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Applications of Aqueous Equilibria Common Ions When we dissolve acetic acid in water, the following equilibrium is established: –CH 3 COOH CH 3 COO -

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Presentation on theme: "Applications of Aqueous Equilibria Common Ions When we dissolve acetic acid in water, the following equilibrium is established: –CH 3 COOH CH 3 COO -"— Presentation transcript:

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2 Applications of Aqueous Equilibria

3 Common Ions When we dissolve acetic acid in water, the following equilibrium is established: –CH 3 COOH CH 3 COO - + H + If sodium acetate were dissolved in solution, which way would this equilibrium shift?

4 Common Ions When we dissolve acetic acid in water, the following equilibrium is established: –CH 3 COOH CH 3 COO - + H + If sodium acetate were dissolved in solution, which way would this equilibrium shift? –Adding acetate ions from a strong electrolyte would shift this equilibrium left (Le Chateliers principle).

5 Common Ion Effect Whenever a weak electrolyte (acetic acid) and a strong electrolyte (sodium acetate) share a common ion, the weak electrolyte ionizes less than it would if it were alone. (Le Chateliers) This is called the common-ion effect.

6 Steps for Common-Ion Problems 1. Consider which solutes are strong electrolyte and weak electrolytes. 2. Identify the important equilibrium (weak) that is the source of H + and therefore determines pH. 3. Create an ICE chart using the equilibrium and strong electrolyte concentrations. 4. Use the equilibrium constant expression to calculuate [H + ] and pH.

7 What is the concentration of silver and chromate in a solution with silver chromate in 0.1 M silver nitrate Ag 2 CrO 4 (s) 2Ag +1 + CrO 4 -2 I C +2x +x E x x Ksp = 9.0 x = [Ag +1 ] 2 [CrO 4 -2 ] = 9.0 x = [.1 +2x] 2 [x] 9.0 x = x [CrO 4 -2 ] = x = 9.0 x M [Ag +1 ] =.1 + 2x =.1M [Ag 2 CrO 4 ] = 9.0 x What are the sources of Ag +1 ? Why is this a plus sign?

8 Sample Problem What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? K a for acetic acid = 1.8 x pH = 4.74

9 Sample Problem Calculate the fluoride concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl K a for HF = 6.8 x [F - ] = 1.2x10 -3 M pH = 1.00

10 Buffers Solutions that resist pH change when small amounts of acid or base are added. Two types weak acid and its salt weak base and its salt HX(aq) + H 2 O(l) H + (aq) + X - (aq) Add OH - Add H + shift to right shift to left Based on the common ion effect.

11 ml HCl added pH buffered unbuffered Buffers

12 Buffers and blood Control of blood pH Oxygen is transported primarily by hemoglobin in the red blood cells. CO 2 is transported both in plasma and the red blood cells. CO 2 (aq) + H 2 OH 2 CO 3 (aq) H + (aq) + HCO 3 - (aq)

13 Composition and Action of Buffered Solutions The K a expression is A buffer resists a change in pH when a small amount of OH - or H + is added. Buffers

14 Buffered Solutions Addition of Strong Acids or Bases to Buffers With the concentrations of HX and X - (note the change in volume of solution) we can calculate the pH from the Henderson-Hasselbalch equation Equation not necessary, since you know K a = [X - ] [H + ] [HX]

15 Buffers Action of Buffered Solutions

16 Buffers Buffer Capacity and pH Buffer capacity is the amount of acid or base neutralized by the buffer before there is a significant change in pH. The greater the amounts(molarity) of the conjugate acid-base pair, the greater the buffer capacity. The pH of the buffer depends on K a

17 Buffered Solutions Addition of Strong Acids or Bases to Buffers

18 Buffer Break the calculation into two parts: stoichiometric and equilibrium. The amount of strong acid or base added results in a neutralization reaction: X - + H + HX + H 2 O HX + OH - X - + H 2 O. By knowing how much H + or OH - was added (stoichiometry) we know how much HX or X - is formed.

