1. Applications of Aqueous Equilibria
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2. Solutions of Acids or Bases Containing a Common Ion
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3. Solution containing weak acid HA and its salt NaA
Salt dissolves in water and breaks up completely into its ions-it is a strong electrolyte
NaA(s)  Na+(aq) + A-(aq)
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4. Common Ion Effect
When AgNO3 is added to a saturated solution of AgCl, it is often described as a source of a common ion, the Ag+ ion.
By definition, a common ion is an ion that enters the solution from two different sources.
Solutions to which both NaCl and AgCl have been added also contain a common ion; in this case, the Cl- ion.
There is an effect of common ions on solubility product equilibria.
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5. The common-ion effect can be understood by considering the following question: What happens to the solubility of AgCl when we dissolve this salt in a solution that is already 0.10 M NaCl?
As a rule, we can assume that salts dissociate into their ions when they dissolve.
A 0.10 M NaCl solution therefore contains 0.10 moles of the Cl- ion per liter of solution.
Because the Cl- ion is one of the products of the solubility equilibrium, LeChatelier's principle leads us to expect that AgCl will be even less soluble in an 0.10 M Cl- solution than it is in pure water.
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6. Calculate the solubility of AgCl in 0.10 M NaCl.
In pure water:Cs = 1.3 x 10-5 M
In 0.10 M NaCl: Cs = 1.8 x 10-9 M
These calculations show how the common-ion effect can be used to make an "insoluble" salt even less soluble in water.
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7. HA(aq) + H2O(l) !H3O+(aq) + A-(aq) 2 H2O(l)! H3O+(aq) + OH-(aq)
The common ion effect can be applied to other equilibria, as well. Consider what happens when we add a generic acid (HA) to water. We now have two sources of a common ion ? the H3O+ ion.
HA(aq) + H2O(l) !H3O+(aq) + A-(aq)
2 H2O(l)! H3O+(aq) + OH-(aq)
Thus, it isn't surprising that adding an acid to water decreases the concentration of the OH- ion in much the same way that adding another source of the Ag+ ion to a saturated solution of AgCl decreases the concentration of the Cl- ion.
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8. The solubility of a solid is lowered if the solution already contains ions common to the solid
Dissolving silver chloride in a solution containing silver ions
Dissolving silver chloride in a solution containing chloride ions
The common-ion effect can also be used to prevent a salt from precipitating from solution.
Instead of adding a source of a common ion, we add a reagent that removes the common ion from solution.
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9. Ksp (Solubility Product Constant, Solubility Product)
Ksp = [Ca2+][F-]2
Experimentally determined solubility of an ionic solid can by used to calculate its Ksp value
The solubility of an ionic solid can be calculated if its Ksp value is known
Relative Solubilities
IF the salts being compared produce the same number of ions in solution, Ksp can be used to directly compare solubility
NaCl(s), KF(s)
Ksp = [cation][anion] = x2
IF the salts being compared produce different numbers of ions, Ksp cannot be directly compared
Ag2S(s)
Ksp = [2x]2[x]
Bi2S3(s) = [2x]2[3x]3
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10. Common Ion-ion produced by both acid and its salt
Ion provided in solution by an aqueous acid (or base) as well as a salt
HF(aq) and NaF (F- in common)
HF(aq)  H+(aq) + F-(aq)
NaF(s) (in water) Na+(aq) + F-(aq)
Excess F- added by NaF
Equilibrium shifts away from added component.
Fewer H+ ions present, making solution less acidic
pH is higher than expected.
NH4OH and NH4Cl (NH4+ in common)
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
NH4Cl(s) (in water) NH4+(aq) + Cl-(aq)
Equilibrium shifts to the left.
pH of solution decreases due to decrease in OH- conc.
Common ion effect-shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction
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11. 3/25/2017

12. Equilibrium Calculations
Consider initial concentration of ion from salt when calculating values for H+ and OH-
Calculate the pH of a solution that contains 0.10 M HC2H3O2 and M NaC2H3O2. (The pH of 0.10 M HC2H3O2 is The addition of the common ion would shift the equilibrium to the left. As a result the [H3O+] would decrease and the pH would rise.)
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13. Ka = 1.8 x 10-5 HC2H3O2(aq) + H2O(l) ⇋ H3O+(aq) + C2H3O2-(aq)
Substance
[HC2H3O2]
[H3O+]
[C2H3O2-]
Initial conc.
0.10
0.00
0.050
Change in conc. -x +x
Equilibrium conc.
0.10 – x x x
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14. Ka = [H3O+][C2H3O2-]/[HC2H3O2]
1.8 x 10-5 = x( x)/0.10 – x ≈0.050x/0.10
x = [H3O+] = 3.6 x 10-5 M
pH = -log[H3O+] = -log(3.6 x 10-5) = 4.4
as we predicted, the pH rose from 2.9 to 4.4.
