+ Aqueous Equilibria/ Buffers/ Ksp AP Chemistry Pages

+ Solutions of Acids or Bases Containing a Common Ion Solutions that contain weak acid (HA) and its salt (NaA); for example, a solution of hydrofluoric acid (HF, K a = 7.2 x ) and its salt sodium fluoride. NaF(s)  Na + (aq) + F - (aq) strong electrolyte common ion is F -, produced by both HF and NaF Compare 1.0 M HF solution to solution of 1.0 M HF and 1.0 M NaF HF(aq) ↔ H + (aq) + F - (aq) Common Ion Effect: We would expect equilibrium of second solution to be driven to the left because of the addition of F- ions from the NaF. Therefore a solution containing both NaF and HF is less acidic than a solution of HF alone. Other examples: adding solid NH 4 Cl to a 1.0 M NH 3 solution. NH 3 (aq) + H 2 O(l) ↔ NH 4 + (aq) + OH - (aq)

+ Example Problem Calculate [H + ], pH and the percent dissociation of HF in a solution containing 1.0 M HF (K a = 7.2 x ) and 1.0 M NaF. (note: a 1.0 M solution of only HF has [H + ] = 2.7 x (pH = 1.57) and a percent dissociation of 2.7%.)

+ Practice Problem Calculate the pH of each of the following solutions. A M HONH 2 (K b = 1.1 x ) B M HONH 3 Cl C. Pure H 2 O D. A mixture containing M HONH 2 and 0.100M HONH 3 Cl Compare the percent ionization of the base in part A with the percent ionization of the base in part D. Explain any differences.

+ Buffered Solutions A buffer is a solution that contains a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid.) A buffered solution resists drastic changes in its pH when either OH - ions or H + ions are added. Our blood is buffered (pH = 7.42) The most important buffering system in the blood involves HCO 3 - and H 2 CO 3. so is Country Time Lemonade

+ Example 1 (pH of a buffered solution) A buffered solution contains 0.50 M acetic acid (K a = 1.8 x ) and 0.50 M sodium acetate. Calculate the pH of this solution.

+ Example 2 (a) Calculate the change in pH that occurs when mol solid NaOH is added to 1.0 L of the buffered solution in the previous example.(b) Calculate the pH after the addition of mol HCl to 1.0 L of the buffered solution. Note: when a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations.

+ Examples of Buffers: HC 2 H 3 O 2 and… H 2 CO 3 and… HF and… NH 4 + and… H 3 PO 4 and… H 2 PO 4 - and… Why do HCl and Cl - not make a buffer?

+ The Henderson-Hasselbalch Equation For any weak acid (HA + H 2 O  H 3 O + + A - ) K a = [H 3 O + ] = If the amounts of HA and A - are large compared to the amount of OH - added, the change in the ratio [HA] / [A - ] will be small. The pH will remain essentially constant.

+ The Henderson-Hasselbalch Equation -log[H 3 O + ] = -logK a – log[HA]/[A - ] pH = pK a + log[A - ]/[HA] Thus, pH = pK a + log([base]/[acid]) The Henderson-Hasselbalch Equation

+ Example Problem (pH of a buffered solution) Calculate the pH of a solution containing 0.75 M lactic acid (K a = 1.4 x ) and 0.25 M sodium lactate. Lactic acid (HC 3 H 5 O 3 ) is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion.

+ Example Problem A buffered solution contains 0.25 M NH 3 (K b = 1.8 x ) and 0.40 M NH 4 Cl. Calculate the pH of this solution.

+ Example Problem (Adding a strong acid to a buffered solution) Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution from the previous example.

