Louisiana Tech University, Chemistry 102 Group Exercises
Chapter 17. Acid and Bases: Additional Aqueous Equilibria: Buffers,
Titrations and Solubility product.
Why?
There are many buffer solutions, acid/base titrations and solubility product
calculations which find many applications in day to day life. E.g. enzyme reactions are
carried out in buffers, standardizing acid/base solutions using indicators, and precipitation
reactions that are used in analysis. Therefore, it is important to understand the types,
buffers, titrations and solubilities of various salts and calculation involving them. Water
plays a major role as solvent for acids/bases, buffers and salts. Buffers involve Ka and Kb
and easy to calculate the pH of buffers using an equation called Henderson-Hesslebach.
Indicators (weak organic acid) for titrations involve Ka and easy to find which indicator is
best for a given titration. Salt also dissociate in water giving a salt dissociation constant
called Ksp, which useful predicting amount of ions in solutions and who to precipitate
them using common ion effect. You need a conceptual understanding of these processes
in order to predict their relative strength and improve applications.
Learning Objectives
Understand hydrolysis of salts, buffers, acid/base titrations and solubility equilibria
of insoluble salts.
1. Describe the hydrolysis of salts in aqueous solution (Section 16.8).
2. Explain how buffers maintain pH, how to calculate their pH, how they are
prepared, and the importance of buffer capacity (Section 17. 1).
3. Use the Henderson-Hasselbalch equation to calculate the pH of a buffer and the
pH change after acid or base has been added to the buffer (Section 17. 1).
4. Interpret acid-base titration curves and calculate the pH of the solution at various
stages of the titration (17.2).
5. Explain how acid rain is formed and its effects on the environment (Section 17-3).
6. Relate a Ksp expression to its chemical equation (Section 17.4).
7. Use the solubility of a slightly soluble solute to calculate its solubility product,
(Section 17.4).
8. Describe the factors affecting the aqueous solubility of ionic compounds (17.5)
9. Apply Le Chatelier's principle to the common ion effect (Section 17.5).
10. Use the solubility product to calculate the solubility of a sparingly soluble solute
in pure water and in the presence of a common ion (Section 17-5).
Success Criteria Students should be able to predict the pH of salt solution, calculate pH
of buffer solutions using Henderson-Hesslebach equation, select an indicator for
acid/base titrations and perform solubility product calculations which find many
applications in day to day.
Resources
Chemistry: The Molecular Science 1st Edition, John W. Moore, Conrad L. Stanitski and
Peter C. Jurs.
Prerequisites
Nomenclature of acid and bases, Acid/base reactions, Strong/weak acids/bases and
strong/weak electrolyte solutions, Acid and base dissociation equilibria and equilibrium
constants (Ka, Kb), % dissociation, and molarity (M), shifting a Chemical Equilibrium:
Le Chatelier's Principle
New Concepts
Hydrolysis of salts in water
The generic chemical reaction may be written for hydrolysis of ions:
Salt of weak acids and strong bases: Produces acidic solution
+
MA(s) + H2O(l) --> HA(aq) + M (aq) + OH¯(aq)
+
E. g. NaC2H3O2 (s) + H2O(l) --> HC2H3O2(aq) + Na (aq) + OH¯(aq)
C2H3O2¯ is a strong conjugate base of weak acid HC2H3O2
Produces basic solution
Salt of weak bases and strong acids: Produces acidic solution
+
HB+ + H2O --> B + H3O
+
E. g. NH4Cl + H2O --> NH3 + H3O + Cl¯ (aq)
+
NH4 is a strong conjugate acid of weak base NH3
Relationship between Ka and Kb of Conjugate Acid-Base Pairs
(Ka)(Kb) = Kw
Salt of strong acid + strong base = neutral
Salt of weak acid + strong base = basic
Salt of strong acid + weak base = acidic
Salt of weak acid + weak base = neutral,
Calculating pH of a aqueous salt solutions
E.g. What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5)
An aqueous solution of NH4Cl is acidic as discussed previously.