19 Buffers The final [HX] and [X - ] after the neutralization reaction are used as the initial concentrations for the equilibrium reaction. HXH + X - Initial conc., M Change, M Eq. Conc., M Then the equilibrium constant expression is used to find [H + ] and pH = - log [H + ] K a = [H + ] [X - ] [HX]

20 Buffer Example Determine the initial pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 o C. K a = 6.5 x HBz(aq) + H 2 O(l) H + (aq) + Bz - (aq) HBzH + Bz - Initial conc., M Change, DM -x +x +x Eq. Conc., M xx x

21 Buffer Example Solve the equilibrium equation in terms of x K a = 6.5 x = x = (6.5 x )(0.10) / (0. 20) (assuming x<<0.10) = 3.2 x M H + pH= - log (3.2 x M) = 4.5 initial pH x ( x) x

22 Buffer Example Determine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 o C after 0.05 moles of HCl is added. HBzH + Bz - Initial conc., M Change, DM -x +x +x Eq. Conc., M xx x The 0.05 mol HCl reacts completely with 0.05 mol Bz - (aq) to form 0.05 mol HBz(aq) Then equilibrium will be re-established based on the new initial concentrations of 0.15 M HBz(aq) and 0.15 M Bz - (aq). First, find the concentrations of First, find the concentrations of the HBz and Bz - after HCl is added.

23 Buffer Example Solve the equilibrium equation in terms of x K a = 6.5 x = x = (6.5 x )(0.15) / (0. 15) (assuming x<<0.15) = 6.5 x M H + pH= - log (6.5 x M) = 4.2 after 0.05 mole HCl added x ( x) x

24 Buffer Example Determine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 o C after 0.05 moles of NaOH is added. HBzH + Bz - Initial conc., M Change, DM -x +x +x Eq. Conc., M xx x The 0.05 mol NaOH reacts completely with 0.05 mol HBz(aq) to form 0.05 mol Bz - (aq) Then equilibrium will be re-established based on the new initial concentrations of 0.05 M HBz(aq) and 0.25 M Bz - (aq). First, find the concentrations of First, find the concentrations of HBz and Bz - after NaOH is added.

25 Buffer Example Solve the equilibrium equation in terms of x K a = 6.5 x = x = (6.5 x )(0.05) / (0. 25) (assuming x<<0.05) = 1.3 x M H + pH= - log (1.3 x M) = 4.9 after 0.05 mole NaOH added x ( x) x

26 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations The plot of pH versus volume during a titration is a titration curve.

27 Remember the Acid-Base Titration Curves? HCl HC 2 H 3 O 2 Buffer Zone Zone

28 For this experiment, you will prepare a buffer that contains acetic acid and its conjugate base, the acetate ions. The equilibrium equation for the reaction is shown below: HC 2 H 3 O 2 (aq) + H 2 O(l) H + (aq)+ C 2 H 3 O 2 - (aq) The equilibrium expression for this reaction, K a, has a value of 1.8 x at 25ºC. Acetic Acid/Acetate Ion Buffer Lab Acetic Acid/Acetate Ion Buffer Lab

29 The ratio between the molarity of the acetate ions to the molarity of the acetic acid in your buffer must equal the ratio between the K a value and 10 - assigned pH. This ratio should be reduced, so that either the [HC 2 H 3 O 2 ] or [C 2 H 3 O 2 - ] has a concentration of 0.10 M, and the concentration of the other component must fall within a range from 0.10 M to 1.00 M. only Complete the calculations only that are needed to prepare mL of your assigned buffer solution that has these specific concentrations. Can you predict the final pH when a strong acid or base is added to the buffer solution?

30 Acetic Acid/Acetate Ion Buffer Lab Acetic Acid/Acetate Ion Buffer Lab

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32 Making a Buffer Calculations Making a Buffer Calculations You want to prepare mL of a buffer with a pH = 10.00, with both the acid and conjugate base having molarities between 0.10 M to 1.00 M. You may choose from any of the acids listed below: You must select an acid with a K a value close to 10 - assigned pH. The only two options are ammonium or the hydrogen carbonate ions.