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15. Buffered Solutions
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16. A buffer is a solution with a very stable pH
Until now, we have considered solutions containing only pure weak acids, weak bases, or their salts dissolved in distilled water
A buffer is a solution with a very stable pH
You can add acid or base to a buffer solution without greatly affecting the pH of the solution
The pH of a buffer will also remain unchanged if the solution is diluted with water or if water is lost through evaporation
A mixture that contains a conjugate acid-base pair is known as buffer solution because the pH changes by a relatively small amount if a strong acid or base is added to it
Buffer systems are used to control pH in many biological and chemical reactions
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17. A weak base and its conjugate acid (A+) will do the same
A buffer is created by placing a large amount of a weak acid or base into a solution along with its conjugate
A weak acid and its conjugate base (B-) will remain in solution together without neutralizing each other
A weak base and its conjugate acid (A+) will do the same
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18. When both the acid and the conjugate base are together in the solution, any hydrogen ions that are added will be neutralized by the base while any hydroxide ions that are added will be neutralized by the acid without having much of an effect on the solution’s pH
A buffer system is a chemical “sponge” for H3O+ or OH- ions that may be produced within a solution
When additional H3O+ ions are added, they react with the conjugate base component of the buffer system to produce H2O and a weak acid that is only slightly ionized, so the pH drops only minimally
H3O+(aq) + conjugate base-(aq)  weak acid(aq) + H2O(l)
When additional OH- ions are added, the basic OH- ions react with the acid component of the buffer system to produce H2O and a weak base, so the pH rises only minimally
OH-(aq) + conjugate acid (aq)  weak base-(aq) + H2O(l)
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19. These solutions are governed by the same equilibrium law as are weak acids and weak bases
Buffered solutions-Equilibrium systems that resist changes in acidity and maintain constant pH when acids or bases are added to them
Most effective pH range for any buffer is at or near the pH where the acid and salt concentrations are equal (pKa)
[H3O+] of weak acid/conjugate base buffer equals the Ka of the weak acid
[OH-] of weak base/conjugate acid buffer equals the Kb of the weak base
In general the most effect pH range of a buffer system is (optimum pH  1.0 pH unit
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20. Convert the buffer relationship into its logarithmic form
Since Ka for a weak acid is small, equilibrium concentrations of the weak acid and its conjugate base are very nearly their initial concentrations
All weak acid-conjugate base buffer systems have general expression
[H3O+] = Ka x [weak acid]0/[conjugate base]0
Convert the buffer relationship into its logarithmic form
pH for buffer-pH = pKa + log[A-]/[HA] = pKa + log [conjugatebase]0/[weak acid]0-obtained from equation for weak acid equilibrium (Ka = [H+][A-]/[HA])
this relationship is the Henderson-Hasselbalch equation
[OH-] = Kb x [weak base]0/[conjugate acid]0
pOH = pKb + log x [conjugate acid]0/[weak base]0
Buffered solutions contain either:
A weak acid and its salt
A weak base and its salt
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21. Preparing a buffer
A buffer can be prepared by dissolving mole of HF and mole of NaF in water to make 1.00 liter of solution
Dissociation reaction of HF is
HF(aq) ⇋ F-(aq) + H+(aq)
Equilibrium law is
Ka = [F-][H+]/[HF] = 6.8 x 10-4
Set up an equilibrium table using the given information
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22. Reaction HF ⇋ F- + H+ Initial conc. 0.200 0.100 0.00 Change -x +x
Equilibrium
0.200 – x x x
Solution
0.199
0.101
1.36 x 10-3
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23. Substituting the information from the equilibrium line into the equilibrium law yields
Ka = ( x)(x)/0.200 – x = 6.8 x 10-4
Assuming that x is small compared to both and 0.200, we simplify the equation to
Ka = (0.100)(x)/0.200 = 6.8 x 10-4
x = 1.36 x 10-3
This value of x satisfies the assumptions made, and the last line of the equilibrium table can be completed
The last column gives us [H+] = 1.36 x 10-3 M
The pH of the buffer solution is calculated as 2.87
For almost all buffer solutions the assumptions hold true, and this type of problem is solved by simply entering the given concentrations of the conjugate acid and conjugate base directly into the equilibrium law
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24. Calculate the pH of the following buffer solutions
0.250 M acetic acid and M sodium acetate
Ka = [C2H3O2-][H+]/[HC2H3O2]
1.8 x 10-3 = (0.150)[H+]/0.250
[H+] = 3.0 x 10-3, and pH = 2.52
A solution of 10.0 g each of formic acid and potassium formate dissolved in 1.00 L of H2O
Ka = [CHO2-][H+]/[HCHO2]
1.8 x 10-4 = (0.417)[H+]/0.217
[H+] = 9.4 x 10-5, and pH = 4.03
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25. 0.0345 M ethylamine and 0.0965 M ethyl ammonium chloride
Kb = [C2H5NH3+][OH-]/[C2H5NH2]
4.3 x 10-4 = (0.0965)[OH-]/0.0345
[OH-] = 1.5 x 10-4, pOH = 3.82 and pH = 10.18
0.125 M hydrazine and M hydrazine hydrochloride
Kb = [N2H5+][OH-]/[N2H4]
9.6 x 10-7 = (0.321)[OH-]/0.125
[OH-] = 3.7 x 10-7, pOH = 6.43 and pH = 7.57