NH4Cl dissociates completely in water:
NH4Cl ---> NH4+ (strong conjugate acid) + Cl- (weak conjugate base)
NH4 dissociates further in water producing an acid solution.
NH4+ + H2O
H 3+O + NH3 ;this is a acid dissociation equilibrium or K a
For a weak base (NH3) or acid (HC2H3O2) with Kb or Ka ,respectively, the value of K a or K b for the
corresponding conjugate acid (NH 4+ ) or base (C2H3O2-) can be calculated from the following expression:
Kb x K a = K w
We need the Ka of NH4+ (conjugate acid) before we can do the calculation: using K bx K a=K w (1.0x10-14)
Kb(NH3) x Ka (NH4+)
= 1.0 x 10-14
-5
+
1.8 x 10 x Ka (NH4 )
= 1.0 x 10-14
-14
1.0 x 10
Ka (NH4+) = ------------ = 5.56 x 10-10
1.8 x 10-5
[H 3+O ] [NH3 ]
x2
Ka(NH4+) = -------------------- = ----------- =
Initial Concentration
Change in Concentration
Equilibrium Concentration
x2
NH4+ (aq)
H3+O (aq)
NH3(aq)
0.5 M
0M
0M
-x
+x
+x
0.5 M - x
x
x
x2
--------- = 5.56 x 10 -10 = -----; x2 =5.56 x 10-10 x 0.5 = 2.78 x 10 -10
0.5 -x
0.5
0.5
x = sqrt 2.78 x 10 -10 = 0.01342 =1.34 x 10 -2
[H+ ] = x = 1.34 x 10 -2 M
pH = -log [H+ ] = - log 1.34 x 10 -2
pH = 4.78
pH of 0.5 M NH4Cl solution is 4.78 (acidic)
Common ion effect
The common ion effect is an application of LeChatelier's Principle. If we mix a soluble
salt containing an ion common to a weak acid/base equilibria we will affect the position
of the equilibrium of the slightly soluble salt system.
E. g. NaC2H3O2 (soluble salt)/ HC2H3O2: Common ion for salt and weak acid is C2H3O2¯
+
NaC2H3O2 (s) + H2O(l)
HC2H3O2(aq) + Na (aq) + OH¯(aq)
NaC2H3O2 (soluble salt)/ HC2H3O2 (weak acid) solution can act as buffer maintaining the pH of
the solution if H3O+ or OH¯ are added to the mixture based on La Chaterlier’s Principle.
Definition of buffers
Mixture of chemicals [soluble salt(conjugate base)/ weak acid or soluble salt (conjugate acid)/
weak base] solution that maintain pH values of a solution within narrow limits by
absorbing or releasing hydrogen ions. This takes place though the common ion effect.
pH for buffer solutions
Henderson-Hasselbalch Equation
According to the Brønsted-Lowry theory of acids and bases, an acid (HA) is capable of
donating a proton (H+) and a base (B) is capable of accepting a proton. After the acid
(HA) has lost its proton, it is said to exist as the conjugate base (A-). Similarly, a
protonated base is said to exist as the conjugate acid (BH+).
Consider the case of acetic acid (CH3COOH) and acetate anion (CH3COO-):
CH3COOH(aq) + H2O(l)
H3O+(aq) +
CH3COO¯(aq), for which the equiilibrium constant is
[H3+O ][ CH3COO¯]
CH3COOH(aq); Ka = ----------------------[HC2H3O2]
Acetate [A-] is the conjugate base of acetic acid [HA].
Acetic acid and acetate is a
conjugate acid/base pair. Use to Kb
x Ka = Kw get Ka when working
with weak bases.
E.g. Calculate the pH of a solution prepared by adding 50 g of NH4Cl and 32.5g of NH3
and diluted to a liter. (Kb(NH3); 1.8 x 10 -5
This is a problem where weak base/salt is making a buffer solution. We can apply
Henderson-Hesselbalch equation for this condition.
