: A Molecular Approach 1 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2 (aq) + H 2 O (aq) after equivalence.

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Presentation on theme: "Tro, Chemistry: A Molecular Approach 1 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2 (aq) + H 2 O (aq) after equivalence."— Presentation transcript:

1 Tro, Chemistry: A Molecular Approach 1 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2 (aq) + H 2 O (aq) after equivalence point added 30.0 mL NaOH 5.0 x 10 -4 mol NaOH xs

2 Tro, Chemistry: A Molecular Approach 2 added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 Adding NaOH to HCHO 2 added 12.5 mL NaOH 0.00125 mol HCHO 2 pH = 3.74 = pK a half-neutralization initial HCHO 2 solution 0.00250 mol HCHO 2 pH = 2.37 added 5.0 mL NaOH 0.00200 mol HCHO 2 pH = 3.14 added 10.0 mL NaOH 0.00150 mol HCHO 2 pH = 3.56 added 15.0 mL NaOH 0.00100 mol HCHO 2 pH = 3.92 added 20.0 mL NaOH 0.00050 mol HCHO 2 pH = 4.34 added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 25.0 mL NaOH equivalence point 0.00250 mol CHO 2 [CHO 2 ] init = 0.0500 M [OH ] eq = 1.7 x 10 -6 pH = 8.23 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52

3 Tro, Chemistry: A Molecular Approach 3

4 4 Titration of 25.0 mL of 0.100 M HCHO 2 with 0.100 M NaOH The 1st derivative of the curve is maximum at the equivalence point Since the solutions are equal concentration, the equivalence point is at equal volumes pH at equivalence = 8.23

5 Tro, Chemistry: A Molecular Approach 5 at pH 3.74, the [HCHO 2 ] = [CHO 2 ]; the acid is half neutralized half-neutralization occurs when pH = pK a

6 Tro, Chemistry: A Molecular Approach 6 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 HA init and mol A init using reaction stoichiometry – calculate pH with Henderson-Hasselbalch using mol HA init and mol A init half-neutralization pH = pK a

7 Tro, Chemistry: A Molecular Approach 7 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 – [H 3 O + ][OH ]=1 x 10 -14

8 Tro, Chemistry: A Molecular Approach 8 Ex 16.7a – A 40.0 mL sample of 0.100 M HNO 2 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 HNO 2 + KOH NO 2 + H 2 O

9 Tro, Chemistry: A Molecular Approach 9 Ex 16.7b – A 40.0 mL sample of 0.100 M HNO 2 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 HA before & moles of added B Make a stoichiometry table and determine the moles of HA in excess and moles A made HNO 2 + KOH NO 2 + H 2 O HNO 2 NO 2 - OH mols Before 0.004000 0 mols added -- 0.00100 mols After 0 0.00300 0.00100

10 Tro, Chemistry: A Molecular Approach 10 Ex 16.7b – A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. Write an equation for the reaction of HA with H 2 O Determine K a and pK a for HA Use the Henderson- Hasselbalch Equation to determine the pH HNO 2 + H 2 O NO 2 + H 3 O + HNO 2 NO 2 - OH mols Before 0.004000 0 mols added -- 0.00100 mols After 0.003000.00100 0 Table 15.5 K a = 4.6 x 10 -4

11 Tro, Chemistry: A Molecular Approach 11 Ex 16.7b – A 40.0 mL sample of 0.100 M HNO 2 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 HA before & moles of added B Make a stoichiometry table and determine the moles of HA in excess and moles A made HNO 2 + KOH NO 2 + H 2 O HNO 2 NO 2 - OH mols Before 0.004000 0 mols added -- 0.00200 mols After 0 0.00200 at half-equivalence, moles KOH = ½ mole HNO 2

12 Tro, Chemistry: A Molecular Approach 12 Ex 16.7b – A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. Write an equation for the reaction of HA with H 2 O Determine K a and pK a for HA Use the Henderson- Hasselbalch Equation to determine the pH HNO 2 + H 2 O NO 2 + H 3 O + HNO 2 NO 2 - OH mols Before 0.004000 0 mols added -- 0.00200 mols After 0.00200 0 Table 15.5 K a = 4.6 x 10 -4

13 Tro, Chemistry: A Molecular Approach 13 Titration Curve of a Weak Base with a Strong Acid

14 Tro, Chemistry: A Molecular Approach 14 Titration of a Polyprotic Acid if K a1 >> K a2, there will be two equivalence points in the titration – the closer the K a s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH

15 Tro, Chemistry: A Molecular Approach 15 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 [H 3 O + ] – using a probe that specifically measures just H 3 O + 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

16 Tro, Chemistry: A Molecular Approach 16 Monitoring pH During a Titration

17 Tro, Chemistry: A Molecular Approach 17 Indicators many dyes change color depending on the pH of the solution these dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l) Ind (aq) + H 3 O + (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