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26. Shortcuts in buffer calculations
Equilibrium law used for buffers involves a ratio of the concentrations of the conjugate acid and conjugate base
We may use the moles of conjugate acids and moles of conjugate base instead of the acid and base molarities
If the concentration of the conjugate acid and conjugate base are equal, their ratio is exactly 1.00
With equal molarities or equal numbers of moles of conjugate acid and conjugate base
Ka = [H+] and pKa = pH for a buffer made from a weak acid and its conjugate base
Kb = [OH-] and pKb = pOH for a buffer made from a weak base and its conjugate acid
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27. pH changes in buffers
Addition of a strong acid to a buffer will decrease the pH slightly
Addition of a strong base to a buffer will increase the pH slightly
To determine the amount that the pH changes when a strong acid or base is added to a buffer, we must calculate the change in concentration of the conjugate acid or base in the buffer
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28. Steps
Determine either the molarity or the number of moles of the conjugate acid and conjugate base in the original buffer
Determine the amount of strong acid or base added in the same units as the conjugate acid and base in step 1
If a strong acid is added to the buffer, add the value from step 2 to the conjugate acid and subtract it from the conjugate base
If a strong base is added to the buffer, add the value from step 2 to the conjugate base and subtract it fro the conjugate acid
Substitute the new values for the conjugate acid and conjugate base into the equilibrium law and calculate the pH as shown before
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29. In an acetate buffer, acetic acid, HC2H3O2, is the conjugate acid, and the acetate ion, C2H3O2-, is the conjugate base
Addition of a strong acid increases the concentration of acetic acid and decreases the concentration of acetate ions.
Addition of a strong base to this buffer increases the acetate ion concentration and decreases the acetic acid concentration
HC2H3O2 + OH-  C2H3O2-
An acetate buffer is prepared with M acetic acid and M NaC2H3O2. If mol of solid NaOH is added to 100 mL of this buffer, calculate the change in pH of the buffer due to the addition of the NaOH.
To calculate the change in pH, the pH of the original buffer is needed, along with the final pH after the base is added
Dissociation reaction for acetic acid
HC2H3O2 ⇋ H+ + C2H5O2-
Equilibrium law
Ka = [H+][ C2H5O2-]/[ HC2H3O2]
Value for Ka and the acetate and acetic acid concentrations are substituted into the equilibrium law
1.8 x 10-5 = [H+])(0.100)/0.250
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30. Hydrogen ion concentration is calculated as
[H+] = 4.5 x 10-5 M, and the pH is 4.35
To calculate the pH of the buffer after the NaOH is added, we must first convert either the molarities of the conjugate acids and bases to moles or the moles of NaOH to molarity so that all the units are the same
Molarity NaOH = mol NaOH/0.100 L = M
Adding this value to the acetate concentration gives us M C2H3O2-
Subtracting it from the acetic acid concentration yields M HC2H3O2
Substituting these new values into the equilibrium law
1.8 x 10-5 = [H+](0.120)/0.230
[H+] = 3.45 x 10-5 M
pH = 4.46
There is an increase of 0.11 pH units when the NaOH is added
The answer is reasonable since the pH is expected to rise slightly when a base is added
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31. Calculations Involving Buffered Solutions Containing Weak Acids
"Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH calculations on buffered solutions require exactly the same procedures introduced in Chapter 14. This is not a new type of problem."
“When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations. This procedure can be presented as follows:
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32. 3/25/2017

33. weak acid conjugate base
OH- + HA  A H2O
weak acid conjugate base
OH- ions are not allowed to accumulate but are replaced by A- ions
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34. Ka = [H+][A-] / [HA] [H+] = Ka x [HA] / [A-]
If the amounts of HA and A- originally present are very large compared with the amount of OH- added, the change in [HA]/[A-] will be small. Therefore the pH change will be small
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35. Buffering also works for addition of protons instead of hydroxide ions
H+ + A-  HA
Conjugate base Weak acid
Buffering with a Weak Base and Its Conjugate Acid
Weak base B reacts with any H+ added
B + H+  BH+
Base Conjugate acid
Conjugate acid BH+ reacts with any added OH
BH OH-  B + H2O
Conjugate acid Base
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36. Summary
Buffered solutions contain relatively large concentrations of a weak acid and the corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base and the conjugate acid BH+
When H+ is added to a buffered solution, it reacts essentially to completion with the weak base present
When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present
The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and the weak base. The pH remains relatively unchanged as long as the concentrations of buffering materials are large compared with the amounts of H+ or OH- added
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37. Buffer Capacity
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38. Buffering Capacity
The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH.