+ Individual Practice Problems Beginning on page 740 # 21,23, 25, 27, 31, 33, 35, 37, and 38

+ Buffering Capacity The buffering capacity of a buffered solution represents the amount of H + or OH - ions the buffer can absorb without a significant change in pH. A buffer is most effective when the concentrations of the acid and its conjugate base are equal; if the concentrations of the acid and conj. base are different by more than a factor of 10, the buffer will not be reasonably effective. The more concentrated the components of a buffer are, the greater the buffer capacity. (A buffer made from 1.0 M acetic acid and 1.0 M sodium acetate has the same pH as a buffer made from 0.1 M solutions of each, but has a much greater ability to resist pH changes.)

+ Example Problem (Adding Strong Acid to a Buffered Solution) Calculate the change in pH that occurs when mol gaseous HCl is added to 1.0 L of each of the following solutions: Solution A: 5.00 M HC 2 H 3 O 2 and 5.00 M NaC 2 H 3 O 2 Solution B: M HC 2 H 3 O 2 and M NaC 2 H 3 O 2 For acetic acid, K a = 1.8 x 10 -5

+ Preparing a Buffer Large changes in the ratio [A - ] / [HA] will produce large changes in pH. We want to avoid this situation. Optimal buffering occurs when [A - ] = [HA]. So pH = pK a. (Most effective range for a buffer is within ±1 pH units of the pK a of the weak acid.)

+ Steps to preparing a buffer: 1.Choose the conjugate acid-base pair. It is best to use a conjugate acid-base pair whose pK a of the weak acid is within ±1 unit of the desired pH. For a buffer at a pH of 3.90, formic acid, HCO 2 H (pK a = 3.74) would be a good choice. 2.Calculate the ratio of buffer components that gives the desired pH (using the Henderson-Hasselbalch equation).

+ Steps to preparing a buffer continued… 3.Determine the buffer concentration. For many laboratory applications, concentrations of 0.50 M are suitable. From the concentration of the acid solutions in the available solutions, calculate the mass of conjugate base necessary to make the buffer. (e.g. we have 0.40 M formic acid and we want to make 1.0 liters of the buffer solution.) 4.Mix the solution and adjust the pH by adding small amounts of strong acid or base.

+ Example Problem A chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts). Which system would work best, and why? 1.chloroacetic acid (K a = 1.35 x ) 2.propanoic acid (K a = 1.3 x ) 3.benzoic acid (K a = 6.4 x ) 4.hypochlorous acid (K a = 3.5 x )

+ Titrations and pH Curves Equivalence Point: (stoichiometric point) The point in the titration where an amount of base has been added to exactly react with all acid originally present in the solution. End Point: Point at which the indicator changes color.

+ Strong Acid-Strong Base Titrations (H + + OH -  H 2 O) Case Study – titrating 50.0 mL of M HNO 3 with M NaOH. a.No NaOH has been added b.10.0 mL of M NaOH has been added. c.20.0 mL (total) of M NaOH has been added. d.50.0 mL (total) of M NaOH has been added. e mL (total) of M NaOH has been added f mL (total) of M NaOH has been added.

+ Titration Curve

+ Titrations of Weak Acids with Strong Bases Calculating the pH curve for a weak acid-strong base titration is a two-step procedure: 1.A stoichiometry problem: The reaction of hydroxide ion with the weak acid is assumed to run to completion, and the concentration of acid remaining and the conjugate base formed are determined. 2.An equilibrium problem: The position of the weak acid equilibrium is determined, and the pH is calculated.

+ Case Study – titrating 50.0 mL of 0.10 M acetic acid (Ka = 1.8 x ) with 0.10 M NaOH. a.No NaOH has been added. b mL of M NaOH has been added. c mL (total) of M NaOH has been added. d.40.0 mL (total) of M NaOH has been added. e mL (total) of M NaOH has been added. f mL (total) of M NaOH has been added. g.75.0 mL (total) of M NaOH has been added.

+ Titration Curve

+ Example Problem (Titration of a Weak Acid) Hydrogen cyanide gas (HCN) is highly toxic. It is a very weak acid (Ka = 6.2 x ) when dissolved in water. If a 50.0-mL sample of HCN is titrated with M NaOH, calculate the pH of the solution a.after 8.00 mL of M NaOH has been added. b.at the halfway point of the titration c.at the equivalence point of the titration.

