Chapter 16, applications of buffers, common ions, titrations, and solubility.
Buffer Solutions
Buffers are solutions with the ability to resist the addition of strong acids or strong bases,
within limits.
They play an important role in chemical processes where it is essential that a fairly
constant pH is maintained. In many industrial and physiological processes, specific
reactions occur at some optimum pH value. When the pH varies to any extent from the
optimum value, undesirable reactions and effects may occur. For example, the pH of your
blood lies at about 7.35. If this value drops below 7.0 (acidosis) the results are fatal. Also if
it rises above 7.7 (alkalosis) the results are as well fatal. Fortunately our blood contains a
buffering system which maintains the acidity at the proper level. If it were not for the
protection of the buffering system, we could not eat and adsorb many of the acidic fruit
juices and foods in our diet.
A typical lab buffer is CH3COOH and its salt NaCH3COO. Most buffer solutions are made
up using a weak acid and its sodium salt! When a strong base such as NaOH is added to the
buffer, the acetic acid reacts with and consumes the excess OH- ion. The OH- reacts with
the H3O+ ion from the acid in the following reaction:
H2O + CH3COOH <-------> H3O+ + CH3COOH3O+ + OH- <----------> H2O
The OH- reduces the H3O+ ion concentration, which causes a shift to the right, forming
additional CH3COO- and H3O+ ions. For practical purposes each mole of OH- added
consumes a mole of CH3COOH and produces a mole of CH3COO-.
OH- + CH3COOH <------------> CH3COO- + H3O+
When a strong acid such as HCl is added to the buffer, the hydronium ions react with the
CH3COO- ions of the salt and form more undissociated CH3COOH.
H3O+ + CH3COO- <-----------> CH3COOH + H2O
As you would expect, there is a limit to the quantity of H+ or OH- that a buffer can absorb
without undergoing a significant change in pH. If a mole of HCl is added to a litre of buffer
solution containing 0.5 moles of sodium acetate/acetic acid buffer the H+ completely
consumes the buffer and results in a drastic change in pH.
The blood buffer is made up from the dissolved carbon dioxide in the plasma.
CO2(g) + H2O <------->
H2CO3 <------> HCO3- + H3O+
When a base is added it reacts with the carbonic acid.
OH- + H2CO3 <--------> HCO3- + H2O
When an acid is added it reacts with the bicarbonate ion.
H3O+ + HCO3- <---------> H2CO3 + H2O
Because there are both a base-neutralizer and an acid-neutralizer then we have a buffer.
Buffer Components
A buffer has two components.
HA
a weak acid &
NaA ---> Na+ + Aa soluble salt of the acid (salt of the conj. base)
Therefore any extra H3O+ will be neutralized by the A- in the buffer.
H3O+ + A- <-------> HA + H2O
And any extra OH- that is added will be neutralized by the acid.
HA + OH- <------> A- + H2O
There are 3 basic types of calculations that can be done with buffer system and these will
be covered next.
The pH of a Buffered Solution
Because HA is a weak acid, very little of it is dissociated at equilibrium. Even if HA were
the only solute. But the solution also contains A- from the dissolved salt. The presence of Asuppresses the already slight ionization of HA by shifting the following equilibrium to the
left. You should recognize this for the common ion effect which it is.
HA <-------->
H+ + A-
ie. The value of [HA] at equilibrium = [HA] initially
If we take the -log of both sides:
log[H+] = -log Ka + (-log [ acid] )
[anion]
pH = pKa - log [acid]
[anion]
Since
we end up with the following:
(Henderson-Hasselbach Equation)
Two factors govern the pH in a buffered solution.
1) pKa of the weak acid;
2) ratio of the initial molar []'s of the acid and it's salt.
If we prepare a solution where [anion] = [acid] then the
log [anion] = log 1 = 0
[acid]
Therefore the pH of the solution will turn out equal to the pKa of the weak acid.
What mostly determines where on a pH scale a buffer can work best is the pKa of the weak
acid. Then by adjusting the ratio of [anion] to [acid], we can cause shifts so that the pH of
the buffered solution comes out on one side or the other of this value of pH.
Calculating the pH of a Buffered Solution
To study the effect of a weakly acidic culture medium on the growth of a certain strain of
bacterium, a microbiologist prepared a buffer solution by making it with 0.11 mol/L
NaCH3COO, sodium acetate, and also 0.09 mol/L CH3COOH, acetic acid, What is the final
pH?
Ka for acetic acid = 1.8 x 10-5
Therefore pKa = 4.74
[acid] = 0.090 mol/L
[anion] = 0.11 mol/L
pH = 4.74 + log (0.11) = 4.74 + log 1.2 = 4.74 + 0.079 = 4.82
(0.09)
You may sometimes see the Henderson-Hasselbach equation given in some chemistry
sources as pH = pKa + log [salt]/[acid]. DON'T USE IT! It only works for monovalent salts.
