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AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS

THE COMMON ION EFFECT The common ion effect occurs when the addition of an ion already present in the system causes the equilibrium to shift away from that ion. Addition of NaCH3COO will cause this reaction to shift:

CH3COOH ↔ H+ + CH3COO-

EXAMPLE: A solution is 0.120 M in acetic acid and 0.0900 M in sodium acetate. Calculate the [H+] at equilibrium. (The Ka of acetic acid is 1.8 x 10-5) BUFFERED SOLUTIONS A buffered solution is one which resists changes in pH when hydroxide ions or hydrogen ions are added. A buffer solution must have a species which will react with any added hydrogen ions. This usually consists of a solution of one of the following: A Weak Acid and its Salt (Example: The “acetic acid / acetate” buffer)

CH3COOH ↔ H+ + CH3COO-

Addition of strong acid: Addition of strong base:

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A Weak Base and its Salt (Example: The “ammonia / ammonium” buffer)

NH3 + H2O ↔ NH4 + + OH-

Addition of strong acid: Addition of strong base: EXAMPLE: Give the complete reaction that will occur initially in the following system: Hydrochloric acid is added to a solution of nitrous acid (HNO2) and potassium nitrite. Potassium hydroxide is added to a solution of nitrous acid (HNO2) and potassium nitrite. Buffer Capacity – the amount of acid or base that can be absorbed by a buffer system without a significant change in pH. In order to have a large buffer capacity, a solution should have large concentrations of both buffer components. Optimum buffering occurs when: 1. [HA] = [A-] or [B] = [BH+] 2. The pKa of the weak acid (or the pKb of the base) should be as close as possible to the desired pH of the buffer system SOLVING EQUILIBRIA PROBLEMS WHEN STRONG ACIDS OR BASES ARE ADDED (CBAC-RICE) 1. Determine the Completion reaction that occurs. 2. Determine the amount of substances present Before the reaction occurs in moles. 3. Determine the amount of substances present After the reaction occurs. 4. Change the amount of substances back to molarity. 5. Determine the equilibrium reaction to begin RICE using these concentrations.

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EXAMPLE: Determine the pH of a system when 100 mL of 0.100 M HCl is added to 455 mL of a solution that is 0.120 M in acetic acid and 0.0900 M in sodium acetate. (The Ka of acetic acid is 1.8 x 10-5.)

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AP CHEMISTRY NOTES 10-2 AQUEOUS EQUILIBRIA: PREPARATION OF A BUFFER

PREPARATION OF A BUFFER SYSTEM The “Henderson-Hasselbach” Equation:

𝐩𝐇 = 𝐩𝐊𝐚 + 𝐥𝐨𝐠 [𝐀−]

[𝐇𝐀]

or

𝐩𝐎𝐇 = 𝐩𝐊𝐛 + 𝐥𝐨𝐠 [𝐁𝐇+]

[𝐁]

EXAMPLE: A buffer solution of pH 5.30 is to be prepared from propionic acid and sodium propionate. The concentration of sodium propionate must be 0.50 mol/L. What should be the concentration of the acid? (Ka = 1.3 x 10-5 for CH3CH2COOH)

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EXAMPLE: You wish to prepare a buffer solution to maintain the pH of a solution at 4.30. A list of possible acids (and their conjugate bases) is shown below:

Acid Conjugate Base Ka

HSO4- SO4

2- 1.2 x 10-2

CH3COOH CH3COO- 1.8 x 10-5

HCN CN- 4.0 x 10-10

Which combination should be selected, and what should the ratio of acid to conjugate base (salt) be?

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AP CHEMISTRY NOTES 10-3 AQUEOUS EQUILIBRIA: TITRATION

Titration involves the use of a known concentration of a base to neutralize an acid of unknown concentration (or vice versa) in order to determine the concentration of that acid. Titrant – a solution of known concentration; this solution is placed in a buret so that it can be easily measured as it is added to another solution Analyte (or Titrand) – the solution to be tested; this solution is of unknown concentration and is placed in a

beaker or Erlenmeyer flask for analysis Equivalence Point (or End Point) – the point at which just enough titrant has been added to another solution to completely react Indicator – substance which changes color with changes in pH; usually weak acids (HIn) They have one color in their acidic form (HIn) and another color in their basic form (In-)

Useful indicator range = pKa + 1 When choosing an indicator, determine the pH at the equivalence point of the titration and choose an indicator with a pKa close to that value. Titration Curve – a plot of pH as a function of the amount of titrant added

STRONG ACID – STRONG BASE TITRATIONS The net ionic reaction for this type of titration is always: H+ + OH- → H2O The pH is easy to calculate because all reactions go to completion. At the equivalence point, the solution is neutral.

