Ionic Equilibria

Ionic Equilibria

AP Chemistry

Chapter 19

Mr. Solsman

• Goals:

• Acid-base buffers, common ion effect, titrations

• Slightly soluble salts

• Complex ions

• I Acid-Base Equilibria

• A. Solutions of Acids or Bases Containing a Common Ion

• A buffer is something that lessens the impact of an external force.

• An acid-base buffer is a solution that lessens changes in [H3O+] resulting from the addition of an acid or base.

• When a small amount of H3O+ or OH- is added to an unbuffered solution, the change in pH is much larger than the change in a buffered solution.

• Buffers work through the common ion effect.

• Suppose we dissolve acetic acid in water and then add sodium acetate as well.

• CH3COOH(aq) + H2O(l) CH3COO-(aq) +

H3O+(aq)

• But adding CH3COONa shifts the equilibrium left (Le Chatelier) in effect decreasing the [H3O+] and lowering the acetic acid dissociation.

• The acetate ion is called the common ion because it is common to both solutions.

• The common-ion effect occurs when a reactant containing a given ion is added to an equilibrium mixture that already contains that ion and the position of equilibrium shifts away from forming more of it.

• Features of a buffer—a buffer consists of high concentrations of the acidic (HA) and basic (A-) components. When small amounts of H3O+ or OH- ions are added to the buffer, they cause a small amount of one buffer component to convert into the other changing the relative concentrations of the two components.

• As long as the H3O+ or OH- added is small compared to HA and A-, the added ions have little effect on the pH because they are consumed by one or the other buffer components.

• A- consumes H3O+ and HA consumes OH-

• Calculate the pH:

• (a) Of a buffer solution of 0.50 M acetic acid and 0.50 M sodium acetate.

• (b) after adding 0.020 mol solid NaOH to 1.0 L of the buffer in part a.

• (c) after adding 0.020 mol HCl to 1.0 L of the buffer in part a.

Calculate the pH of a buffer of 0.50 M HF and 0.45 M F- (a) before and (b) after the addition of 0.40 g NaOH to 1.0 L of the buffer. (Ka of HF = 6.8 x 10-4)

• How does a buffer work?

• Consider the reaction:

• HA + H2O A- + H3O+

• Adding OH- gives:

• HA + OH- A- + H2O

• We know Ka = [H3O+] [A-] / [HA]

• and [H3O+] = Ka [HA] / [A-]

• Adding OH- ions causes HA to be converted to A-, the ratio of [HA] / [A-] decreases, but if the original amounts of HA and A- are large, the change is very small. The pH and [H3O+] remain nearly constant.

• Similar reasoning applies when protons are added to a buffered solution of a weak acid and a salt of its conjugate base. Because the A- ion has a high affinity for H3O+, the added ions react with A- to form the weak acid: H3O+ + A- HA

• There is a net change of A- to HA. Again if [A-] and [HA] are large, little change in pH occurs.

• Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate.

• Henderson-Hasselbalch equation:

• For any weak acid:

• HA + H2O H3O+ + A-

• [H3O+] = Ka [HA] / [A-]

• Or pH = pKa + log([Base]/[Acid)]

• Using Henderson-Hasselbalch calculate the pH from the last problem.

• A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.40 M NH4Cl. Find the pH of the solution.

• Adding strong acid to a buffered solution

• Calculate the pH of the solution that results when 0.10 mol of gaseous HCl is added to 1.0 L of the buffered solution in the previous example.

• Buffer capacity

• The buffer capacity represents the amount of protons or hydroxide ions the buffer can absorb without significantly changing the pH.

• The pH of a buffered solution is determined by the ratio of [A-] to [HA].

• The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-].

• Calculate the change in pH that occurs when 0.0100 mol of gaseous HCl is added to each of the following substances:

• Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2

• Solution B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2

• A chemist needs a solution buffered at pH 4.30 and can choose from the following acids and their salts: Which works best?

• (a) chloroacetic acid (Ka = 1.35 x 10-3)

• (b) propanoic acid (Ka = 1.3 x 10-5)

• (c) benzoic acid (Ka = 6.4 x 10-5)

• (d) hypochlorous acid (Ka = 3.5 x 10-8)

• Describe how you would prepare a “phosphate buffer” with a pH of about 7.40.

• Hint: the concentration of the acid component must be roughly equal to the conjugate base component. Or when pH pKa.