18 18 Phenolphthalein

19 Tro, Chemistry: A Molecular Approach 19 Methyl Red

20 Tro, Chemistry: A Molecular Approach 20 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 – pK a of HInd pH at equivalence point

21 21 Acid-Base Indicators

22 Tro, Chemistry: A Molecular Approach 22 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

23 Tro, Chemistry: A Molecular Approach 23 Solubility Product the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp for an ionic solid M n X m, the dissociation reaction is: M n X m (s) nM m+ (aq) + mX n (aq) the solubility product would be K sp = [M m+ ] n [X n ] m for example, the dissociation reaction for PbCl 2 is PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) and its equilibrium constant is K sp = [Pb 2+ ][Cl ] 2

24 Tro, Chemistry: A Molecular Approach 24

25 Tro, Chemistry: A Molecular Approach 25 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 M n X m (s) nM m+ (aq) + mX n (aq)

26 Tro, Chemistry: A Molecular Approach 26 Ex 16.8 – Calculate the molar solubility of PbCl 2 in pure water at 25 C Write the dissociation reaction and K sp expression Create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Cl ] Initial00 Change+S+2S EquilibriumS2S PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) K sp = [Pb 2+ ][Cl ] 2

27 Tro, Chemistry: A Molecular Approach 27 Ex 16.8 – Calculate the molar solubility of PbCl 2 in pure water at 25 C Substitute into the K sp expression Find the value of K sp from Table 16.2, plug into the equation and solve for S [Pb 2+ ][Cl ] Initial00 Change+S+2S EquilibriumS2S K sp = [Pb 2+ ][Cl ] 2 K sp = (S)(2S) 2

28 Tro, Chemistry: A Molecular Approach 28 Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25 C is 1.05 x 10 -2 M

29 Tro, Chemistry: A Molecular Approach 29 Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25 C is 1.05 x 10 -2 M Write the dissociation reaction and K sp expression Create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Br ] Initial00 Change+(1.05 x 10 -2 )+2(1.05 x 10 -2 ) Equilibrium(1.05 x 10 -2 )(2.10 x 10 -2 ) PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq) K sp = [Pb 2+ ][Br ] 2

30 Tro, Chemistry: A Molecular Approach 30 Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25 C is 1.05 x 10 -2 M Substitute into the K sp expression plug into the equation and solve K sp = [Pb 2+ ][Br ] 2 K sp = (1.05 x 10 -2 )(2.10 x 10 -2 ) 2 [Pb 2+ ][Br ] Initial00 Change+(1.05 x 10 -2 )+2(1.05 x 10 -2 ) Equilibrium(1.05 x 10 -2 )(2.10 x 10 -2 )

31 Tro, Chemistry: A Molecular Approach 31 K sp and Relative Solubility molar solubility is related to K sp but you cannot always compare solubilities of compounds by comparing their K sp s in order to compare K sp s, the compounds must have the same dissociation stoichiometry

32 Tro, Chemistry: A Molecular Approach 32 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 PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) addition of Cl shifts the equilibrium to the left

33 Tro, Chemistry: A Molecular Approach 33 Ex 16.10 – Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25 C Write the dissociation reaction and K sp expression Create an ICE table defining the change in terms of the solubility of the solid [Ca 2+ ][F ] Initial00.100 Change+S+2S EquilibriumS0.100 + 2S CaF 2 (s) Ca 2+ (aq) + 2 F (aq) K sp = [Ca 2+ ][F ] 2

34 Tro, Chemistry: A Molecular Approach 34 Ex 16.10 – Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25 C Substitute into the K sp expression assume S is small Find the value of K sp from Table 16.2, plug into the equation and solve for S [Ca 2+ ][F ] Initial00.100 Change+S+2S EquilibriumS0.100 + 2S K sp = [Ca 2+ ][F ] 2 K sp = (S)(0.100 + 2S) 2 K sp = (S)(0.100) 2

35 Tro, Chemistry: A Molecular Approach 35 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) M n+ (aq) + nOH (aq) for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M 2 (CO 3 ) n (s) 2 M n+ (aq) + nCO 3 2 (aq) H 3 O + (aq) + CO 3 2 (aq) HCO 3 (aq) + H 2 O(l)

36 Tro, Chemistry: A Molecular Approach 36 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 K sp, we can determine if precipitation will occur – Q = K sp, the solution is saturated, no precipitation – Q < K sp, the solution is unsaturated, no precipitation – Q > K sp, the solution would be above saturation, the salt above saturation will precipitate some solutions with Q > K sp will not precipitate unless disturbed – these are called supersaturated solutions

37 Tro, Chemistry: A Molecular Approach 37 precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added

38 Tro, Chemistry: A Molecular Approach 38 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 K sp values are significantly different

