Common Ion Effect

Chapter 15Applications of Aqueous Equilibria

Copyright2000 by Houghton Mifflin Company. All rights reserved. 1Video clip of buffersCopyright2000 by Houghton Mifflin Company. All rights reserved. 2Common Ion EffectThe shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction.AgCl(s) Ag+(aq) + Cl(aq)

The common ion effect is an application of Le Chateliers principle.

The procedures for finding the pH of a solution containing a weak acid or base plus a common ion are very similar to the problems in Ch. 14.

In this example when a solution of NaCl is added it shifts the direction of the equilibrium to the left. This will reduce the concentration of Cl- ionsAcid Solutions containing Common IonsIn section 14.5 we found that the equilibrium concentrations of H+ in a 1.0 M HF solution is 2.7 X 10-2 M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka= 7.2 x 10-4) and 1.0 M NaF.Copyright2000 by Houghton Mifflin Company. All rights reserved. 3What are the major species involved?Hf, F-, Na+, and water

Na+ is neutral and water is a very weak acid or base. Neither will play a part in the reaction

The important species are HF and F-, which participate in the acid dissociation equilibrium that controls the [H+] in this solution.

Look at the equilibrium equation and Ka expression

Na= has neither acidic or basic properties and water is a very weak acid or base. So the important species are HF and F-Copyright2000 by Houghton Mifflin Company. All rights reserved. 4A Buffered Solution. . . resists change in its pH when either H+ or OH are added.1.0 L of 0.50 M H3CCOOH + 0.50 M H3CCOONapH = 4.74Adding 0.010 mol of solid NaOH raises the pH of the solution to 4.76, a very minor change.The most important practical example of a buffered solution is our blood, which can absorb the acids and bases produces in biologic reactions without changing its pH. A constant pH for blood is vital because cells can survive only in a very narrow pH range.

A buffered solution may contain a weak acid and its salt( for example HF and NaF) or a weak base and its salt( NH3 and NH4Cl.The pH of a Buffered SolutionA buffered solution contains 0.50 M acetic acid HC2H3O2, Ka= 1.8 x 10-5) and 0.50 M of sodium acetate(NaC2H3O2). Calculate the pH of this solution.

What are the major species?HC2H3O2, C2H3O2-, Na+, and H2OCopyright2000 by Houghton Mifflin Company. All rights reserved. 5Na+ is neutral and water is very weak acid or base so they will not play a part.

So the acetic acid dissociation which involves the acetic acid and acetate ion will control the pH of the solutionCopyright2000 by Houghton Mifflin Company. All rights reserved. 6Key Points on Buffered Solutions1.They are weak acids or bases containing a common ion.2.After addition of strong acid or base, deal with stoichiometry first, then equilibrium.pH changes in Buffered SolutionsCalculate the change in pH that occurs when 0.010 mol of solid NaOH is added to 1.0 L of the buffered solution from the previous example. Compare this pH change with that which occurs when 0.010 mol of solid NaOH is added to 1.0 L of water.Copyright2000 by Houghton Mifflin Company. All rights reserved. 7The major species are: HC2O3O2, Na+, C2H3O2-, OH-, and H2OHow does a Buffer WorkOH- + HA A- + H2O

[H+] = Ka [HA]/[A-]Copyright2000 by Houghton Mifflin Company. All rights reserved. 8Suppose a buffered solution contains large amounts of a weak acid(HA) and its conjugate base(A-). When hydroxide ions from a strong base are added, the weak acid (HA) is the best source of protons and the reaction above happens:

The net result is the the OH- ions are not allowed to accumualate but are replaced by A- ions

If we rearrange the equilibrium expressium to solve for [H+] we get the equation above:

In other words, the equilibrium concentration of [H+] and thus the pH is determined by the ratio of [HA]/[A-]

The essence of buffering then is the [HA] and [A-]are large compared the the amount of OH- added. Thus when the OH- is added the concentrations of HA and A- change but only by small amounts.

Another useful equation can be obtained by taking the negative log of both sides and inverting the log term reverses the sign.

This new equation is called the Henderson-Hasselbalch equation which is on the next slideCopyright2000 by Houghton Mifflin Company. All rights reserved. 9Henderson-Hasselbalch EquationUseful for calculating pH when the [A]/[HA] ratios are known.

The pH of a Buffered Solution(weak acid and conj. base)Calculate the pH of a solution containing .75 M lactic acid (Ka= 1.4 x 10-4) and .25 M sodium lactate. Lactic acid is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion.Copyright2000 by Houghton Mifflin Company. All rights reserved. 10The pH of a buffered Solution(weak base and its conj. acid)A buffered solution contains .25 M NH3 (Kb= 1.8 x 10-5) and .40 M NH4Cl. Calculate the pH of this solutionCopyright2000 by Houghton Mifflin Company. All rights reserved. 11Adding Strong Acid to a Buffered SolutionCalculate the pH of the solution that results when .10 mol of gaseous HCl is added to 1.0 L of a buffered solution that contains .25 M NH3 (Kb= 1.8 x 10-5) and .40 M NH4Cl.Copyright2000 by Houghton Mifflin Company. All rights reserved. 12Copyright2000 by Houghton Mifflin Company. All rights reserved. 13Buffered Solution CharacteristicsBuffers contain relatively large amounts of weak acid and corresponding base.Added H+ reacts to completion with the weak base.Added OH reacts to completion with the weak acid.The pH is determined by the ratio of the concentrations of the weak acid and weak base.Buffer Capacitythe amount of H+ or OH- a buffer can absorb without a significant change in the pHa buffer with a large capacity can absorb a large amount of H+ or OH- with only a little change in the pHa buffer with a large capacity contains a large concentration of the weak acid and its saltAdding Strong acid to a Buffered Solution #2Calculate the change in pH that occurs when .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: .050 M HC2H3O2 and .050 M NaC2H3O2 Copyright2000 by Houghton Mifflin Company. All rights reserved. 15For both solutions the initial pH can be determined by the H-H equation.How to Prepare a BufferpH of a buffered solutiondetermined by the ratio of [A-]/[HA]optimal buffering (least change in pH) will occur when [A-] = [HA]so pH = pKa + log (1) so pH = pKa + 0so pH = pKaPick an acid with a Ka closest to the desired pH of the buffered solution.