33 Making a Buffer Calculations Making a Buffer Calculations

34 Place ~250 mL of distilled water in a 500 mL volumetric flask. Add 2.68 g NH 4 Cl and dissolve. Then mix in 18.9 mL of 14.8 M NH 3. Fill with distilled water to the 500 mL mark on the flask. If there is no concentrated NH 3 available, the NH 3 can be produced by neutralizing additional NH 4 Cl with 1.00 M NaOH. Dissolve g NH 4 Cl (2.68 g g) in 280. mL of 1.00 M NaOH. Then dilute with distilled water and fill to the 500 mL mark on the flask.

35 Making a Buffer Calculations Making a Buffer Calculations If there is no NH 4 Cl available, the NH 4 + can be produced by neutralizing additional NH 3 with 1.00 M HCl. Place ~250 mL distilled water in volumetric flask. Add 50.0 mL of 1.00 M HCl and mix. Then add 22.3 mL of concentrated NH 3 (18.9 mL mL). Mix and fill with distilled water to the 500 mL mark on the flask. Place ~250 mL of distilled water in a 500 mL volumetric flask. Add 2.68 g NH 4 Cl and dissolve. Then mix in 18.9 mL of 14.8 M NH 3. Fill with distilled water to the 500 mL mark on the flask.

36 Making a Buffer Calculations Making a Buffer Calculations Prepare 500. mL of the buffer that has [CO 3 2- ] = M and [HCO 3 1- ] = M.

37 Making a Buffer Calculations Making a Buffer Calculations Prepare 500. mL of the buffer that has [CO 3 2- ] = M and [HCO 3 1- ] = M. Place ~250 mL distilled water in a 500 mL volumetric flask. Add 5.30 g Na 2 CO 3 and 7.52 g NaHCO 3 and dissolve. Fill with distilled water to the 500 mL mark on the flask.

38 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations The plot of pH versus volume during a titration is a titration curve.

39 Indicator examples Acid-base indicators are weak acids that undergo a color change at a known pH. phenolphthalein pH

40 Indicator examples methyl red bromthymol blue Select the indicator that undergoes a color change closest to the pH at the equivalence point, where all of the acid has been neutralized by the base.

41 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl). –Before any base is added, the pH is given by the strong acid solution. Therefore, pH < 7. –When base is added, before the equivalence point, the pH is given by the amount of strong acid in excess. Therefore, pH < 7. –At equivalence point, the amount of base added is stoichiometrically equivalent to the amount of acid originally present. Therefore, the pH is determined by the salt solution. Therefore, pH = 7.

42 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl).

43 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations We know the pH at equivalent point is To detect the equivalent point, we use an indicator that changes color somewhere near –Usually, we use phenolphthalein that changes color between pH 8.3 to –In acid, phenolphthalein is colorless. –As NaOH is added, there is a slight pink color at the addition point. –When the flask is swirled and the reagents mixed, the pink color disappears. –At the end point, the solution is light pink. –If more base is added, the solution turns darker pink.

44 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations The equivalence point in a titration is the point at which the acid and base are present in stoichiometric quantities. The end point in a titration is the observed point. The difference between equivalence point and end point is called the titration error. The shape of a strong base-strong acid titration curve is very similar to a strong acid-strong base titration curve.

45 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations

46 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations Initially, the strong base is in excess, so the pH > 7. As acid is added, the pH decreases but is still greater than 7. At equivalence point, the pH is given by the salt solution (i.e. pH = 7). After equivalence point, the pH is given by the strong acid in excess, so pH < 7.

47 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Consider the titration of acetic acid, HC 2 H 3 O 2 and NaOH. Before any base is added, the solution contains only weak acid. Therefore, pH is given by the equilibrium calculation. As strong base is added, the strong base consumes a stoichiometric quantity of weak acid: HC 2 H 3 O 2 (aq) + NaOH(aq) C 2 H 3 O 2 - (aq) + H 2 O(l)

48 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations

49 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations There is an excess of acetic acid before the equivalence point. Therefore, we have a mixture of weak acid and its conjugate base. –The pH is given by the buffer calculation. First the amount of C 2 H 3 O 2 - generated is calculated, as well as the amount of HC 2 H 3 O 2 consumed. (Stoichiometry.) Then the pH is calculated using equilibrium conditions. (Henderson-Hasselbalch.)