39 Tro, Chemistry: A Molecular Approach 39 Ex 16.13 What is the minimum [OH ] necessary to just begin to precipitate Mg 2+ (with [0.059]) from seawater? precipitating may just occur when Q = K sp

40 Tro, Chemistry: A Molecular Approach 40 Ex 16.14 What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH ] = 1.9 x 10 -6 M

41 Tro, Chemistry: A Molecular Approach 41 Ex 16.14 What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH ] = 1.9 x 10 -6 M precipitating Ca 2+ begins when [OH ] = 2.06 x 10 -2 M when Ca 2+ just begins to precipitate out, the [Mg 2+ ] has dropped from 0.059 M to 4.8 x 10 -10 M

42 Tro, Chemistry: A Molecular Approach 42 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

43 Tro, Chemistry: A Molecular Approach 43 Qualitative Analysis

44 44

45 Tro, Chemistry: A Molecular Approach 45 Group 1 group one cations are Ag +, Pb 2+, and Hg 2 2+ all these cations form compounds with Cl that are insoluble in water – as long as the concentration is large enough – PbCl 2 may be borderline molar solubility of PbCl 2 = 1.43 x 10 -2 M precipitated by the addition of HCl

46 46 Group 2 group two cations are Cd 2+, Cu 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, and Hg 2+ all these cations form compounds with HS and S 2 that are insoluble in water at low pH precipitated by the addition of H 2 S in HCl

47 47 Group 3 group three cations are Fe 2+, Co 2+, Zn 2+, Mn 2+, Ni 2+ precipitated as sulfides; as well as Cr 3+, Fe 3+, and Al 3+ precipitated as hydroxides all these cations form compounds with S 2 that are insoluble in water at high pH precipitated by the addition of H 2 S in NaOH

48 Tro, Chemistry: A Molecular Approach 48 Group 4 group four cations are Mg 2+, Ca 2+, Ba 2+ all these cations form compounds with PO 4 3 that are insoluble in water at high pH precipitated by the addition of (NH 4 ) 2 HPO 4

49 49 Group 5 group five cations are Na +, K +, NH 4 + all these cations form compounds that are soluble in water – they do not precipitate identified by the color of their flame

50 Tro, Chemistry: A Molecular Approach 50 Complex Ion Formation transition metals tend to be good Lewis acids they often bond to one or more H 2 O molecules to form a hydrated ion – H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag + (aq) + 2 H 2 O(l) Ag(H 2 O) 2 + (aq) ions that form by combining a cation with several anions or neutral molecules are called complex ions – e.g., Ag(H 2 O) 2 + the attached ions or molecules are called ligands – e.g., H 2 O

51 Tro, Chemistry: A Molecular Approach 51 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(H 2 O) 2 + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) + 2 H 2 O (l) – generally H 2 O is not included, since its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq)

52 Tro, Chemistry: A Molecular Approach 52 Formation Constant the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) the equilibrium constant for the formation reaction is called the formation constant, K f

53 Tro, Chemistry: A Molecular Approach 53 Formation Constants

54 Tro, Chemistry: A Molecular Approach 54 Ex 16.15 – 200.0 mL of 1.5 x 10 -3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Write the formation reaction and K f expression. Look up K f value Determine the concentration of ions in the diluted solutions Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 2 2+ (aq)

55 55 Ex 16.15 – 200.0 mL of 1.5 x 10 -3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Create an ICE table. Since K f is large, assume all the Cu 2+ is converted into complex ion, then the system returns to equilibrium [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] Initial6.7E-40.110 Change-6.7E-4-4(6.7E-4)+ 6.7E-4 Equilibriumx0.116.7E-4 Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 2 2+ (aq)

56 Tro, Chemistry: A Molecular Approach 56 Ex 16.15 – 200.0 mL of 1.5 x 10 -3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 2 2+ (aq) Substitute in and solve for x confirm thex is small approximation [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] Initial6.7E-40.110 Change-6.7E-4-4(6.7E-4)+ 6.7E-4 Equilibriumx0.116.7E-4 since 2.7 x 10 -13 << 6.7 x 10 -4, the approximation is valid

57 Tro, Chemistry: A Molecular Approach 57 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) K sp = 1.77 x 10 -10 Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7 adding NH 3 to a solution in equilibrium with AgCl (s) increases the solubility of Ag +

58 58

59 Tro, Chemistry: A Molecular Approach 59 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 Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sb 2+

60 Tro, Chemistry: A Molecular Approach 60 Al 3+ Al 3+ is hydrated in water to form an acidic solution Al(H 2 O) 6 3+ (aq) + H 2 O (l) Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) addition of OH drives the equilibrium to the right and continues to remove H from the molecules Al(H 2 O) 5 (OH) 2+ (aq) + OH (aq) Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH (aq) Al(H 2 O) 3 (OH) 3(s) + H 2 O (l)

61 Tro, Chemistry: A Molecular Approach 61


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