Another way to determine the best buffer is to determined the ratio of the [HA]/[A-]. The ratio closest to 1 is the best choicePreparing a BufferA chemist needs a solution buffered at pH 4.30 and can choose from the following acids ( and their sodium salts):Chloroacetic acid (Ka= 1.35 x 10-3)Propanoic acid (Ka = 1.3 x 10-5)Benzoic acid (Ka = 6.4 x 10-5)Hypochlorus acid (Ka = 3.5 x 10-8)Copyright2000 by Houghton Mifflin Company. All rights reserved. 17Titration and pH curvesTitrationused to determine the concentration or amount of an unknown acid or baseuses a solution of known concentration (the titrant)uses a buret (for precision)uses an indicator to show the endpoint can be monitored by plotting pH vs. amount of titrant addedCopyright2000 by Houghton Mifflin Company. All rights reserved. 19Titration (pH) Curve con. Equivalence (stoichiometric) point: Enough titrant has been added to react exactly with the solution being analyzed.Strong Acid-Strong Base Titrationsmillimoles may be used because of the small quantities used in a titration1000 millimoles = 1000mmoles = 1 moleMolarity = moles/liter = mmoles/mLmmoles = Volume (in mL) x MolarityStrong Acid-Strong Base Titration Example Titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOHWhat is the pH at various stages of the titration?Initially...No NaOH has been addedpH is determined by H+ from the 0.200 M HNO3pH = - log (0.200) = 0.69950.0 mL x 0.200 M H+ = 10.0 mmol H+ presentCon.Still the beginning of the reaction10.0 mL of 0.100 M NaOH has been addedThe added OH- will neutralize an equivalent amount of H+, I.e., 10.0 ml x 0.100 M or 1.00 mmole.How much H+ is left? 10.0 mmole - 1.00 mmole = 9.0 mmole H+What is the concentration of H+ now? 9.0 mmole/ (50.0 mL + 10.0 mL) = 0.15 MpH = -log (0.15) = 0.82pH is increasing Con.20.0 mL of NaOH has been addedpH = 0.94250.0 mL of NaOH as been addedpH = 1.301 Con.Strong Acid-Strong Base TitrationAt the equivalence point100.0 mL of NaOH has been added100.0 mL x 0.100 M = 10.0 mmole OH-This is enough OH- to completely react with the H+ in solution. At the equivalence point, the pH is 7, the solution is neutralCon.150.0 mL of NaOH has been addednow OH- is in excess, the pH is determined by the excess OH-150.0 mL x 0.100 M = 15.00 mmoles OH-15.00 mmoles OH- - 10.0 mmoles H+ = 5.0 mmoles OH- in solution5.0 mmoles/ (50.0 + 150.0 mL) = 0.025 M OH-.so [H+] = 4.0 x 10-13pH = 12.40Con.200.0 mL of NaOH has been addedstill excess OH-pH = 12.60pH Curve

Copyright2000 by Houghton Mifflin Company. All rights reserved. 27Applications of Aqueous EquilibriaStrong Acid-Strong Base TitrationpH changes very little initially until close to the equivalent point (lots of H+, added OH- doesnt change the pH much)Near the equivalence point, there is less H+, so added OH- changes the pH a lotAt the equivalence point, the pH = 7.00Just after the equivalence point, added OH- also changes the pH a lot

Weak Acid-Strong Base TitrationCalculating the titration curve is like solving a series of buffer problems.Involves a stoichiometry problem where the reaction goes to completion and the concentrations of the weak acid and the conjugate base are calculatedInvolves an equilibrium problem, calculate pH from thisCon.pH curve for this titration is different before the equivalence point from the strong acid-strong base titrationafter the equivalence point, the titration curves are the same

Con.For a weak acid-strong base titration, the pH rises more rapidly in the beginning of the titration, then levels off at the halfway point due to buffering effects.pH at the equivalence point is higher than for a strong acid-strong base titrationWeak acid-Strong Base titrationThe amount of acid determines the equivalence pointThe pH value at the equivalence point depends on the acid strengththe weaker the acid, the higher the pH at the equivalence pointWeak Acid-Strong Base TitrationLets consider the titration of 50.0 ml of 0.10 M acetic acid(HC2H3O2, Ka= 1.8 x 10-5) with 0.10 M NaOH.

We will calculate the pH at various pointsRepresenting volumes of added NaOH.Copyright2000 by Houghton Mifflin Company. All rights reserved. 33pH curve for weak acid-strong base

Copyright2000 by Houghton Mifflin Company. All rights reserved. 34Acid-Base IndicatorsTwo common methods for determining the equivalence point of a titrationUse a pH meter and then plot the titration curve. The center of the vertical region of the pH curve indicates the equivalence point.Use an acid-base indicator to mark the end point with a change in color.Acid-Base Indicators con.The endpoint (when the indicator changes color) may not be the same as the equivalence point.An indicator must be chosen based on the acid and base used in the titration so that the endpoint is as close to the equivalence point as possible.Acid-Base Indicators