CHAPTER 17 SI WORKSHEET
17.1 : THE COMMON-ION EFFECT
Common-Ion Effect: Whenever a weak electrolyte and a strong electrolyte contain a common ion, the weak electrolyte ionizes less than it would if it were alone in solution.
What is the common ion in the following
1) Hydrochloric acid and Hydrofluoric acid ______________________
2) Nitrous Acid and Potassium Nitrite __________________________
Calculating the pH when a common ion is involved
3) Calculate the pH of a solution made from equal amounts of 0.30 M hydrofluoric acid and 0.70 M sodium fluoride. K a
= 7.1 x 10 -4
4) A buffer solution was prepared which had a concentration of 0.20 M in acetic acid and 0.10 M in sodium acetate. If the K a
for acetic acid is 1.74 x 10 -5 , calculate the theoretical hydrogen ion concentration and pH of the buffer solution.
Calculating Ion concentrations when a common ion is involved
5) Calculate the fluoride ion concentration and pH of a solution that is 0.25M in HF and 0.10M in HCl
Ka HF = 6.8 x10 -4
6) Calculate the formate ion concentration of a solution that is 0.025M in formic acid ( HCOOH; Ka = 1.8 x 10 -4 ) and 0.15M in HNO
3
.

17.2 BUFFERED SOLUTIONS
Solutions which contain a weak conjugate acid-base pair, can resist drastic changes in pH upon the addition of small amounts of strong acid or strong base
Buffer Capacity is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree.
pH range of any buffer is the pH range over which the buffer acts effectively. Range of pH = pKa +/- 1.
Which of the following conjugate acid – base pairs will or will not function as a buffer
1) C
2
H
5
COOH AND C
2
H
5
OO 2) HCl AND Cl -
3) HClO AND ClO 4) NaC
2
H
3
O
2
AND C
2
H
3
O
2
5) HNO
3
AND NaNO
3
CALCULATING THE pH OF A BUFFER
USE THE HENDERSON-HASSELBACH EQUATION
6) The Ka of acetic acid is 1.76 10

5
. The pH of a buffer prepared by combining 50.0 mL of 1.00 M potassium acetate and 50.0 mL of 1.00 M acetic acid is __________.
7) Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO H) and 0.230 mol of sodium formate (NaCO H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is
1.77 10

4
.
CALCULATING CONCENTRATIONS OR MOLES OF A BUFFER SOLUTION
8) How many moles of sodium hypobromite, NaBrO, should be added to 1.00 L of 0.200 M hypobromous acid, HBrO (K a
= 2.5 x 10 -9 ), to form a buffer solution of pH 8.80? Assume that no volume change occurs when the NaBrO is added.

9) You need to make a buffer with a pH of 3.50 from citric acid and sodium dihydrogen citrate. If you ignore all higher ionizations of the dihydrogen citrate ion, what concentration of sodium dihydrogen citrate needs to be added to 0.40 M acid (K a
= 7.4 x 10 -4 )?
CALCULATING pH CHANGES IN BUFFERS
10) What is the final pH if 0.020 mol HCl is added to 0.500 L of a 0.28 M NH
3
and 0.22 M NH
4
Cl buffer solution (K b
(NH
3
) = 1.8 x 10 -5 )?
11) A buffer contains 0.30 M acetic acid and 0.20 M sodium acetate. What is the pH of the buffer after
0.030 mol/L of a strong acid or 0.030 mol/L of a strong base are added (K a
= 1.8 x 10 -5 )?
17.3 ACID-BASE TITRATIONS
Titration: The process of reacting a solution of unknown concentration with one of known concentration
Equivalence Point: The point in a titration at which the added solute reacts completely with the solute present in the solution.
Titration Curve: A graph of pH as a function of added titrant
Draw the titration curve for the following:
Strong Acid- Strong Base Strong Base to a Weak Acid Strong Acid to a Weak Base
Equivalence Point: pH________ pH=__________________ pH=___________________