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EXAMPLE: 100. mL of 1.00 M HCl is titrated with 0.500 M NaOH. Calculate the pH after 50.0 mL of base have been added. Calculate the pH after 200. mL of base have been added. Calculate the pH after 300. mL of base have been added.

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WEAK ACID – STRONG BASE TITRATIONS The reaction of a strong base with a weak acid is assumed to go to completion. Notice the difference in the titration curve below:

The vertical portion at the equivalence point is much shorter, the curve starts at a higher pH, and the equivalence point is slightly basic (the strong base is the dominant species). NOTE For a weak acid – strong base titration: At half-way to the equivalence point, pH = pKa Equilibrium calculations must be done to determine the pH of these solutions throughout the titration process. EXAMPLE: 30.0 mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HF. (Ka of HF = 7.2 x 10-4) Determine the pH of the final solution.

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50.0 mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HF. Determine the pH of the final solution. 60.0 mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HF. Determine the pH of the final solution.

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WEAK BASE – STRONG ACID TITRATIONS The reaction of a strong acid with a weak base is assumed to go to completion. Notice the difference in the titration curve below:

The vertical portion at the equivalence point is much shorter, the curve starts at a basic pH, and the equivalence point is slightly acidic (the strong acid is the dominant species). Equilibrium calculations must be done to determine the pH of these solutions throughout the titration process as well. EXAMPLE: Calculate the pH when 100. mL of 0.050 M NH3 is titrated with 10.0 mL of 0.10 M HCl. (Ka of NH3 = 1.8 x 10-5)

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Calculate the pH when 100. mL of 0.050 M NH3 is titrated with 50.0 mL of 0.10 M HCl. Calculate the pH when 100. mL of 0.050 M NH3 is titrated with 60.0 mL of 0.10 M HCl.

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Note the titration curve for a polyprotic acid:

.

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AP CHEMISTRY NOTES 10-4 AQUEOUS EQUILIBRIA: SOLUBILITY PRODUCT

Saturated solutions of “insoluble” salts are another type of chemical equilibria. For a saturated solution of AgCl, the equation would be:

AgCl(s) ↔ Ag+ + Cl-

The solubility product constant (Ksp) would be:

Ksp = [Ag+] [Cl-]

The molar solubility of a substance is the number of moles of a solute that dissolve to produce a liter of saturated solution. (This is “x” in a RICE problem.) The Ksp of AgCl is 1.6 x 10-10. This means that: a. If the product of [Ag+] [Cl-], also known as “Q” at points other than equilibrium, is less than this value, the solution is unsaturated and no precipitate will form. Q < Ksp → unsaturated, no precipitate forms b. If the product of [Ag+] [Cl-] is equal to this value, the solution is exactly saturated, and no precipitate will form. Q = Ksp → saturated, no precipitate forms c. If the product of [Ag+] [Cl-] is greater than this value, the solution is supersaturated, and a precipitate will form. Q > Ksp → supersaturated, precipitate forms EXAMPLE: The molar solubility of silver sulfate is 1.5 x 10-2 mol/L. Calculate the solubility product of the salt.

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EXAMPLE: Calculate the molar solubility of calcium phosphate as well as its solubility in g/L. In addition,

calculate the concentration of each ion in a saturated solution. (The Ksp of calcium phosphate is 1.2 x 10-26.)

EXAMPLE: What is the molar solubility of lead (II) iodide in a 0.50 M solution of sodium iodide? EXAMPLE: Exactly 200 mL of 0.040 M BaCl2 are added to exactly 600 mL of 0.080 M K2SO4. Will a precipitate form?