Applications of Aqueous Equilibria Or ICE Problems Part III

Common Ion Produced by both ___________________- while in solution Effect of common ion can be predicted by ____________ Will have an impact of pH Polyprotic acids can also experience the common ion effect. What is the common ion? Problems will solve with ICE, but the initial concentration of A- will not be zero b/c of the common ion from the salt

Example: A solution of __________. What is the common ion and how does it impact equilibrium? What happens to the pH of this solution in comparison to a solution of HF only? NaF  Na+ + F- HF  H+ + F- F- is the common ion. The equil. shifts left and the pH would be less acidic. A solution of NH4Cl and NH3

Buffered Solutions Resists a change in ___ when either __________ is added Common ion ______ is a buffered solution that maintains the necessary pH for cells for survive Can be a weak ____ and salt (HF and NaF) OR a weak ____and a salt (NH3 and NH4Cl)

A buffered solution contains 0. 50 M acetic acid and 0 A buffered solution contains 0.50 M acetic acid and 0.50 M sodium acetate. Calculate the pH of this solution.

How Does a Buffer Work? Buffered soln of HA and A- to which OH- is added --- OH- + HA  A- + H2O ***OH- do not accumulate, they are replaced by A- pH is governed by [HA]/[A-] ratio and b/c such large quantities have been used in comparison to the [OH-] – the ratio changes very little [HA] and [A-] >> [OH-] added, thus they change very little

Calculate the change in pH when 0. 010 mol of solid NaOH is added to 1 Calculate the change in pH when 0.010 mol of solid NaOH is added to 1.0 L of the buffered solution from the last example. Compare this to the pH change when 0.010 mol of NaOH is added to 1.0 L of water.

Buffered Soln of HA and A- is which H+ is added - - *** H+ does not accumulate Same as w/OH-  [HA] & [A-] >> [H+] so the ratio changes very little

Henderson-Hasselbalch Equation Useful when [A-]/[HA] ratio is known, but not necessary to solve the problem. pH = pKa + log [A-] = pKa + log [base] [HA] [acid] **Assume that _______ and _______ are the initial concentrations.

Calculate the pH of a solution containing 0. 75 M lactic acid (Ka – 1 Calculate the pH of a solution containing 0.75 M lactic acid (Ka – 1.4 x 10-4) and 0.25 M sodium lactate.

A buffered solution contains 0. 25 M NH3 (Kb = 1. 8 x 10-5) and 0 A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.

Calculate the pH of the solution that results when 0 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 from the previous example.

Few more buffer items For a particular buffering system, all solutions that have the same ratio [A-]/[HA] will have the same pH. All buffered solution have a ________________ present.

Buffer Capacity Represents the amount of H+ or OH- the buffer can absorb w/o a significant change in pH To have a large capacity, the buffer must contain a large concentration of buffering components Determined by the magnitudes of [HA] and [A-]

Last little buffer stuff Large changes in the ratio [A-]/[HA] will produce large changes in pH. Optimal buffering occurs when [HA] = [A-], most resistant to addition of H+ or OH- Best buffering when [A-]/[HA] = 1 or pka = desired pH

Calculate the change in pH that occurs when 0 Calculate the change in pH that occurs when 0.010 mol of gaseous HCl is added to 1.0 L of each of the following solutions. Solution A – 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 Solution B – 0.050 M HC2H3O2 and 0.050 M NaC2H3O2

Is is a buffer. A. 1 M KOH and 0. 1 M CH3NH3Cl B. 1 M KOH and 0 Is is a buffer??? A. 0.1 M KOH and 0.1 M CH3NH3Cl B. 0.1 M KOH and 0.2 M CH3NH2 C. 0.2 M KOH and 0.1 M CH3NH3Cl D. 0.1 M KOH and 0.2 M CH3NH3Cl

Online Lecture #2 -Titrations

Titration and pH Curves _________ – method used to determine the amount of acid or base in solution Progress of the titration can be monitored by plotting the pH of the solution as a function of the titrant added _________________ Titration – H+ + OH-  H2O To determine the pH or [H+] at any point, the amount of [H+] that remains must be divided by the total volume of solution