Calculating Equivalence Point
1) A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was
__________ M.
2) A 50.0 mL sample of an aqueous H SO
4
solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62.5 mL of the base. The concentration of H SO
4
is __________
M.
Calculating the pH at the Equivalence Point
3) Calculate the pH at the equivalence point when (A) 40.0mL of 0.025M benzoic acid (Ka= 6.3 x 10 -5 ) is titrated with 0.050M NaOH; (B) 40.0mL of 0.100M NH
3
is titrated with 0.100M HCl.
Calculating pH for a Strong Acid –Strong Base Titration
4) A 30.0mL sample of 0.150M KOH is titrated with 0.125M HClO
4
solution. Calculate the pH after the following volumes of acid have been added: A) 30.0mL B) 35.0mL C) 36.0mL D) 37.0mL E) 40.0mL

Calculating pH for a Weak Acid- Strong Base Titration
5) A 35.0mL sample of 0.150M acetic acid is titrated with 0.150M NaOH solution. Calculate the pH after the following volumes of bases has been added: A) 0mL B) 17.5mL C) 34.5mL D) 35.0mL E) 35.5mL F)
50.0mL

17.4 SOLUBILITY EQUILIBRIA
Solubility-Product Constant: (Ksp) An equilibrium constant related to the equilibrium between a solid salt and its ions in solution. It provides a quantitative measure of the solubility of a slightly soluble salt.
The solubility product of a compound equals the product of the concentration of the ions involved in the equilibrium, each raised to the power of its coefficients in the equilibrium equation.
The magnitude of Ksp is a measure of how much of the solid dissolves to form a saturated solution.
The larger the Ksp, the greater is the molar solubility
Calculating Ksp from Solubility
1) If the molar solubility of CaF
2
at 35.0
˚
C is 1.24 x 10 -3 mol/L, what is Ksp at this temperature?
2) The molar solubility of PbBr
2
at 25.0
˚
C is 1.0 x 10 -2 mol/L . Calculate Ksp.
3) A 1.00-L solution saturated at 25.0
˚
C with calcium oxalate ( CaC
2
O
4
) contains 0.0061g of calcium oxalate. Calculate the solubility-product constant for this salt at 25.0
˚
C.
Calculating Solubility from Ksp
4) The Ksp for LaF
3
is 2.0 x 10 -19 . What is the solubility of LaF
3
in water in moles per liter?
5) The concentration of iodide ions in a saturated solution of lead (II) iodide is __________ M. The solubility product constant of PbI
2
is

8
1.4 10 .
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17.5 FACTORS THAT AFFECT SOLUBILITY
Common-Ion Effect: The solubility of a slightly soluble salt is decreased by the presence of a second solute that furnishes a common ion.
Solubility and pH: In general, if a compound contains a basic anion( the anion of a weak acid), its solubility will increase as the solution becomes more acidic.
The solubility of slightly soluble salts containing basic anions increases as [H+] increases( as pH is lowered).
Calculating the Effect of a Common Ion on Solubility
1) Calculate the solubility of LaF
3
in grams per liter in A) 0.010M KF solution B) 0.050M LaCl
3
solution.
2) Calculate the solubility of AgBr in A) 3.0 x 10 -2 M AgNO
3
solution B) 0.10M NaBr solution.
Ksp for AgBr 5.0 x 10 -13
Calculating the effect of pH on Solubilty
3) Calculate the solubility of Mn(OH)
2
in moles per liter when buffered at pH ( A) 7.0 (B) 9.5 (C) 11.8
Ksp for manganese (II) hydroxide: 1.6 x 10 -13
4) Calculate the molar solubility of Fe(OH)
2
when buffered at pH (A) 8.0 (B) 10.0 (C) 12.0
Ksp for iron(II) hydroxide: 7.9 x 10 -16
Concept Questions
5) The molar solubility of __________ is not affected by the pH of the solution.
A) Na PO
3 4
B) NaF C) KNO
3
D) AlCl
3
E) MnS

6) In which of the following aqueous solutions would you expect AgCl to have the lowest solubility?
Highest?
A) pure water B) 0.020 M BaCl
2
C) 0.015 NaCl D) 0.020 AgNO
3
E) 0.020 KCl
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17.6 Precipitation and Separation of Ions
If Q > Ksp, precipitation occurs until Q = Ksp
If Q = Ksp, equilibrium exists (saturated solution)
If Q < Ksp, solid dissolves until Q = Ksp
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17.7 Read and Summarize
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