[ACID] = [NH4Cl] conjugate acid, [NH3]= [BASE]
[NH4Cl] = (50.0/53.3)/1= 0.9346 M;
[NH3] =(32.5/17.04)/1 = 1.9072 M mol/L)
Ka= Kw/Kb = (1x10-14/1.8 x 10-5)= 5.56 x 10-10; pKa = 9.26
pH = 9.26 + log (1.9072/0.9346)= 9.26 + log2.04 =9.26 + 0.3097 =9.57
pH indicators titrations:
Acid-Base indicators are dyes that are themselves weak acids and bases. However, the
conjugate acid-base forms of the dye have different colors. The actual chemical
structures of the dyes are often quite complex; however, we can use the generic symbol
for the indicator as HIn. The Brönsted-Lowry equation for the indicator is:
HIn + H2O
H3O+ + InColor of an indicator at certain pH
Methyl violet is an indicator used in acid-base titrations. Its acid form, HIn, is red, while
its conjugate base from, In-, is yellow. The color change occurs in the pH range 0.00-1.6.
In a solution of pH = 6.00, what would be the
color of the indicator? Yellow
Selecting an indicator for a acid/base titration
Strong acid and strong base titrations produce neutral salt solutions at the end point
therefore any indicator that shows a color change around pH 7 is suitable.
Weak acid and strong base titrations produce a salt solution that is basic at the end
point therefore first we need to calculate the pH of the salt solution from the
concentration of the salt at the end point then find the indicator which change color close
to that pH.
E.g. Assume that an indicator works BEST when the equivalent point of a titration
comes in the middle of the indicator range. Which of the following indicators would
be the best for a titration of 35.00 mL of 0.10 M HC2H3O2 (Ka=1.8 x 10-5) with 35.00
mL 0.10 M NaOH?
methyl violet
0.0-1.6
bromocresol green
3.8-5.4
methyl orange
3.2-4.4
phenolphthalein
8.2-10.0
What is the molarity of salt solution (NaC2H3O2) at the end point?
Answer: HC2H3O2 : 0.10 M NaOH, therefore produce 0.10x 0.035= 3.5 x 10 -3 moles of Molarity of
NaC2H3O2 = 3.5 x 10-3/0.070 = 0.05 M
PH of the 0.05 M NaC2H3O2solution
NaC2H3O2 (s) + H2O(l)
HC2H3O2(aq) + Na+ (aq) + OH¯(aq)
-14
-5
Kb = (1x 10 /1.8 x 10 ) = 5.56 x 10-10 = x2/0.05 –x
x= sqrt of 5.56 x 10-10 x 0.05 = 5.27 x 10-6
x= [OH-] = 5.27 x 10-6 therefore pOH = 5.27 and pH = 8.73
phenolphthalein (8.2-10.0) is the indicator for this titration.
Solubility product of salts: Ksp
The solubility product Ksp for a salt is the product of the ion concentrations of the salt.
For example, the ionization of the salt Li2CO3 in a solution is,
Li2CO3(s) + H2O(l) = 2 Li+(aq) + CO32-(aq); Ksp = [Li+]2 [CO32-]
Less soluble salts have small Ksp value.
Qsp (ionic product (quotient)) and Ksp and Saturation
Qsp < Ksp Unsaturated solution
Qsp = Ksp Saturate solution
Qsp > Ksp Oversaturated solution
Common-ion effect in solubility of salts- The decrease in the solubility of a salt that
occurs when the salt is dissolved in a solution that already contains another source of one
of its ions. For example, if AgCl is added to a NaCl solution (which contains the
common ion, Cl-) the solubility of the AgCl decreases.
GHW#9
Printed Name:_____________________
Group Name:__________
Key Questions
1) If the following substance is dissolved in pure water, will the solution be acidic,
neutral, or basic?
a) Solid sodium carbonate-(Na2CO3):
b) Sodium chloride- (NaCl):
c) Sodium acetate- (NaC2H3O2):
d) Ammonium sulfate-((NH4)2SO4):
2) What is the pH of a 0.05 M aqueous NH4Cl solution? (Kb (NH3) = 1.8 x 10-5)
a) equilibrium reaction for the hydrolysis of salt:
b) Ka for the conjugate acid NH4+:
c) ICE set-up:
I:______________________________________________.
