Chapter 17 Additional Aspects of Aqueous Equilibria

The Common Ion Effect

The common ion effect is an event that occurs when an ionic equilibrium shifts due to the addition of a solute that contains an ion that is the same as an existing ion in the equilibrium.

An Example of the Common-Ion Effect HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)

Example Problem The Common-Ion Effect Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Ka for HF is 6.8  10−4. [H3O+] [F−] [HF] Ka = = 6.8  10-4

The Common-Ion Effect HF(aq) + H2O(l) H3O+(aq) + F−(aq) Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M. [HF], M [H3O+], M [F−], M Initially 0.20 0.10 Change −x +x At Equilibrium 0.20 − x  0.20 0.10 + x  0.10 x

The Common-Ion Effect (0.10) (x) 6.8  10−4 = (0.20) (0.20) (6.8  10−4) (0.10) = x 1.4  10−3 = x

The Common-Ion Effect Therefore, [F−] = x = 1.4  10−3 [H3O+] = 0.10 + x = 0.10 + 1.4  10−3 = 0.10 M So, pH = −log (0.10) pH = 1.00

Buffers

A buffer is an acid-base solution that resist changes in pH by neutralizing the effects of external acids and bases. NH4Cl and NH3 is a buffer system.

Examples of five buffers. Three solutions and two powders.

SEM of a red blood cells passing through a small artery Blood is a suspension and the serum of part of blood is buffered at 7.35-7.45

Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

Buffers If acid is added, the F− reacts to form HF and water.

Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

Derivation of the Henderson-Hasselbalch Equation

A specific equation exists for calculating the pH of a buffer ..i.e. the Henderson-Hasselbalch (H-H) equation.

Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H2O H3O+ + A− [H3O+] [A−] [HA] Ka =

Buffer Calculations Rearranging slightly, this becomes [HA] Ka = [H3O+] Taking the negative log of both side, we get base [A−] [HA] −log Ka = −log [H3O+] + −log pKa pH acid

Buffer Calculations So pKa = pH − log [base] [acid] Rearranging, this becomes pH = pKa + log [base] [acid] This is the Henderson–Hasselbalch equation.

Practice Problem: Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  10−4.

Henderson–Hasselbalch Equation pH = pKa + log [base] [acid] pH = −log (1.4  10−4) + log (0.10) (0.12) pH = 3.85 + (−0.08) pH = 3.77

pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a pKa close to the desired pH.

When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.

Addition of Strong Acid or Base to a Buffer Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

Calculating pH Changes in Buffers Before the reaction, since mol HC2H3O2 = mol C2H3O2− pH = pKa = −log (1.8  10−5) = 4.74

Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC2H3O2(aq) + OH−(aq)  C2H3O2−(aq) + H2O(l) HC2H3O2 C2H3O2− OH− Before reaction 0.300 mol 0.020 mol After reaction 0.280 mol 0.320 mol 0.000 mol

Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0. 200) pH = 4.74 + 0.06 pH = 4.80

Acid-Base Titrations

Titration A lab method used to determine the concentration of unknown solutions.

Titration For an acid-base titration the equivalence point is the state where the moles of acid in the solution is equal to the moles base in the solution.

Titration For an acid-base titration the endpoint is when the indicator changes color.

Strong Acid-Strong Base Titration

Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly.

Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base

Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

Weak Acid-Strong Base Titration

Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. The pH at the equivalence point will be >7. Phenolphthalein is commonly used as an indicator in these titrations.

Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

Strong Acid-Weak Base Titration

Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.

Titration of Polyprotic Acids

A polyprotic acid is a Bronsted acid with two or more ionizable protons. H2SO4 is the only polyprotic acid that is a strong acid.

Titrations of Polyprotic Acids Polyprotic acids will have An equivalence point for each ionizable proton.