The pH of a buffered solution is determined by the ratio [A-]/[HA]
The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-]
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39. Preparing a Buffer
Preparation of a buffer starts with the selection of the desired pH
Then a table of Ka and Kb values for weak acids and bases is consulted to find an appropriate conjugate acid-base pair to use
These 2 steps determine the ratio of the salt to the weak acid or base
Next, the moles of acid or base that need to be buffered are estimated
The total number of moles of the conjugate acid-base pair should be at least 20 times the amount of the acid or base that needs to be buffered
The volume of buffer solution needed is determined next
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40. Method for preparing the buffer is decided on
Conjugate acid and conjugate base are measured and dissolved to the desired volume
Measuring the desired amount of conjugate acid and then adding the appropriate amount of a strong base to convert some of the conjugate acid into its conjugate base
Conjugate base is measured and the necessary amount of strong acid is added to convert some of the conjugate base into its conjugate acid
Optimal buffering occurs when [HA] is equal to [A-] ( [A-]/[HA] = 1 )
The pKa of the weak acid to be used should be as close as possible to the desired pH
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41. Titration
Titration technique used for chemical analysis utilizing reactions of 2 solutions
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42. One reactant solution is placed in a beaker and the other in a buret (long, graduated tube with stopcock)
The stopcock (valve that allows chemist to add controlled amounts of solution from buret to beaker)
An indicator is added to solution in beaker (substance which undergoes a color change in the pH interval of the equivalence point)
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43. Classic chemical reaction-Fe2+ and permanganate ion MnO4-
Chemist reads volume of solution in buret at start of experiment and again at point where indicator changes color
Difference in these volumes represents the volume of reactant delivered from the buret
Controlled addition of a solution of known concentration (titrant) in order to determine concentration of solution of unknown concentration
The crucial point about the titration experiment is that the indicator is designed to change color when the amount of reactant delivered from the buret is exactly the amount needed to react with the solution in the beaker (Equivalence Point (Stoichiometric Point)-point in a titration at which the reaction between titrant and unknown has just been completed.)
Classic chemical reaction-Fe2+ and permanganate ion MnO4-
5Fe2+ + MnO4- + 8H+  Mn2+ + 5Fe3+ + 4H2O
The purple permanganate ion is the indicator of the point where the correct amount has been added to completely react all of the Fe2+ ions in the sample
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44. 3/25/2017

45. Titration Curve (pH Curve)-plotting of the pH of the solution as a function of the volume of titrant added
Monitors the pH of the solution as an acid-base titration proceeds from beginning to well beyond the equivalence point
pH is plotted on the y-axis and volume of base (or acid) delivered is plotted on the x-axis
Shape of titration curve makes it possible to identify the equivalence point and is an aid in selecting an appropriate indicator
It can also be used to calculate the Ka or Kb of a weak acid or base
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46. There are four major points of interest during a titration experiment
Start of the titration, where the solution contains only one acid or base
pH is calculated as previously described
Region where titrant is added up to the end point, and the solution now contains a mixture of unreacted sample and products
Mixture of a conjugate acid and its conjugate base
If weak acid or base is involved, this is a buffer solution
If only strong acids and bases are used, this region is unbuffered
End point, where all the reactant has been converted into product
Solution is the salt of the acid or base
titration curve used to determined pH during titration
Region after end point, where solution contains product and excess titrant
pH depends on excess titrant used
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47. Points on titration curve
Midpoint of most vertical part of graph corresponds to exact endpoint.
This will also correspond to the equivalence point, or the point at which the equivalents of acid equals the equivalents of base.
In addition, the midpoint will also determine the pH of the salt that was formed during the titration.