50 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations At the equivalence point, all the acetic acid has been consumed and all the NaOH has been consumed. However, C 2 H 3 O 2 - has been generated. –Therefore, the pH is given by the C 2 H 3 O 2 - solution. –This means pH > 7. More importantly, pH 7 for a weak acid-strong base titration. After the equivalence point, the pH is given by the strong base in excess.

51 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations For a strong acid-strong base titration, the pH begins at less than 7 and gradually increases as base is added. Near the equivalence point, the pH increases dramatically. For a weak acid-strong base titration, the initial pH rise is more steep than the strong acid-strong base case. However, then there is a leveling off due to buffer effects.

52 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations The inflection point is not as steep for a weak acid- strong base titration. The shape of the two curves after equivalence point is the same because pH is determined by the strong base in excess. Two features of titration curves are affected by the strength of the acid: –the amount of the initial rise in pH, and –the length of the inflection point at equivalence.

53 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations The weaker the acid, the smaller the equivalence point inflection. For very weak acids, it is impossible to detect the equivalence point.

54 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Titration of weak bases with strong acids have similar features to weak acid-strong base titrations.

55 Acid-Base Titrations Titrations of Polyprotic Acids Titrations of Polyprotic Acids In polyprotic acids, each ionizable proton dissociates in steps. Therefore, in a titration there are n equivalence points corresponding to each ionizable proton. In the titration of Na 2 CO 3 with HCl there are two equivalence points: –one for the formation of HCO 3 - –one for the formation of H 2 CO 3.

56 Acid-Base Titrations Titrations of Polyprotic Acids Titrations of Polyprotic Acids

57 Acid-Base Titrations Titrations of Polyprotic Acids Titrations of Polyprotic Acids K a = [H + ] [X - ] [HX] At the equivalence point, [H + ] = [OH - ] At ½ the equivalence point, [X - ] = [HX] SOOOO…. At ½ equivalence point, K a = [H + ] So pK a = pH

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59 Solubility What happens when an ionic compound is dissolved in water? What forces does water have to overcome? Are all ionic compounds soluble? –Remember the solubility rules? –Its time to review them!

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61 Properties of aqueous solutions There are two general classes of solutes.Electrolytic ionic compounds in polar solvents dissociate in solution to make ions conduct electricity may be strong (100% dissociation) or weak (less than 10%)Nonelectrolytic do not conduct electricity solute is dispersed but does not dissociate

62 Heterogeneous Equilibria A (s) + H 2 O B (aq) + C (aq) K sp = [B] [C] Why is this not divided by [A]? Why is [H 2 O] not included? Known as the Solubility Product = K sp

63 Solubility Products, K SP K SP expressions are used for ionic materials that are only slightly soluble in water. Their only means of dissolving is by dissociation. AgCl(s) Ag + (aq) + Cl - (aq) K SP = [ Ag + ] [ Cl - ]

64 Solubility Products, K SP At equilibrium, the system is a saturated solution of silver and chloride ions. The only way to know that it is saturated it to observe some AgCl at the bottom of the solution. As such, [AgCl] is a constant and K SP expressions do not include the solid form in the equilibrium expression. The [H 2 O] for solvation process is also excluded from the K SP expression.

65 Solubility Products, K SP Determine the solubility of AgCl in water at 20 o C in terms of grams / 100 mL K SP = [Ag + ] [Cl - ] = 1.7 x At equilibrium, [Ag + ] = [Cl - ] so 1.7 x = [x] 2 [Ag + ] = 1.3 x M g AgCl = 1.3 x mol/L * 0.10 L * g/mol Solubility= 1.9 x g / 100 mL

66 Calculating K sp Calculate the Ksp for Bismuth Sulfide which has a solubility of 1.0 x Bi 2 S 3 2Bi S -2 Ksp = For every Bi 2 S 3 dissolved 2 Bi +3 and 3S -2 are formed Ksp = [2 x 1.0 x ] 2 [3 x 1.0 x ] 3 = 1.1 x [Bi +3 ] 2 [S -2 ] 3

67 Calculating Solubility Calculate the concentration of ions for copper (II) iodate. k sp = 1.4 x at 25 °C Cu(IO 3 ) 2(s) Cu IO x = [x] [ 2x] 2 = 4x 3 x = 3.3 x mol/L = [Cu +2 ] 2x = 6.6 x mol/L = [IO 3 -1 ] What is the molarity of a saturated solution of Cu(IO 3 ) 2 ? 3.3 x mol/L = [Cu(IO 3 ) 2 ]

