Aqueous Equilibria Follow-up

Titration Curves To identify weak acids we can use pKa which can be determined from the graph. Ka = [H3O+][A-] [HA] We can get Ka = [H3O+] if we measure when the [A-] = [HA] At that point the pKa = pH Strong base will remove the stoichiometric equivalent of acid, so the equivalence point will be where the amount of base added is equal to the amount of acid at the start.

Figure 17.12 Titration curve for a diprotic acid.

pKa The point at which the [A-] = [HA] is when we have reacted half the acid with the strong base. So this is half way to the equivalence point. So if you get the pH when half of the volume of base is added that was necessary to get to the equivalence point, that is equal to the pKa. To find the equivalence point on the graph you can find the inflection point (2nd derivative) to see where the graph changes is direction of the curve. The second inflection point is going to require the same volume of base that the first inflection point required.

Common Ion Problem Calculate the pH of a solution that is 0.60M HF and 1.00M KF. Ka = 7.2x10-4 for HF.

Solution Major species: HF, K+, F-. There will be very little H+ since HF is weak. The reaction is HF + H2O ↔ H3O+ + F- 7.2x10-4 = (1+x)(x)/(0.60-x) Assume both x’s are small in reference to the 0.60M and 1.00M. HF H3O+ F- I 0.60 1.00 C -x +x E 0.60-x 1.00+x

Completion of Problem So 7.2x10-4 = 1x/0.60 x = 4.3x10-4 = [H3O+] pH = -log [H3O+] = 3.37

You try! What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.20 mol of sodium acetate to enough water to make 1.0 L of solution? Ka for acetic acid = 1.8 x 10-5 pH = 4.56

Buffers Buffered solutions can be made in 2 ways. 1) Adding a salt that has the common conjugate ion to the weak acid or base. 2) Adding a strong acid/base to the weak base/acid solution to create the common ion. The buffer capacity is related to how much can be added to the solution before the pH is effected. The capacity is increased as the concentration is increased, and increases as the ratio of the conjugates gets closer to 1.

HBz(aq) + H2O(l) H+(aq) + Bz-(aq) 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 oC. Ka = 6.5 x 10-5 HBz(aq) + H2O(l) H+(aq) + Bz-(aq) HBz H+ Bz- Initial conc., M 0.10 0.00 0.20 Change, DM -x +x +x Eq. Conc., M 0.10 - x x 0. 20 + x

Buffer Example Solve the equilibrium equation in terms of x Ka = 6.5 x 10-5 = x = (6.5 x 10-5 )(0.10) / (0. 20) (assuming x<<0.10) = 3.2 x 10-5 M H+ pH = - log (3.2 x 10-5 M) = 4.5 initial pH x (0. 20 + x) 0.10 - x

Buffered Solutions Addition of Strong Acids or Bases to Buffers

Buffer Example Determine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC after 0.05 moles of HCl is added. Total volume of solution is 1L. First, find the concentrations of the HBz and Bz- after HCl is added. HBz H+ Bz- Initial conc., M 0.15 0.00 0.15 Change, DM -x +x +x Eq. Conc., M 0.15 - x x 0. 15 + 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).

Buffer Example Solve the equilibrium equation in terms of x Ka = 6.5 x 10-5 = x = (6.5 x 10-5 )(0.15) / (0. 15) (assuming x<<0.15) = 6.5 x 10-5 M H+ pH = - log (6.5 x 10-5 M) = 4.2 after 0.05 mole HCl added x (0. 15 + x) 0.15 - x

Buffer Example Determine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC after 0.05 moles of NaOH is added.

Solution First, find the concentrations of HBz and Bz- after NaOH is added. 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). HBz H+ Bz- I 0.05 0.00 0.25 C -x +x E 0.05 - x x 0. 25 + x

Buffer Example x (0. 25 + x) 0.05 - x Solve the equilibrium equation in terms of x Ka = 6.5 x 10-5 = x = (6.5 x 10-5 )(0.05) / (0. 25) (assuming x<<0.05) = 1.3 x 10-5 M H+ pH = - log (1.3 x 10-5 M) = 4.9 after 0.05 mole NaOH added x (0. 25 + x) 0.05 - x

Making a Buffer Calculations You want to prepare 500.0 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 Ka value close to 10- assigned pH. The only two options are ammonium or the hydrogen carbonate ions.

Making a Buffer Calculations Lets use concentrated NH3 which is 14.8M and ammonium chloride.

Finish Place ~250 mL of distilled water in a 500 mL volumetric flask. Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M NH3. Fill with distilled water to the 500 mL mark on the flask.

Making a Buffer Calculations If there is no concentrated NH3 available, the NH3 can be produced by neutralizing additional NH4Cl with 1.00 M NaOH.

Solution Dissolve 17.66 g NH4Cl (2.68 g + 14.98 g) in 280. mL of 1.00 M NaOH. Then dilute with distilled water and fill to the 500 mL mark on the flask.

