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Solubility and Buffers

Equilibrium 3

Solubilityand BuffersEquilibrium 3

The common ion effect-the key to understanding buffers.

Buffers, the Acid Rain Slayer: Crash Course Chemistry #31

The acetic acid/acetate equilibrium

  • Consider a solution of acetic acid.



  • If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.


“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”


Practice

1. (a) Consider the equilibrium B (aq) + H2O (l) ↔ HB+ (aq) + OH- (aq). Using LeChatlier's principle, explain the effect or the presence of a salt of HB+ on the ionization of B. (b) Give an example of a salt that can decrease the ionization of NH3 in solution.


2. Use information from Appendix D to calculate the pH of a solution that is 0.250 M in sodium formate (HCOONa) and 0.100 M in formic acid (HCOOH).

buffers

  • Buffers are solutions of a weak conjugate acid–base pair.

  • They are particularly resistant to pH changes, even when strong acid or base
    is added.

henderson-hasselbach equation

pH Range for Buffers

  • 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.

Example

1. What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 x 10−4.


2. (a) A buffer is prepared by adding 10.0 g of ammonium chloride (NH4Cl) to 250 mL of 1.00 M NH3 solution. (a) What is the pH of this buffer? (b) Write the ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the ionic equation for the reaction that occurs when a few drops of potassium hydroxide is added to the buffer.


3. A buffer consisting of H2PO4- and HPO42-, helps control the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the pH of a soft drink in which the major buffer ingredients are 6.5 g of NaH2PO4 and 8.0 g of Na2HPO4 per 355 mL of solution?

Calculating pH changes in buffers

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Species present in buffer

Using the example pictured above, write down the concentration of the acetic acid AND the acetate ion.

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Adding strong base

The addition of strong base will react with which species in the buffer system? Base will react with acid, so the base and acetic acid will react in a nonequilibrium double replacement reaction-step 1 in the calculation is stoichiometry.

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Different concentrations for the buffer species.

The amount of acetic acid will decrease as given by the stoichiometry step AND the acetate ion concentration will increase as given by the stoichiometry step. Set us an ICE table OR use the Henderson Hasselbach to calculate the new pH. Unless the amount of strong acid or base is very large, the pH should not change much.

Practice

1. 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.



2. Explain why a mixture formed by mixing 100 mL of 0.100 M CH3COOH and 50 mL of 0.100 M NaOH will act as a buffer.

titrations

Acid Base Titrations are the most common

There are other types of titrations besides acid/base (e.g., redox titrations) but titrations are simply a lab technique used to quantitativiely determine the concentration of an unknown solution. In acid-base titrations, the reaction is usually a simple double replacement reaction. Most often, the standard solution (the one you know the concentration of) goes into the buret as the figure above inidicates. The unknown solution (a measured volume) goes into the flask with an indicator. Indicators are chosen (see below) to determine the equivalence point (when stoichiometric amount of acid = that of the base). A simple stoichiometric calculation will yield the unknown concentration.

Indicators

The strength of an acid and base determines the pH of the equivalence point

As the titration curves below indicate, the term neutralization is often misleading in acid-base titrations. Only in a strong acid + strong base titration will the equivalence point = 7. For other types of acid-base titrations, the equivalence point pH is determined by the pH of the salt produced. Because of salt hydrolysis, the pH of a weak acid + strong base is in the basic range due to the anion of the salt hydrolyzing water to produce hydroxide ions. The opposite is true for weak base + strong acid.
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Strong Acid + Strong Base

Note the choice of indicators that change color near the inflection point which corresponds to the equivalence point for the titration.
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Weak Acid + Strong Base

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Weak Base + Strong Acid

Polyprotic acids yield multiple equivalence points corresponding to the number of protons.

Practice

1. Predict whether the equivalence point of each of the following titrations is below, above, or at pH = 7: (a) formic acid titrated with NaOH, (b) calcium hydroxide titrated with perchloric acid , (c) pyridine titrated with nitric acid.


2. Select an appropriate indicator for each titration in #1.


3. A 20.0 mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: (a) 20.0 mL, (b) 23.0 mL, (c) 24.0 mL, (d) 25.0 mL, (e) 30.0 mL.

Will a precipitate form?

Compare Q and K

  • In a solution,

    • If Q = Ksp, the system is at equilibrium and the solution is saturated.

    • If Q < Ksp, more solid can dissolve until Q = Ksp.

    • If Q > Ksp, the salt will precipitate until Q = Ksp.

Using different solubilities to selectively precipiate ions: Qual Schemes

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Saturated solutions are equilibrium systems

No matter how soluble a salt is, if there is solid present at the bottom of a saturated solution, there is a dynamic equilibrium between dissolution (dissolving) and precipitation. The magnitude of the Ksp is an indication of how soluble a salt is.
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The common ion effect affects solubility

If a common ion is introduced to a saturated solution, the equilibrium will shift to the left decreasing the concentration of the dissolved salt.
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Solubility and pH

  • If a substance has a basic anion, it will be more soluble in an acidic solution.

  • Substances with acidic cations are more soluble in basic solutions.

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Complex Ion formation: LeChatlier's principle

Many metal cations (the positive half of these ionic salts we have been discussing) form complex ions with solvent molecules like water, ammonia, and hydroxide ion. The equilibrium constants for the formation of many of these complex ions are quite large as compared to the magnitude of the Ksp's. This "side reaction" effectively removes one of the products from the solubility equilibrium which is an example of LeChatlier's principle in action. Removing a product pushes the equilibrium in the direction of the removal to put that species back. This forces the solubility equilibrium to the right and, therefore, a higher solubility of the metal salt results.

Practice

1. (a) The molar solubility of PbBr2 at 25°C is 1.0 x 10-2 mol/L. Calculate Ksp. (b) If 0.0490 g of AgIO3 dissolves per liter of solution, calculate the solubility product constant. (c) Using the appropriate Ksp value from Appendix D, calculate the pH of a saturated solution of Ca(OH)2. (17.52)


2. Calculate the solubility of LaF3 in grams per liter in (a) pure water, (b) 0.010 M KF solution, (c) 0.050 M LaCl3 solution. (17.56)


3. Calculate the molar solubility of Ni(OH)2 when buffered at pH (a) 8.0, (b) 10.0, (c) 12.0.


4. Using the value of Ksp for Ag2S, Ka1 and Ka2 for H2S, and Kf for AgCl2-, calculate the equilibrium constant for the following reaction:


Ag2S (s) + 4Cl- (aq) + 2H+ (aq) ↔ 2AgCl2- (aq) and H2S (aq)


5. A 1.0 M Na2SO4 solution is slowly added to 10.0 mL of a solution that is 0.20 M in Ca2+ and 0.30 M in Ag+. (a) Which compound will precipitate first: CaSO4 (Ksp = 2.4 x 10-5) or Ag2SO4 (Ksp = 1.5 x 10-5)? (b) How much Na2SO4 solution must be added to initiate the precipitation?
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