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48. Half-equivalence point is at the center of the buffer region
pH increases more quickly at first, then levels out into buffer region
At the half-equivalence point, enough base has been added to convert exactly half of the acid into conjugate base
The concentration of the acid is equal to the concentration of the conjugate base
The curve remains fairly flat until just before the equivalence point, when the pH increases sharply
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49. 3/25/2017

50. 3/25/2017

51. Strong Acid-Strong Base Titration-titration curve can be divided into three segments
Prior to the equivalence point (0-x-amount at equivalence point)
The pH rises slowly at first because the solution contains a sizeable excess of H3O+ ions (buffering effect created by relatively high concentration of H+ in solution)
[H+] (thus the pH) can be calculated
We begin to see significant rise in pH close to the equivalence point
At the equivalence point (x)
All excess H3O+ ions have been removed
[H3O+] is now x 10-7 M, which corresponds to pH of 7.000
Beyond the equivalence point (x – total amount used)
Once the endpoint has been passed, the rate of pH change diminishes again-resembles first part of graph except at higher pH
The solution now contains an excess of OH- ions and as pH rises, it approaches that of the titrant
[OH-] can be calculated from excess OH- and volume of solution
[H+] and pH can be calculated from [OH-] and Kw
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52. 3/25/2017

53. Strong Base with Strong Acid
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54. Weak Acids-Strong Bases Titration-segments of titration curve
Before titrant added
pH of solution from given
Prior to the equivalence point (0-x)
As NaOH is added below the equivalence point
HC2H3O2(aq) + OH-(aq)  C2H3O2-(aq) + H2O(l)
Solution is now a buffer system that contains both the weak acid (HC2H3O2) and its conjugate base (C2H3O2-)
At the point of half-neutralization, [HC2H3O2] = [C2H3O2-] and [H3O+] = Ka
At the equivalence point (x)
All of the HC2H3O2 has been exhausted and the total volume is now double the original volume of the base
The equivalence point corresponds to the pH-always > 7
The stronger the basic anion, the higher the pH of the equivalence point
Beyond the equivalence point (x-total)
Solution now contains excess of OH- ions and as the pH rises, it approaches that of the titrant
If the acid being titrated is weak, then the graph will not be nearly as vertical at the endpoint.
Weaker acid is, the more the graph deviates from being vertical.
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55. 3/25/2017

56. Weak Bases-Strong Acids Titration
Before HCl has been added-pH of solution calculated from given
Prior to the equivalence point (0-x)
As HCl is added below the equivalence point
NH3(aq) + H3O+(aq)  NH4+(aq) + H2O(l)
The solution is now a buffer system that contains both the weak base (NH3) and its conjugate acid (NH4+)
At the point of half-neutralization, [NH3] = [NH4+] and [OH-] = Kb
At the equivalence point (x)
All of the NH3 has been exhausted and the total volume is now double the original volume of the acid
The equivalence point corresponds to pH (always < 7.00)
The conjugate acid of the weak base lowers the pH
Beyond the equivalence point (x- total)
The solution now contains an excess of H3O+ ions and as the pH falls, it approaches that of the titrant
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57. 3/25/2017

58. Polyprotic Acid-Strong Base Titration
When titrating a polyprotic acid, the graph will show a steep rise, or endpoint, for each of the protons in the acid.
An acid with 2 protons will have two endponts, one for each hydrogen.
An acid with 3 protons will have three endpoints, one for each proton.
With each successive ionization, the characteristic shape of the titration curve becomes less distinct
First endpoint is fairly obvious, the second endpoint is not as well defined, the third endpoint is worse still, and so on.
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59. 3/25/2017

60. When a monoprotic acid is completely neutralized by a monohydroxide base, equal numbers of moles of H3O+ and OH- ions react
H3O+(aq) + OH-(aq)  2H2O(l)
At neutralization, nH30+ = nOH-
M = n/V
N = MV
(MV)acid = (MV)base
When 25.0 mL of HCl is neutralized by M NaOH, 35.0 mL of the base are consumed. Calculate the molarity of the acid.
MVacid = MVbase
M(25.0 mL) = (0.100 M)(35.0 M)
Macid = M
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61. If the acid contains more than one ionizable proton, and/or the base contains more than one dissociable hydroxide ion, then the equation given above must be modified in order to include these numbers
(M x V x #H3O+)acid = (M x V x #OH-)base where “#” indicates the number of acidic hydrogens or dissociable hydroxide ions in chemical formula
What volume of M H2SO4 is needed to be completely neutralized by 300. mL of M KOH?