• How would one prepare a benzoic acid/benzoate buffer with pH 4.25, starting with 5.0 L of 0.050 M sodium benzoate (C6H5COONa) solution and adding the acidic component? (The Ka of benzoic acid is 6.3 x 10-5.)

Summary, Buffered Sol’ns

• 1. They contain relatively large concentrations of a weak acid and corresponding weak base.

• 2. When H3O+ is added, it reacts essentially to completion with the weak base present.

• 3. When OH- is added, it reacts essentially to completion with the weak acid present.

• 4. The pH is determined by the ratio of the concentrations of the weak acid and base. As long as this ration is constant, the pH will remain virtually constant. This will be true as long as the concentrations of the buffering mtl’s are large compared to the amounts of H3O+ or OH- added.

• The pKa of the weak acid to be used in the buffer should be as close as possible to the pH of the solution.

• Acid-Base Titration Curves

• Two common devices for measuring pH are a pH meter and an acid-base indicator.

• An acid-base indicator is a weak organic acid (HIn) that has a different color than its conjugate base (In-).

• The color change occurs over a specific and narrow pH range.

• The amount of indicator used is small enough that the solution pH is not affected.

• The indicator changes color over a 2 pH range.

• 1. Strong Acid-Strong Base Titrations

• A titration curve is a plot of pH versus the volume of added titrant.

• There are distinct regions of the curve.

• The equivalence point is the point at which the number of moles of OH- added equals the number of moles of H3O+ originally present. This occurs at the nearly vertical portion of the curve where the solutions consists of the anion of the SA and cation of the SB.

• These ions do not react with water, so the pH is neutral; pH = 7.00.

• Since we are usually titrating with milliliters, an alternative way of looking at molarity is often used:

• Molarity = mmol / mL

• Strong Acid-Strong Base Titrations

• 50.0 mL of 0.200 M HNO3 are titrated with 0.100 M NaOH. Calculate the pH after the additions of 0, 10, 20, 50, 100, 150, and 200 mL of NaOH. Then construct a titration curve and label it properly.

• Titrating a strong base with a strong acid is very similar to the last example.

• However OH- is in excess before the equivalence point and H+ is in excess after the equivalence point.

• The curve is also flipped.

• Weak Acid-Strong Base Titrations

• 50.0 mL of 0.10 M acetic acid (Ka = 1.8 x 10-5) are titrated with 0.10 M NaOH. Calculate the pH after the additions of 0, 10, 25, 40, 50, 60, and 75 mL samples of NaOH. Then construct a titration curve and label it properly.

• Weak Base-Strong Acid Titration

• Use the same procedures as in the last problems. Decide which major species are present in solution and decide which reactions run to completion. Choose the dominant equilibrium and calculate the pH!

• 20.0 mL of 0.10 M triethylamine, (CH3CH2)3N, (Kb = 5.2 x 10-4) are treated with 0.100 M HCl. Calculate the pH after 0.0, 10, 15, 19, 19.95, 20.05, and 25 mL additions of HCl. Construct a titration curve.

The pH Curves for the Titrations of 50.0-mL

Samples of 0.10 M Acids with Various Ka

Values with 0.10 M NaOH

• The equivalence point occurs in each case when the same volume of 0.10 M NaOH has been added but the curve shape differs dramatically. The weaker the acid, the greater the pH at the equivalence point. The vertical region of the curve becomes shorter as the acid becomes weaker.

• SA-SB at the equivalence pH = 7.00

• WA-SB pH > 7.00

• WB-SA pH < 7.00

• In the beginning, in an acid-base titration, the pH increases more rapidly than it does in the strong acid case.

Polyprotic Acids

• Titrating a polyprotic acid with a base leads to two (or more) buffer regions and two (or more) equivalence points.

• Amino acids for example are polyprotic acids. Amino acids have a basic amino group and an acidic carboxylic acid group.

• A structure that has positive and negative sites is called a zwitterion.

Slightly Soluble Ionic Compounds

• Most solutes, even those called “soluble” have a limited solubility in a particular solvent.

• When a soluble compound dissolves in water, it dissociates into ions. It is usually assumed that it dissociates completely into ions.

• Such is not usually the case however; transition metals and heavy main group metals have a significant covalent character in their metal-nonmetal bonding, and their solutions often contain other species as well. PbCl2(s) : Pb2+

(aq) Cl-(aq) PbCl2(aq)

PbCl+(aq)

• The relationship between solid ionic solutes and their aqueous ions can be expressed as follows:

• PbSO4(s) Pb2+(aq) + SO4

2-(aq)

• Qc = [Pb2+][SO42-] / [PbSO4]

• Qsp = Qc [PbSO4]

• So Qsp = [Pb2+][SO42-] = Ksp

• This new equilibrium is the solubility product constant, Ksp.