68 Solubility Products, K SP Another example Calculate the silver ion concentration when excess silver chromate is added to a M sodium chromate solution. K SP Ag 2 CrO 4 = 1.1 x Ag 2 CrO 4 (s) 2Ag + + CrO 4 2-

69 Solubility Products, K SP K SP = 1.1 x = [ Ag + ] 2 [ CrO 4 2- ] [CrO 4 2- ] = [CrO 4 2- ] Ag 2 CrO 4 + [CrO 4 2- ] Na 2 CrO 4 With such a small value for K SP, we can assume that the [CrO 4 2- ] from Ag 2 CrO 4 is negligible. If were wrong, our silver concentration will be significant (>1% of the chromate concentration.) Then youd use the quadratic approach.

70 Solubility Products, K SP K SP = 1.1 x = [ Ag + ] 2 [ CrO 4 2- ] [CrO 4 2- ] = M [ Ag + ] = ( K SP / [ CrO 4 2- ] ) 1/2 = (1.1 x / M ) 1/2 = 1.1 x M [ Ag + ] << [ CrO 4 2- ] so our assumption was valid.

71 Relative Solubility Be careful when comparing solubility products (Ksp) Compare –AgI Ksp = 1.5 x –CuI Ksp = 5.0 x Which is more soluble? Copper (I) iodide is more soluble. Each compound produces the same number ions

72 Relative Solubility CuS Ksp = 8.5 x Ag 2 S Ksp = 1.6 x Bi 2 S 3 Ksp = 1.1 x Which is more soluble? Each has a different number of ions, so calculate the solubility of each. Estimate (guess) three groups and calculate the solubility

73 Factors influencing solubility Common ion and salt effects. As with other equilibria weve discussed, adding a common ion will result in a shift of a solubility equilibrium. AgCl (s) Ag + (aq) + Cl - (aq) K SP = [Ag + ] [Cl - ] Adding either Ag + or Cl - to our equilibrium system will result in driving it to the left.

74 Factors That Affect Solubility Common-Ion Effect Common-Ion Effect

75 Factors influencing solubility Complex ion formation. The solubility of slightly soluble salts can be increased by complex ion formation. Example. Addition of excess Cl - AgCl (s) Ag + (aq) + Cl - (aq) + 2Cl - (aq) AgCl 2 - (aq) A large excess of chloride results in the formation of the complex. More AgCl will dissolve as a result. A large excess of chloride results in the formation of the complex. More AgCl will dissolve as a result.

76 Factors That Affect Solubility Formation of Complex Ions Formation of Complex Ions

77 Factors That Affect Solubility Formation of Complex Ions Formation of Complex Ions Consider the addition of ammonia to AgCl (white precipitate): The overall reaction is Effectively, the Ag + (aq) has been removed from solution. By Le Châteliers principle, the forward reaction (the dissolving of AgCl) is favored. K net = K spK f = (1.8 x )(1.7 x 10 7 ) = 3.1 x at 25ºC + +

78 Factors That Affect Solubility Amphoterism Amphoterism Amphoteric oxides will dissolve in either a strong acid or a strong base. Examples: hydroxides and oxides of Al 3+, Cr 3+, Zn 2+, and Sn 2+. The hydroxides generally form complex ions with four hydroxide ligands attached to the metal: Hydrated metal ions act as weak acids. Thus, the amphoterism is interrupted:

79 Factors That Affect Solubility Amphoterism Amphoterism Hydrated metal ions act as weak acids. Thus, the amphoterism is interrupted:

80 Factors influencing solubility Hydrolysis. If the anion of a weak acid, or cation of a weak base is part of a K SP, solubilities are greater than expected. AgCN (s) Ag + (aq) + CN - (aq) + H 2 O (l) HCN (aq) + OH - (aq) This competing equilibrium causes the CN - to be lower than expected. More AgCN will dissolve as a result. This competing equilibrium causes the CN - to be lower than expected. More AgCN will dissolve as a result.