Strong Acid-Strong Base Titration Millimole – mmol, a thousandth of a mole 1 mmol = 1mol/1000 = 10-3 mol Molarity = mol = mmol solute L mL solution # of mmol = volume(mL) X molarity Before Equivalence Point - [H+] = mmol H+ mL of soln Equivalence Point – [H+] = [OH-] pH = 7.00 After Equivalence Point – [OH-] = mmol OH- Then [H+] from Kw

Example – Get some paper Titration of 50.0 mL of .200 M HNO3 with .100 M NaOH. Calculate the pH @ the following intervals: 0 mL NaOH 100 mL NaOH 20 mL NaOH 200 mL NaOH Make sure I show you the titration curve.

______ Acid-_____ Base Titrations Important – Even weak acids react to completion w/ OH- When using a weak acid, the calculation of [H+] after a strong base has been added requires an ICE problem. The pH of the equivalence point of the weak acid – strong base titration will always be greater than 7. Equivalence point is defined by stoichiometry not by the pH.

____ Acid-______ Base Titrations Halfway to the equivalence point the [HA] = [A-] It is the amount of acid, not its strength, that determines the equivalence point. (mol OH- = mol H+) pH value at the equivalence point is affected by acid strength, the weaker the acid the greater the pH

Weak Acid-Strong Base Titrations ________ Problem Solving 1) ____________ Problem Determine the acid remaining and the conjugate base Before, Change, After (BCA problem) 2) __________ Problem Weak acid equilibrium to determine pH ICE

Weak Acid-Strong Base Titrations Example Problem – Get some paper Titration of 50.0 mL of .10 M HC2H3O2 with .10 M NaOH. Calculate the pH @ the following intervals: 0.0 mL NaOH 50.0 mL NaOH 25.0 mL NaOH 75.0 mL NaOH Ka of HC2H3O2 = 1.8 x 10-5 Make sure I show you the titration curve!

Weak Acid-Strong Base Titrations Example Problem – Get some paper Titration of 50.0 mL of .100 M HCN with .100 M NaOH. Calculate the pH @ a. 8.00 mL of NaOH added b. Halfway point of titration c. Equivalence point of titration Ka of HCN = 6.2 x 10-10

Weak Base w/ Strong Acid Titration Example Problem – More paper Titration of 100.0 mL of .050 M NH3 with .10 M HCl. Calculate the pH at the following intervals: 0.0 mL HCl 50.0 mL HCl 25.0 mL HCl 60.0 mL HCl Make sure I show you the titration curve. Ask me about a polyprotic curve as well!!

Acid – Base _________ Marks the endpoint by changing color Endpoint and equivalence point are not necessarily the same (but will be close) Complex molecules that are weak acids One color w/ H+ attached and another color when H+ is removed Phenolpthalein – HIn – clear (acid) In- - pink (base)

Online Lecture #3 – Solubility Equilibria

Solubility Equilibria Ionic solids when put into water dissociate into cations and anions – BaSO4(s)  Ba+2 + SO4-2 As the ion concentrations increase, they collide and reform the solid – SO4-2 + Ba+2  BaSO4(s) An ________ is eventually reached – BaSO4(s)  Ba+2 + SO4-2 For which an equilibrium expression can be written – Ksp = [Ba+2][SO4-2] (leave out solid) Ksp - solubility product constant

Calculate the Ksp of 1.0 x 10-15 M Bi2S3.

Calculate the solubility of Cu(IO3)2 that has a Ksp of 1.4 x 10-7.

Common Ion Effect and Solubility The solubility of a solid is _______ if the solution already contains ions common to the solid Relative Solubilities If the salts being compared have the ____ number of ions, use the ksp to compare the solubilities. If the salts being compared have _______numbers of ions, calculate the solubilities from ksp and then compare.

Calculate the solubility of solid CaF2 (Ksp = 4. 0 x 10-11) in 0 Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in 0.25 M NaF.

For each of the following pairs of solids, determine which solid has the smaller molar solubility. A. CaF2(s), Ksp = 4.0 x 10-11 or BaF2(s), Ksp = 2.4 x 10-5 B. Ca3(PO4)2(s), Ksp = 1.3 x 10-32 or FePO4(s), Ksp = 1.0x 10-22