• Ksp depends only on the temperature, not individual ion concentrations.

• In general, for MpXq

• Qsp = [Mq+]p [Xp-]q = Ksp

• Cu(OH)2(s) Cu2+(aq) + 2 OH-

(aq)

• Ksp = [Cu2+][OH-]2

• MnS(s) Mn2+(aq) + S2-

(aq)

• S2-(aq) + H2O(l) HS-

(aq) + OH-(aq)

• MnS(s)+H2O(l) Mn2+(aq) + HS-

(aq) + OH-(aq)

• Ksp = [Mn2+][HS-][OH-]

• Write Ksp for the following:

• Magnesium carbonate

• Iron(II) hydroxide

• Calcium phosphate

• Silver Sulfide

• Copper(I) bromide has a measured solubility of 2.0 x 10-4 M at 25o C. Calculate its Ksp value.

• Copper(I) bromide has a measured solubility of 2.0 x 10-4 M at 25o C. Calculate its Ksp value.

• CuBr(s) Cu+(aq) + Br-(aq)

• Ksp = 4 x 10-8 mol2/L2

• Calculate the Ksp value for bismuth sulfide which has a solubility of 1.0 x 10-15 M at 25oC.

• The Ksp for copper(II) iodate is 1.4 x 10-7 at 25o C. Calculate its solubility at 25o C.

• When powdered CaF2 is shaken with pure water at 180C, 1.5 x 10-4 g dissolves for every 10.0 mL of solution. Calculate the Ksp at this temperature.

• A suspension of Mg(OH)2 in water is sold as milk of magnesia for minor stomach disorders by neutralizing acid. The [OH-] is too low to harm the mouth and throat. What is the molar solubility of Mg(OH)2 (Ksp = 6.3 x 10-10) in pure water?

Relative Solubilities

• Relative solubilities can only be predicted by comparing the Ksp values for salts that produce the same total number of ions.

• For salts that produce different numbers of ions when dissolved, the Ksp values cannot be compared directly to determine relative solubilities.

• In these cases, the higher the Ksp, the greater the solubility.

Common Ion Effect

• The presence of a common ion decreases the solubility of a slightly soluble ionic compound.

• Consider what happens when Na2CrO4 is added to a solution of PbCrO4:

• Calculate the solubility of solid calcium fluoride (Ksp = 4.0 x 10-11) in a 0.025 M NaF solution.

• Determine the solubility of a barium sulfate solution in pure water and after 0.10 M Na2SO4 is added. (Ksp = 1.1 x 10-10)

Effect of pH on Solubility

• The hydronium ion concentration can have a profound effect on the solubility of an ionic compound. If the compound contains the anion of a weak acid, addition of H3O+ (from a strong acid) increases its solubility.

• For example adding HNO3 to CaF2 affects the solubility.

• CaF2(s) Ca2+(aq) + 2 F-

(aq)

• 2 F-(aq) + 2 H3O+(l) 2 HF(aq) + 2 H2O(l)

• Reaction shifts right, increasing the solubility of CaF2.

• CaCO3(s) Ca2+(aq) + CO32-(aq)

• Adding strong acid introduces H3O+, which reacts with CO3

2- to form the weak acid HCO3

-

• CO32- + H3O+ HCO3

- + H2O

• HCO3- + H3O+ H2CO3 + H2O

• H2CO3 CO2 + 2 H2O (acidic solution)

• Consider:

• Mg(OH)2(s) Mg2+(aq) + 2 OH-(aq)

• Additional OH- ions force the equilibrium left, decreasing the solubility of Mg(OH)2.

• Adding H3O+, increases the solubility since OH- reacts with the H3O+ ions, forcing the reaction to the right.

• A general rule is that if the anion X- is an effective base—that is, if HX is a weak acid---the salt MX will show increased solubility in an acidic solution.

• Ex: OH-, S2-, CO32-, C2O4

2-, and CrO42-

• Predict the effect on solubility of adding a strong acid to the following:

• Zinc sulfide

• Silver iodide

• ZnS(s) Zn2+ + HS- + OH-

• HS- + H3O+ H2S + H2O

• OH- + H3O+ 2 H2O

• ZnS(s) Zn2+ + HS- + OH-

• HS- + H3O+ H2S + H2O

• OH- + H3O+ 2 H2O

• Increases solubility

• AgI(s) Ag+ + I-

• AgI(s) Ag+ + I-

• No effect since I- is the conjugate base of a strong acid (HI).