C:_____________________________________________.
E______________________________________________.
d) Calculation of x [H3O]+:
e) pH of the solution:
3) What is the pH of a 0.05 M aqueous NaC2H3O2 solution? (Ka (HC2H3O2) = 1.8 x 10-5)
a) equilibrium reaction for the hydrolysis of salt:
b) Kb for the conjugate base C2H3O2-:
c) ICE set-up:
I:______________________________________________.
C:_____________________________________________.
E______________________________________________.
d) Calculation of x [OH]-:
e) pOH and pH of the solution:
4) A 50.00-mL sample of 0.100 M KOH is being titrated with 0.100 M HNO3. Calculate
the pH of the solution after 52.00 mL of HNO3 is added.
a) acid base reaction:
b) moles of KOH:
c) moles of HNO3:
+
d) [H3O] :
e) pH of the solution:
5) For the (buffer effect) of HC2H3O2/NaC2H3O2
a) Acid dissociation reaction:
b) Salt hydrolysis reaction:
c) Common ions in both equilibria:
d) Which way salt hydrolysis equilibrium move adding H3O+:
e) Which way salt hydrolysis equilibrium move adding OH-:
6) Describe the (buffer effect) of NH3/NH4Cl
a) Buffer type: (weak acid or base)/soluble salt):
b) Base dissociation reaction:
c) Salt hydrolysis reaction:
d) common ions in both equilibria:
e) Which way salt hydrolysis equilibrium move adding H3O+:
f) Which way salt hydrolysis equilibrium move adding OH-:
7) What is the pH of a solution that is 0.2 M in acetic acid (Ka = 1.8 x 10-5) and 0.2 M in
sodium acetate?
a) Is it a acid, base, salt or buffer solution?
b) Henderson-Hesselbalch equation:
c) pKa:
e) pH of the solution:
d)
8) If 50 ml of a 0.01 M HCl solution is titrated with a 0.01 M NaOH solution, what will
be the concentration of salt (NaCl) the pH at the endpoint?
a) NaCl solution acidic, basic or neutral?
b) Concentration of [NaCl]:
c) pH of the solution?
d) Suitable indicator for the titration:
9) If 50 ml of a 0.01 M HCl solution is titrated with a 0.01 M NH3 (Kb = 1.8 x 10-5)
solution, what will be
a) The initial pH (0.01 M NH3):
b) Concentration of NH4Cl at the endpoint:
c) pH at the endpoint:
d) Suitable indicator for the titration:
10) If 50 ml of a 0.01 M HC2H3O2 solution is titrated with a 0.01 M NaOH solution, what
will be the
a) Molarity of NaC2H3O2 at the endpoint:
b) The pH at the endpoint:
c) What indicator would be most suitable for this titration:
11) For a saturated solution of
a) AgCl in water:
i) Solubility equilibrium reaction:
ii) Ksp expression:
b) CaF2 in water:
i) Solubility equilibrium reaction:
ii) Ksp expression
c) Fe2S3 in water
i) Solubility equilibrium reaction:
ii) Ksp expression
12) Which of following has the highest molar solubility (mole/L)?
Salt
Ksp
a) CaCO
3
5 × 10-9
b) PbCO3 1.4 × 10-13
c) Li2CO3 2 × 10-3
d) NiCO3 1.2 × 10-7
13) For Li2CO3, Ksp is 2 × 10-3 M3. What is the concentration of Li+ in a saturated
solution of Li2CO3?
14) Chemical analysis gave [Pb2+] = 0.012 M, and [Br-] = 0.024 M in a solution. From a
table, you find Ksp for PbBr2 has a value of 4 x 10-5 M3.
a) Solubility equilibrium reaction:
b) Qsp:
c) Ksp:
d) Qsp < Ksp, Qsp = Ksp or Qsp > Ksp?
e) Is the solution saturated, oversaturated or unsaturated?