MV#acid = MV#base
(0.200 M)V(2) = (0.450)(300 mL)(1)
Vacid = 338 mL
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62. Acid-Base Indicators
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63. Determination of Equivalence Point
Use a pH meter, find midpoint of vertical line in the titration curve
Use of indicators
Indicators used in titrations must have two important properties
The pH at the end point of the titration curve must change by at least 2 pH units very rapidly
The pK of the indicator must be close to the end-point pH of the titration
Titration curve shows us required large change in pH often occurs
Proper selection of an indicator mandates that the pH at the end point and the pKa of the indicator be close to each other
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64. How Indicators Work
Indicators are weak acids and weak bases whose respective conjugate bases and conjugate acids have different colors
The reason is that the loss or gain of a proton changes the energy of the electrons within their structures
This, in turn, changes the energy of light absorbed, which is observed as a change in color
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65. They change colors within a specified pH range
If conjugate acid has 1 color, its conjugate base has another
The eye will see these colors clearly only if there is 10x more of one color than of the other
If acid is yellow/base is blue, we see yellow if conjugate acid is 10x more concentrated than conjugate base, and blue if conjugate base is 10x more conc. than conjugate acid
[conjugate base]/[conjugate acid] < 0.1 (yellow observed)
[conjugate base]/[conjugate acid] > 10 (blue observed
Between these 2 ratios various shades of green observed
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66. Since they are weak acids/bases, they undergo acid-base equilibrium
HIn(aq) + H2O(l) ⇋ H3O+ + In-(aq)
KIn = [H3O+][In-]/[In]
The species HIn is known as the acid form of the indicator and In- is known as the basic form of the indicator
Each form has its own distinct color
We can rearrange the equilibrium constant expression to yield
[H3O+] = KIn x [HIn]/[In-]
When [HIn] > [In-], indicator has color associated with acid form
When [In-] > [HIn], indicator has color associated with basic form
When [HIn] = [In-], color of indicator changes
At this point, [H3O+] = KIn
Ideally, this condition should represent end point of titration
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67. Since indicators are weak acids/bases, significance of this color phenomenon best shown w/Henderson-Hasselbalch equation
pH = pKa + log([A-]/[HA]) = pKa + log([base]/[acid])
[HA] = molar conc. of undissociated weak acid (M)
[A-] = molar conc. of conjugate base (M)
pOH = pKb + log x [HB+]/[B]
[B] = molar conc. of weak base
[HB+] = molar conc. of conjugate acid
Substituting 0.1 into the log term yields
pH = pKa + log(0.1) = pKa – 1.0
pH = pKa + log(10) = pKa + 1.0
For a particular buffering system (acid-conjugate base pair), all solutions that have same ratio [A-]/[HA] will have same pH
In order to select a suitable indicator for a titration, it is best if the pKIn lies within one pH unit of the equivalence point
pKIn = pHequivalence point  1
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68. If both concentrations are 0.20M
We have a buffer solution with concentrations of 0.20 M HC2H3O2 and 0.50 M C2H3O2-. The acid dissociation constant for HC2H3O2 is 1.8 x Find the pH of the solution.
pH = pKa + log x [C2H3O2-]/[ HC2H3O2]
pH = -log(1.8 x 10-8) + log x (0.50 M)/(0.20)
pH = -log(1.8 x 10-8) + log(2.5)
pH = (4.7) + (0.40) = 5.1
If both concentrations are 0.20M
pH = -log(1.8 x 10-8) + log x (0.20 M)/(0.20)
pH = -log(1.8 x 10-8) + log(1)
pH = (4.7) + (0) = 4.7
Notice that when the concentrations of acid and conjugative base in solution are the same, pH = pKa (and pOH = pKb)
When you choose an acid for a buffer solution, it is best to pick an acid with a pKa that is close to the desired pH
That way you can have almost equal amounts of acid and conjugate base in the solution, which will make the buffer as flexible as possible in neutralizing both added H+ and OH-.
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69. Indicators
Indicator color changes will be sharp, occurring with the addition of a single drop of titrant
Suitable indicators must be selected based on the equivalence point
Strong acid-strong base titrations may use indicators with end points as far apart as pH 5 and pH 9
Titration of weak acids or weak bases requires more careful selection of an indicator with appropriate transition interval
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70. 3/25/2017

71. Solubility Equilibria and the Solubility Product
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72. Why Do Some Solids Dissolve in Water?
Sugar is a molecular solid, in which the individual molecules are held together by relatively weak intermolecular forces.
When sugar dissolves in water, the weak bonds between the individual sucrose molecules are broken, and these C12H22O11 molecules are released into solution.
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73. It takes energy to break bonds between C12H22O11 molecules
It also takes energy to break H bonds in water that must be disrupted to insert one of these sucrose molecules into solution
Sugar dissolves in water because energy is given off when slightly polar sucrose molecules form intermolecular bonds with the polar water molecules.
The weak bonds that form between the solute and the solvent compensate for the energy needed to disrupt the structure of both the pure solute and the solvent.
In the case of sugar and water, this process works so well that up to 1800 grams of sucrose can dissolve in a liter of water.
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74. NaCl(s)  Na+(aq) + Cl-(aq)
Ionic solids (or salts) contain (+)and (-) ions, which are held together by the strong force of attraction between particles w/opposite charges.
When one dissolves in water, ions that form solid are released into solution, where they become associated w/polar solvent molecules.
H2O
NaCl(s)  Na+(aq) + Cl-(aq)
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75. We can generally assume that salts dissociate into their ions when they dissolve in water.
Ionic compounds dissolve in water if the energy given off when the ions interact with water molecules compensates for the energy needed to break the ionic bonds in the solid and the energy required to separate the water molecules so that the ions can be inserted into solution.