• Predicting the Formation of a Precipitate

• Q and K can be used to determine if a precipitate will form in a reaction and, if not, what concentrations of ions will cause one to form.

If Qsp = Ksp the solution is saturated and no change occurs.

• If Qsp > Ksp precipitate forms until the solution is saturated.

• If Qsp < Ksp the solution is unsaturated and no precipitate forms.

• Phosphate in natural waters often precipitates as insoluble salts, such as Ca3(PO4)2. In the Cuyahoga River, [Ca2+]i = [PO4

-3]i = 1.0 x 10-9 M. Will Ca3(PO4)2 precipitate? Ksp of Ca3(PO4)2 = 1.2 x 10-29

• A solution is prepared by mixing 150.0 mL of 1.00 x 10-2 M Mg(NO3)2 and 250.0 mL of 1.00 x 10-1 M NaF. Calculate the concentration of Mg2+ and F- at equilibrium with solid MgF2 (Ksp = 6.4 x 10-9).

Complex Ion Equilibria

• A complex ion is a charged species consisting of a metal ion surrounded by ligands.

• A ligand is a Lewis base having a lone electron pair that can be donated to an empty orbital on the metal ion to form a covalent bond.

• Common ionic ligands are: OH- , Cl-, and CN-.

• Molecular ligands: H2O and NH3

• Ligands are added to the metal ion one at a time in steps. The number of ligands, called the coordination n umber, commonly exists in groups of 2, 4, or 6. Each step has its own equilibrium constant value.

• Kf = Kf1 x Kf2 x Kf3 x …

• Kf values are very large indicating tremendous stability.

• Examples:

• Co(H2O)62+ and Ni(NH3)6

2+

• CoCl42- and Cu(NH3)4

2+

• Ag(NH3)2+

• Calculate the concentrations of Ag+, Ag(S2O3)-, and Ag(S2O3)2

3- in a solution prepared by mixing 150.0 mL of 1.00 x 10-3 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3.

• The stepwise formation equilibria are

• Ag+ + S2O32- Ag(S2O3)-

• Kf1 = 7.4 x 108

• Ag(S2O3)- + (S2O3)2- Ag(S2O3)23-

• Kf2 = 3.9 x 104

• Cyanide ion is toxic because it forms a stable complex ion with the Fe3+ ion certain iron-containing proteins engaged in energy production. To study this effect, a biochemist mixes 25.5 mL of 3.1 x 10-2 M Fe(H2O)6

3+ with 35 mL of 1.5 M NaCN. What is the final [Fe(H2O)6

3+]?

• Kf of Fe(CN)63- = 4.0 x 1043

• We know that H3O+ increases the solubility of a slightly soluble ionic compound if its anion is that of a weak acid. Similarly, a ligand increases the solubility of a slightly soluble ionic compound if it forms a complex ion with the cation.

• For example:

• ZnS(s) + H2O(l) Zn2+ + HS- + OH-

• Ksp = 2.0 x 10-22

• Add: 1.0 M NaCN

• Zn2+ + 4 CN- Zn(CN)42- Kf = 4.2 x 1019

• Koverall = Ksp x Kf = 8.4 x 10-3

• A critical step in black and white film processing is the removal of excess AgBr by “hypo” an aqueous solution of sodium thiosulfate (Na2S2O3), through formation of the complex ion Ag(S2O3)2

3-. Calculate the solubility of AgBr in (a) H2O, (b) 1.0 M hypo, Kf of Ag(S2O3)2

3- is 4.7 x 1013 and Ksp of AgBr is 5.0 x 10-13.

• Selective precipitation is a technique in which metal ions in aqueous solutions are separated by using a reagent whose anion forms a precipitate with only one or a few metal ions in the mixture.

• A solution of precipitating ion is added until the Qsp value of the more soluble compound is almost equal to its Ksp value. This method ensures that the Ksp value of the less soluble compound is exceeded as much as possible. The maximum amount of the less soluble compound precipitates, but none of the more soluble compound does.

• A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4 x 10-8) or CuI (Ksp = 5.3 x 10-12) precipitate first? Specify the concentration of I- necessary to begin precipitation of each salt.