+ Example Problem (Calculating Ka) A chemist has synthesized a monoprotic weak acid and wants to determine its K a value. To do so, the chemist dissolves 2.00 mmol of the solid acid in mL water and titrates the resulting solution with M NaOH. After 20.0 mL NaOH has been added, the pH is What is the K a value for the acid?

+ Titrations of Weak Bases with Strong Acids Case Study – titrating mL of M NH 3 with 0.10 M HCl. (K b = 1.8 x ) Calculate the pH… a.Before the addition of any HCl b.Before the equivalence point. (half-way) c.At the equivalence point. d.Beyond the equivalence point.

+ Titration Curve

+ KHP is a COMMON standardizing agent. Moles KHP = moles of H +

+ Titration Curve for Polyprotic Acids Titration curve for Phosphoric Acid (H 3 PO 4 ) with NaOH

+ Acid-Base Indicators Two ways to determine the equivalence point of an acid-base titration: 1.Use a pH meter to monitor pH and plot the titration curve. The center of the vertical region represents the equivalence point. 2.Use an acid-base indicator, which marks the end point in a titration by changing color. The equivalence point of a titration defined by stoichiometry is not necessarily the same as the end point, but careful selection of the indicator will ensure the error is negligible. We choose an indicator with an end point close to the equivalence point, so that the color change in the indicator corresponds to the point at which the acid has been neutralized.

+ Acid-Base Indicators Continued… Phenolphthalein: HIn (colorless)  H + + In - (pink) How does a molecule function as an indicator? Hypothetical indicator HIn, a weak acid with K a = 1.0 x HIn(aq)  H + (aq) + In - (aq) Red Blue For most indicators, about 1/10 of the initial form must be converted to the other form before a new color is apparent.

+ Example Problem Bromothymol blue, an indicator with a K a value of 1.0 x 10 -7, is yellow in its HIn form. Suppose we put a few drops of this indicator in a strongly acidic solution. If this solution is titrated with NaOH, at what pH will the indicator color change first be visible?

+ Example Problem What indicator should be used for the following titrations: (page 715) a.100 mL of M HCl with M NaOH (equivalence point at pH 7.00) b.100 mL of M HC 2 H 3 O 2 with M NaOH (equivalence point at pH 8.7)

+ Individual Practice Beginning on page 741 #39, 41, 49, 51, 53, 55, 57, 59, 61, 63, 65, 69, 71, and 73

+ Solubility Equilibria and the Solubility Product Constant (Ksp) PbCl 2 (s) ↔ Pb 2+ (aq) + 2 Cl - (aq) what does it mean that PbCl 2 is “insoluble”? Solubility Product Constant (K sp ) PbCl 2 = 1.17 x 10 -5

+ Example Problem Write the solubility product expression for the following species: 1. magnesium carbonate 2. iron (II) hydroxide 3. silver carbonate 4. calcium phosphate

+ Example Problems Copper (I) bromide has a measured solubility of 2.0 x mol/L at 25°C. Calculate the K sp value. Calculate the K sp value for bismuth sulfide (Bi 2 S 3 ), which has a solubility of 1.0 x mol/L at 25°C. The K sp value for copper (II) iodate is 1.4 x at 25°C. Calculate its molar solubility at 25°C.

+ CAUTION: Be careful comparing relative solubilities AgI (K sp = 1.5 x ); CuI (K sp = 5.0 x ); CaSO 4 (K sp = 6.1 x ) CuS (K sp = 8.5 x ); Ag 2 S (K sp = 1.6 x ); Bi 2 S 3 (K sp = 1.1 x )

+ Common Ion Effect What is the solubility of silver chromate (K sp = 9.0 x ) in a M solution of silver nitrate? Calculate the solubility of solid CaF 2 (K sp = 4.0 x ) in a M NaF solution.