81 Factors That Affect Solubility Solubility and pH Solubility and pH Again we apply Le Châteliers principle: –If the F - is removed, then the equilibrium shifts towards the decrease and CaF 2 dissolves. –F - can be removed by adding a strong acid: –As pH decreases, [H + ] increases and solubility increases. The effect of pH on solubility is dramatic.

82 Factors That Affect Solubility Solubility and pH Solubility and pH

83 Solubility Equilibria Solubility-Product Constant, K sp Solubility-Product Constant, K sp Consider for which K sp is the solubility product. (BaSO 4 is ignored because it is a pure solid so its concentration is constant.)

84 Solubility Equilibria Solubility-Product Constant, K sp Solubility-Product Constant, K sp In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers. Solubility is the amount (grams) of substance that dissolves to form a saturated solution. Molar solubility is the number of moles of solute dissolving to form a liter of saturated solution.

85 Solubility Equilibria Solubility and K sp Solubility and K sp To convert solubility to K sp solubility needs to be converted into molar solubility (via molar mass); molar solubility is converted into the molar concentration of ions at equilibrium (equilibrium calculation), K sp is the product of equilibrium concentration of ions.

86 Solubility Equilibria Solubility and K sp Solubility and K sp

87 Precipitation Reactions We will use Q and K sp If Q is bigger than K sp, what will happen? –Smaller The Q in a solubilty problem is called the Ion Product

88 A solution prepared by adding 750 mL of 4.00 x M Ce(NO 3 ) 3 to mL of 2.00 x M KIO 3. Will Ce(IO 3 ) 3 precipitate? (K sp = 1.9 x ) Step one: Calculate the concentration of Ce +3 and IO 3 -1 before a reaction occurs [Ce +3 ] 0 = (750.0 mL)(4.00x10 -3 M) = 2.86 x M ( mL) [IO 3 -1 ] 0 = (300.0 mL)(2.00x10 -3 M) = 5.71 x M ( mL) Q = (2.86 x M) x (5.71 x M) 3 = 5.32 x Q > K sp therefore the precipitation reaction will occur

89 Precipitation and Separation of Ions At any instant in time, Q = [Ba 2+ ][SO 4 2- ]. –If Q < K sp, precipitation occurs until Q = K sp. –If Q = K sp, equilibrium exists. –If Q > K sp, solid dissolves until Q = K sp. Based on solubilities, ions can be selectively removed from solutions. Consider a mixture of Zn 2+ (aq) and Cu 2+ (aq). CuS (K sp = ) is less soluble than ZnS(K sp = ), CuS will be removed from solution before ZnS.

90 Precipitation and Separation of Ions As H 2 S is added to the green solution, black CuS forms in a colorless solution of Zn 2+ (aq). When more H 2 S is added, a second precipitate of white ZnS forms. Selective Precipitation of Ions Ions can be separated from each other based on their salt solubilities. Example: if HCl is added to a solution containing Ag + and Cu 2+, the silver precipitates(K sp for AgCl is ) while the Cu 2+ remains in solution, since CuCl 2. Removal of one metal ion from a solution is called selective precipitation.

91 Qualitative Analysis for Metallic Elements Qualitative analysis is designed to detect the presence of metal ions. Quantitative analysis is designed to determine how much metal ion is present.

92 Qualitative Analysis for Metallic Elements We can separate a complicated mixture of ions into five groups: –Add 6 M HCl to precipitate insoluble chlorides (AgCl, Hg 2 Cl 2, and PbCl 2 ). –To the remaining mix of cations, add H 2 S in 0.2 M HCl to remove acid insoluble sulfides (e.g. CuS, Bi 2 S 3, CdS, PbS, HgS, etc.). –To the remaining mix, add (NH 4 ) 2 S at pH 8 to remove base insoluble sulfides and hydroxides (e.g. Al(OH) 3, Fe(OH) 3, ZnS, NiS, CoS, etc.). –To the remaining mixture add (NH 4 ) 2 HPO 4 to remove insoluble phosphates (Ba 3 (PO 4 ) 2, Ca 3 (PO 4 ) 2, MgNH 4 PO 4 ). –The final mixture contains alkali metal ions and NH 4 +.


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