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76. Solubility Equilibria-based on following assumption:
When solids dissolve in water, they dissociate to give the elementary particles from which they are formed.
Molecular solids dissociate to give individual molecules
H2O
C12H22O11(s)  C12H22O11(aq)
Ionic solids dissociate into solutions of (+)/(-) ions they contain
NaCl(s)  Na+(aq)+ Cl-(aq)
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77. When the salt is first added, it dissolves and dissociates rapidly.
The conductivity of solution therefore increases rapidly at first.
Dissolve
NaCl(s)  Na+(aq) + Cl-(aq)
Dissociate
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78. Na+(aq)+Cl-(aq)  NaCl(s) Precipitate
Concentrations of these ions soon become large enough that the reverse reaction starts to compete with forward reaction, which leads to a decrease in rate at which Na+ and Cl- ions enter solution.
Associate
Na+(aq)+Cl-(aq)  NaCl(s)
Precipitate
Eventually, the Na+ and Cl- ion concentrations become large enough that the rate at which precipitation occurs exactly balances the rate at which NaCl dissolves.
Once that happens, there is no change in the concentration of these ions with time and the reaction is at equilibrium.
AT equilibrium, solution is saturated, because it contains the maximum conc. of ions that can exist in equilibrium w/solid salt.
The amount of salt that must be added to a given volume of solvent to form a saturated solution is called solubility of salt.
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79. Solubility Rules-patterns obtained from measuring solubility of different salts and are based on the following definitions of the terms
A salt is soluble if it dissolves in water to give a solution with a concentration of at least 0.1 moles per liter at room temperature.
A salt is insoluble if the concentration of an aqueous solution is less than M at room temperature.
Slightly soluble salts give solutions that fall between these extremes.
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80. The Solubility Product Expression
Silver chloride is so insoluble in water ( g/L) that a saturated solution contains only about 1.3 x 10-5 moles of AgCl per liter of water.
H2O   
AgCl(s)  Ag+(aq) +Cl-(aq)
Equilibrium constant expressions for this reaction gives following result.
Water isn't included because it is neither consumed nor produced in this reaction, even though it is a vital component of the system.
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81. [AgCl] is more ambiguous.
[Ag+] and [Cl-] terms represent concentrations of the Ag+ and Cl- ions in moles per liter when this solution is at equilibrium.
[AgCl] is more ambiguous.
It doesn't represent conc. of AgCl dissolved in water- assume that AgCl dissociates into Ag+/Cl- ions when it dissolves in water
It can't represent amount of solid AgCl in system-equilibrium is not affected by the amount of excess solid added to the system.
The [AgCl] term has to be translated quite literally as the number of moles of AgCl in a liter of solid AgCl.
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82. Concentration of solid AgCl calculated from density & molar mass of AgCl.
This quantity is a constant, however.
The number of moles per liter in solid AgCl is the same at the start of the reaction as it is when the reaction reaches equilibrium.
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83. Since the [AgCl] term is a constant, which has no effect on the equilibrium, it is built into the equilibrium constant for the reaction.
[Ag+][Cl-] = Kc x [AgCl]
This equation suggests that product of the equilibrium concentrations of Ag+ and Cl- ions in this solution is equal to a constant.
Since this constant is proportional to solubility of the salt, it is called the solubility product equilibrium constant for reaction, or Ksp.
Ksp = [Ag+][Cl-]
Ksp expression for a salt is product of the concentrations of the ions, with each concentration raised to power equal to coefficient of that ion in the balanced equation for the solubility equilibrium.
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84. The Relationship Between Ksp And the Solubility of a Salt
Ksp is called the solubility product because it is literally the product of the solubilities of the ions in moles per liter.
Solubility product of salt can be calculated from its solubility, or vs.
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85. We start with the balanced equation for the equilibrium. H2O
Photographic films are based on the sensitivity of AgBr to light. When light hits a crystal of AgBr, a small fraction of the Ag+ ions are reduced to silver metal. The rest of the Ag+ ions in these crystals are reduced to silver metal when the film is developed. AgBr crystals that do not absorb light are then removed from the film to "fix" the image.
Example: Let's calculate the solubility of AgBr in water in grams per liter, to see whether AgBr can be removed by simply washing the film.
We start with the balanced equation for the equilibrium.
H2O   
AgBr(s)  Ag+(aq) + Br-(aq)
We then write the solubility product expression for this reaction.
Ksp = [Ag+][Br-] = 5.0 x 10-13
One equation can't be solved for two unknowns-Ag+/ Br- ion conc.
We can generate a second equation, however, by noting that one Ag+ ion is released for every Br- ion.
Because there is no other source of either ion in this solution, concentrations of these ions at equilibrium must be the same.