+ pH and Solubility The pH of a solution can greatly affect a salt’s solubility. Milk of Magnesia: Mg(OH) 2 ↔ Mg 2+ (aq) + 2 OH - (aq) K sp = 1.8 x How would the addition of OH - ions (increase in pH) affect the solubility? How would the addition of H + ions (decrease in pH) affect the solubility?

+ Example Problem Saturated solution of Mg(OH) 2. (K sp = 1.8 x ) 1. What is [Mg 2+ ]? 2. What is pH? 3. Now suppose that solid Mg(OH) 2 is equilibrated with a solution buffered at a more acidic pH of 9.0. What is [Mg 2+ ]?

+ pH and Solubility Ag 3 PO 4 (s) ↔ 3 Ag + (aq) + PO 4 3- (aq) Why more soluble in acidic solutions? Is PO 4 3- a strong or weak base? (H + + PO 4 3-  HPO 4 2- ) AgCl(s) ↔ Ag+(aq) + Cl-(aq) Is Cl- a strong or weak base? How does this affect the solubility of AgCl in acidic solutions? General Rule – If the anion X - is an effective base (HX is a weak acid), the salt MX will show an increased solubility in acidic solutions.

+ Precipitation and Qualitative Analysis Ion Product (Q sp ) – just like K sp, but use initial concentrations (Q = [Ca 2+ ] 0 [F - ] 0 2 ) If Q > K sp, precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy K sp. If Q < K sp, no precipitation occurs. If Q = K sp, it is a saturated solution.

+ Example Problems A solution is prepared by adding mL of 4.00 x M Ce(NO 3 ) 3 to mL of 2.00 x M KIO 3. Will Ce(IO 3 ) 3 (K sp = 1.9 x ) precipitate from this solution? Suppose we mix mL of M lead (II) nitrate and mL of M sodium iodide. Determine whether a precipitate occurs and then determine the equilibrium concentrations of Pb 2+ and I - ions.(Ksp = 1.4 x )

+ Example Problem A solution is prepared by mixing mL of 1.00 x M Mg(NO 3 ) 2 and mL of 1.00 x M NaF. Calculate the concentrations of Mg 2+ and F - at equilibrium with solid MgF 2 (K sp = 6.4 x ).

+ Example Problem (Selective Precipitation) A solution contains 1.0 x M Cu + and 2.0 x M Pb 2+. If a source of I - is added gradually to this solution, will PbI 2 (K sp = 1.4 x ) or CuI (K sp = 5.3 x ) precipitate first? Specify the concentration of I - necessary to begin precipitation of each salt.

+ Selective Precipitation 1.Since metal sulfide salts differ dramatically in their solubilities, the sulfide ion is often used to separate metal ions by selective precipitation. For example, consider a solution containing a mixture of Fe 2+ (K sp of FeS = 3.7 x ) and Mn 2+ (K sp of MnS = 2.3 x ). Which will precipitate first as you carefully add the S 2- ? 2.Another advantage of the sulfide ion is that it is basic, so its concentration can be controlled by regulating the pH of the solution. H 2 S ↔ H + + HS - K a1 = 1.0 x HS - ↔ H + + S 2- K a2 = 1.0 x In an acidic solution [S 2- ] will be small, but in basic solutions [S 2- ] will be relatively large. This means that the most insoluble salts (for example, CuS, K sp = 8.5 x ) can be precipitated from an acidic solution, leaving the more soluble ones still dissolved. The remaining cations can be precipitated by making the solution slightly basic.

+ Qualitative Analysis Group I – Insoluble chlorides 1.Adding dilute aqueous HCl will precipitate only Ag +, Pb 2+, and Hg All other chlorides are soluble and remain in solution. 2.Remove the group I precipitate, leaving the other ions in solution. Group II – sulfides insoluble in acid solution 1.After the insoluble chlorides are removed, the solution is still acidic. If H 2 S is added to this solution, only the most insoluble sulfides (Hg 2+, Cd 2+, Bi 3+, Cu 2+, and Sn 4+ ) will precipitate, since [S 2- ] is relatively low. 2.Remove the precipitate and the more soluble sulfides remain dissolved.