Making a Buffer Calculations If there is no NH4Cl available, the NH4+ can be produced by neutralizing additional NH3 with 1.00 M HCl.

Solution 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 NH3 (18.9 mL + 3.4 mL). Mix and fill with distilled water to the 500 mL mark on the flask.

Making a Buffer Calculations Prepare 500. mL of the buffer that has [CO32-] = 0.100 M and [HCO31-] = 0.179 M.

Making a Buffer Calculations Place ~250 mL distilled water in a 500 mL volumetric flask. Add 5.30 g Na2CO3 and 7.52 g NaHCO3 and dissolve. Fill with distilled water to the 500 mL mark on the flask.

Acetic Acid/Acetate Ion Buffer Lab 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: HC2H3O2(aq) + H2O(l) <=> H+ (aq)+ C2H3O2- (aq) The equilibrium expression for this reaction, Ka, has a value of 1.8 x 10-5 at 25ºC.

Acetic Acid/Acetate Ion Buffer Lab 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 Ka value and 10- assigned pH. This ratio should be reduced , so that either the [HC2H3O2] or [C2H3O2- ] 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. Complete the calculations only that are needed to prepare 100.0 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?

Acetic Acid/Acetate Ion Buffer Lab

Acetic Acid/Acetate Ion Buffer Lab

Buffer Answers 1-6 1) 1.36g NaC2H3O2 and 17mL of HC2H3O2 2) 3.76g HC2H3O2 and 17mL of HCl 3) 27.0mL HC2H3O2 and 10mL NaOH 4) 2.45g NaC2H3O2 and 10mL HC2H3O2 5) 3.81g NaC2H3O2 and 10mL HCl 6) 28mL HC2H3O2 and 18mL NaOH

Buffer Answers 7-11 7) 2.45g NaC2H3O2 and 70.0mL HC2H3O2 8) 10.07g NaC2H3O2 and 90.3mL HCl 9) 82.5mL HC2H3O2 and 15.4mL NaOH 10) 7.89g NaC2H3O2 and 12.5mL HC2H3O2 11) 9.25g NaC2H3O2 and 16.1mL HCl

Ksp The equilibrium is established when the solution is saturated. The only way to know when the solution is saturated is to observe solid in the bottom of the beaker. Calculate the silver ion concentrations in saturated solutions of silver chloride (Ksp = 1.8x10-10) and silver sulfate (Ksp = 1.5x10-5)

Answers 1.8x10-10 = [x][x] [Ag+] = 1.34x10-5M 1.5x10-5 = [2x]2[x] [Ag+] = 0.311M

Common Ion Ksp Problem What is the concentration of silver ions and chromate ions in a solution with solid silver chromate is added to a container with 0.1 M silver nitrate? Ksp = 9x10-12

Answer Ag2CrO4(s)  2Ag+1 + CrO4-2 I 0.1 0 C +2x +x E 0.1 + 2x x Ksp = 9.0 x 10-12 = [Ag+1]2 • [CrO4-2] = 9.0 x 10-12 = [.1 +2x]2 • [x] 9.0 x 10-12 = 0.12 • x [CrO4-2] = x = 9.0 x 10-10 M [Ag+1] = .1 + 2x = .1M [Ag2CrO4] = 9.0 x 10-10

A solution prepared by adding 750 mL of 4. 00 x 10-3 M Ce(NO3)3 to 300 A solution prepared by adding 750 mL of 4.00 x 10-3 M Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 precipitate? (Ksp = 1.9 x 10-10) Step one: Calculate the concentration of Ce+3 and IO3-1 before a reaction occurs [Ce+3]0 = (750.0 mL)(4.00x10-3 M) = 2.86 x 10-3 M (750. + 300.mL) [IO3-1]0 = (300.0 mL)(2.00x10-3 M) = 5.71 x 10-3 M (750. + 300.mL) Q = (2.86 x 10-3 M) x (5.71 x 10-3 M)3 = 5.32 x 10-10 Q > Ksp therefore the precipitation reaction will occur

Precipitation and Separation of Ions At any instant in time, Q = [Ba2+][SO42-]. If Q < Ksp, precipitation occurs until Q = Ksp. If Q = Ksp, equilibrium exists. If Q > Ksp, solid dissolves until Q = Ksp. Based on solubilities, ions can be selectively removed from solutions. Consider a mixture of Zn2+(aq) and Cu2+(aq). CuS (Ksp= 610-37) is less soluble than ZnS(Ksp=210-25), CuS will be removed from solution before ZnS.

Precipitation and Separation of Ions As H2S is added to the green solution, black CuS forms in a colorless solution of Zn2+(aq). When more H2S 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 Cu2+, the silver precipitates(Ksp for AgCl is 1.810-10) while the Cu2+ remains in solution, since CuCl2. Removal of one metal ion from a solution is called selective precipitation.

Factors that Affect Solubility Presence of a common ion. pH Formation of complex ions Amphoterism