[Ag+] = [Br-]
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86. Substituting this equation into Ksp expression gives following result.
[Ag+]2 = 5.0 x 10-13
Taking the square root of both sides of this equation gives the equilibrium concentrations of the Ag+ and Br- ions.
[Ag+] = [Br-] = 7.1 x 10-7M
Once we know how many moles of AgBr dissolve in a liter of water, we can calculate the solubility in grams per liter.
The solubility of AgBr in water is only gram per liter. It therefore isn't practical to try to wash the unexposed AgBr off photographic film with water.
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87. Solubility product calculations with 1:1 salts such as AgBr are relatively easy to perform.
In order to extend such calculations to compounds with more complex formulas we need to understand the relationship between the solubility of a salt and concentrations of its ions at equilibrium.
Symbol Cs describes the amount of a salt that dissolves in water.
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88. The Role of the Ion Product (Qsp) In Solubility Calculations
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89. Consider a saturated solution of AgCl in water.
  H2O   
AgCl(s)  Ag+(aq)+Cl-(aq)
Because AgCl is a 1:1 salt, the concentrations of the Ag+ and Cl- ions in this solution are equal.
Saturated solution of AgCl in water:
[Ag+] = [Cl-]
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90. AgNO3(s) Ag+(aq)+NO3-(aq) H2O AgCl(s)  Ag+(aq) + Cl-(aq)
Imagine what happens when a few crystals of solid AgNO3 are added to this saturated solution of AgCl in water.
According to the solubility rules, silver nitrate is a soluble salt.
It therefore dissolves and dissociates into Ag+ and NO3- ions.
As a result, there are two sources of the Ag+ ion in this solution.
AgNO3(s) Ag+(aq)+NO3-(aq)     
  H2O    
AgCl(s)  Ag+(aq) + Cl-(aq)
Adding AgNO3 to a saturated AgCl solution therefore increases the Ag+ ion concentrations.
When this happens, the solution is no longer at equilibrium because product of the concentrations of the Ag+ and Cl- ions is too large.
In more formal terms, we can argue that the ion product (Qsp) for the solution is larger than the solubility product (Ksp) for AgCl.
Qsp = (Ag+)(Cl-) > Ksp
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91. The ion product is literally the product of the concentrations of the ions at any moment in time.
When it is equal to the solubility product for the salt, the system is at equilibrium.
The reaction eventually comes back to equilibrium after the excess ions precipitate from solution as solid AgCl.
When equilibrium is reestablished, however, the concentrations of the Ag+ and Cl- ions won't be the same.
Because there are two sources of the Ag+ ion in this solution, there will be more Ag+ ion at equilibrium than Cl- ions:
Saturated solution of AgCl to which AgNO3 has been added:
[Ag+] > [Cl-]
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92. Now imagine what happens when a few crystals of NaCl are added to a saturated solution of AgCl in water.
There are two sources of the chloride ion in this solution.
  H2O   
NaCl(s)  Na+(aq)+ Cl-(aq)      
AgCl(s)  Ag+(aq)+Cl-(aq)
Once again, the ion product is larger than the solubility product.
Qsp = (Ag+)(Cl-) > Ksp
This time, when the reaction comes back to equilibrium, there will be more Cl- ion in the solution than Ag+ ion.
Saturated solution of AgCl to which NaCl has been added:
[Ag+] < [Cl-]
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93. The figure below shows a small portion of the possible combinations of the Ag+ and Cl- ion concentrations in an aqueous solution.
Any point along the curved line in this graph corresponds to a system at equilibrium, because the product of the Ag+ and Cl- ion concentrations for these solutions is equal to Ksp for AgCl.
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94. Point A represents a solution at equilibrium that could be produced by dissolving two sources of the Ag+ ion such as AgNO3 and AgCl in water.
Point B represents a saturated solution of AgCl in pure water, in which the [Ag+] and [Cl-] terms are equal.
Point C describes a solution at equilibrium that was prepared by dissolving two sources of the Cl- ion in water, such as NaCl and AgCl.
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95. If Qsp < Ksp: AgCl(s) Ag+(aq) + Cl-(aq)
Any point that is not along the solid line in the above figure represents a solution that is not at equilibrium.
Any point below the solid line (such as Point D) represents solution for which the ion product is smaller than the solubility product.
Point D: Qsp < Ksp
If more AgCl were added to solution at Point D, it would dissolve
If Qsp < Ksp:  AgCl(s) Ag+(aq) + Cl-(aq)
Points above the solid line (such as Point E) represent solutions for which the ion product is larger than the solubility product.
Point E:  Qsp > Ksp
The solution described by Point E will eventually come to equilibrium after enough solid AgCl has precipitated.
If Qsp > Ksp: Ag+(aq) + Cl-(aq) AgCl(s)
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96. Submit answers by email:
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