+ Qualitative Analysis Continued… Group III – sulfides insoluble in basic solution 1.The solution is made basic and more H 2 S is added. This will precipitate cations Co 2+, Zn 2+, Mn 2+, Ni 2+, and Fe 2+. If any Cr 3+ and Al 3+ ions are present, they also will precipitate, but as insoluble hydroxides. 2.Remove the precipitate, leaving the other ions in solution. Group IV – insoluble carbonates or phosphates 1.At this point, all the cations have been precipitated except those from Groups 1A and 2A of the periodic table. 2.The Group 2A cations form insoluble carbonates and can be precipitated by the addition of CO 3 2- or (NH 4 ) 2 HPO 4.

+ Qualitative Analysis Continued… Group V – alkali metal and ammonium ions 1.The only ions remaining in solution at this point are Group 1A cations and the NH 4 + ion, all of which form soluble salts with the common anions. 2.The Group 1A cations are usually identified by the characteristic color they produce when heated in a flame. (emission spectra) Additional testing is necessary to determine which ions are present within each of the groups.

+ Equilibria Involving Complex Ions Complex Ion: A charged species consisting of a metal ion surrounded by ligands. Ligand: A Lewis base ( molecule or ion having a lone electron pair that can be donated to an empty orbital on the metal ion to form a covalent bond. (H 2 O, NH 3, Cl -, CN - ) Coordination Number: The number of ligands attached to a metal ion. Most common Coordination Numbers are 6; 4; and 2 Cu(H 2 O) 4 2+ What is the Lewis acid? Cu 2+ What is the Lewis base? H 2 O

+ Equilibria Involving Complex Ions Metal ions add ligands one at a time: Ag + + NH 3 ↔ Ag(NH 3 ) + K f1 = 2.1 x 10 3 (formation constant) Ag(NH 3 ) + + NH 3 ↔ Ag(NH 3 ) 2 + K f2 = 8.2 x 10 3 Consider a solution prepared by mixing mL of 2.0 M NH 3 with mL of M AgNO 3. Calculate the equilibrium concentrations of Ag +, NH 3, Ag(NH 3 ) +, and Ag(NH 3 ) 2 +. After the solutions are mixed, but before and reaction occurs… What are the initial concentrations of [NH 3 ] and [Ag + ]?

+ Example Problem Calculate the concentrations of Ag +, Ag(S 2 O 3 ) -, and Ag(S 2 O 3 ) 2 3- in a solution prepared by mixing mL of M AgNO 3 with mL of 5.00 M Na 2 S 2 O 3. The stepwise formation equilibria are: Ag + + S 2 O 3 2- ↔ Ag(S 2 O 3 ) - K f1 = 7.4 x 10 8 Ag(S 2 O 3 ) - + S 2 O 3 2- ↔ Ag(S 2 O 3 ) 2 3- K f2 = 3.9 x 10 4

+ Complex Ions and Solubility Adding the right ligand can drastically increase the solubility of a slightly soluble ionic compound. For example, AgCl (slightly soluble in water). (1)AgCl(s) ↔ Ag + (aq) + Cl - (aq)K sp = 1.77 x (2)Ag + (aq) + 2 NH 3 (aq) ↔ Ag(NH 3 ) 2 + (aq)K f = 1.7 x 10 7 (3)AgCl(s) + 2 NH 3 (aq) ↔ Ag(NH 3 ) 2 + (aq) + Cl - K overall = 3.0 x K overall = K sp x K f

+ Individual Practice Beginning on page 743: # 75, 77, 79, 81, 85, 89, 91, 97, 99, 101, 103, 105

+ KHP is a COMMON standardizing agent. Moles KHP = moles of H +