Solubility Product Constant (Ksp)

BaSO4(s) Ba2+(aq) + SO42−(aq) The solubility product constant (Ksp) is an equilibrium constant for a slightly soluble salt. BaSO4(s) Ba2+(aq) + SO42−(aq)

Ksp is used in calculating the solubility of a slightly soluble salt. Solubility is given as g/L which can be converted to molar solubility mol/L (M).

Relationship Between Solubility and Ksp

Practice Exercise: Calculating Molar Solubility Calculate the molar solubility of Al(OH)3 if the Ksp is 1.8 x 10-33 .

Factors Affecting Solubility

Three factors affect the solubility of a slightly soluble salt. A Common Ion (solubility decreases) (2) pH (solubility increases for a basic anion in an acidic solution and for an acidic cation in a basic solution) (3) Complex Ion Formation (solubility increases)

Effect of a Common Ion on Solubility An ion that is already dissolved in a solution will shift the equilibrium to the left which decreases the solubility of the salt.

Practice Exercise: Calculating Molar Solubility in the Presence of a Common Ion a.Calculate the molar solubility of AgBr. b. Calculate the molar solubility of AgBr in a 0.030 M AgNO3 solution. c. Calculate the molar solubility of AgBr in a 0.10 M NaBr solution. .

Effect of pH on Solubility

Explain how decreasing the pH increases the solubility of this precipitate. Basic anions react with acid shifting the equilibrium to the right.

Explain how lower pH can cause sinkholes? Limestone is CaCO3. Rainwater has a pH ~ 5-6

Effect of Complex Ions on Solubility Recall metal ions are Lewis acids and can react with Lewis bases to form complex ions. The rxn is an equilibrium and is described by the equilibrium constant Kf.

Effect of Complex Ions on Solubility Write a chemical rxn and give an explanation that accounts for the dissolution of the solid.

Factors Affecting Solubility Amphoterism Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. Examples of such cations are Al3+, Zn2+, and Sn2+.

Predicting the Formation of a Precipitate Q = Ksp, no change Q < Ksp, more solid dissolves until Q = Ksp. Q > Ksp, the salt will precipitate until Q = Ksp.

Qualitative Analysis of Ions by Selective Precipitation

What is selective precipitation (ppt) ? Selective ppt is a qualitative analysis method where ions are separated and identified using a series of precipitation rxns.

General Flowchart for an Analysis of an Unknown Aqueous Solution

Qualitative Analysis of Ions by Selective Precipitation

Practice Problems

Questions 1-3 refer to Equation I and Figure I 1. Which graph represents the change in solubility of BaCO3 if HBr is added to a saturated solution of BaCO3. Figure: 17-22-07UNE7 Title: Exercise 17.7 Caption: Graphs representing the behavior of BaCO3 under different conditions. Graph A Graph B Graph C

Questions 1-3 refer to Equation I and Figure I 2. Which graph represents the change in solubility of BaCO3 if K2SO4 is added to a saturated solution of BaCO3. Figure: 17-22-07UNE7 Title: Exercise 17.7 Caption: Graphs representing the behavior of BaCO3 under different conditions. Graph A Graph B Graph C

Questions 1-3 refer to Equation I and Figure I 3. Which graph represents the change in solubility of BaCO3 if Li2CO3 is added to a saturated solution of BaCO3. Figure: 17-22-07UNE7 Title: Exercise 17.7 Caption: Graphs representing the behavior of BaCO3 under different conditions. Graph A Graph B Graph C

The beaker below contains an aqueous solution of potassium ions and barium ions. Recommend a qualitative analysis strategy that will separate the ions.

Circle the salts that are more soluble in an acidic solution than in pure water. BiI3 CdS CaCO3 AgCN Ba3(PO4)2 Strategy Write the chemical equation and determine if the conjugate acid of the anion is weak.

Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

The beaker below contains an aqueous solution of chromium (III) ions and iron (III) ions. Recommend a qualitative analysis strategy that will separate the ions.