Use the formula given previously in these notes.
The Effectiveness of a Buffer
Suppose we drop 0.01 moles of strong base into our buffer from the last example. What will
be the measured effects?
HA + OH- <-------> A+
start 0.09 M
----0.11 M
H2 O
finish -0.01 M -0.1 M
0.08 M 0.0 M
+0.01 M
0.12 M
pH = pKa + log [anion]
[acid]
= 4.74 + log (0.12)
(0.08)
= 4.74 + log 1.5
= 4.74 + 0.18
= 4.92 The change in pH is small compared to what it would have been in pure water!
Buffer Capacity
No buffer has an unlimited capacity. ie; buffers can only absorb so much abuse
before they are destroyed. The capacity of a buffer is the amount of acid or base it can
handle before the pH of the solution changes drastically.
If you add enough strong acid to neutralize all of the buffers basic component, then
additional strong acid will make the pH drop rapidly. The same applies for a strong base
and the buffers acidic component. The buffer's pH is a function of it's pKa and the ratio of
concentrations of anion and acid, but the buffer's capacity depends upon actual
concentrations.
Preparation of A Buffer
To prepare a buffer first choose an acid with a pKa within a ±1 pH unit of the
desired value. Then manipulate the ratios to get the desired pH. A solution buffered at pH
5.00 is needed in a chemistry experiment. Can we use acetic acid and sodium acetate to
make it? If so, what ratio of acetate ion to acetic acid is needed?
Ka of acetic acid = 1.8 x 10-5 pKa = 4.74
the pKa is within a range of 5 ± 1
ie. between the values of 4 - 6.
therefore, acetic acid is okay to use.
next use pH = pKa + log [anion]
[acid]
5.00 = 4.74 + log [anion]
[acid]
0.26 = log [anion]
[acid]
100.26 = [anion]
[acid]
= 1.8
Therefore the ratio of anion to acid is 1.8 to 1. Therefore, use a ratio of 1.8 moles of acetate
ion to 1.0 mole of acetic acid. The solution will then be buffered at a pH of 5.00
OR
We could use a ratio of 0.18 moles to 0.100 moles if we wanted a smaller buffering
capacity.
Buffer Questions
1. Calculate the pH of a buffered solution made up of 0.015 M sodium acetate and 1 L of
0.10 M acetic acid.
2. A nutrition scientist needed an aqueous buffer for a pH of 3.90. Would formic acid
andit's salt sodium formate, make a good pair for this purpose? If so, what mole ratio
of the anion of this salt, CHO2-1, to the acid, HCHO2, is needed?
3. A buffer solution was prepareed by dissolving 2.5 grams of NH4Cl in 125 mL of 0.24 M
ammonia solution. At what pH will this solution serve as a buffer?
4. a) A formic acid, sodium formate solution is made up by dissolving 0.2 mole of formic
acid and 0.3 mole of sodium formate in 500 mL of water? What pH wil the resulting
solution be at?
b) If 0.5 mL of 12 M HCl is added to this buffer solution what will be the resulting
pH?
c) If 0.0125 mole of NaOH is added to the original solution in part a) what will be the
resulting pH?
5. a) A 0.05 M NH3 solution has 0.1 mole of powdered NH4Cl added to it. What will be
the resulting pH?
b) 200 mL of 0.1 M NH3 is added to 200 mL of 0.25 M NH4Cl. What is the pH of the
resulting solution?
c) 500 mL of 1 M citric acid has 0.1 mole of sodium citrate added to it. What is the
resulting pH?
6. Barbituric acid is an extremely powerful sedative. Most drug manufacturers use equal
amounts, grams that is, of barbuturic acid and sodium barbiturate in their pills. An
average 'Percodan' tm, a barbiturate based pill, is 70% bulk starch and sugar. If the
average pill is 0.5 g in mass what will be the pH of the solution made by dissolving the
above pill in 50 mL of water?
7. A solution of pH = 7.0 is required for an experiment. You decide to use 100 g of sodium
dihydrogen phosphate (NaH2PO4), in 1 L of water. How much sodium monohydrogen
phosphate, (Na2HPO4) is needed in grams to achieve this pH?
Acid-Base More Buffers Worksheet
1.
2.
3.
4.
5.
6.
7.
8.
Write ionic equations that illustrate how each pair of compounds can serve as a
buffer pair.
(a) H2CO3 and NaHCO3 (the "carbonate" buffer in blood)
(b) NaH2PO4 and Na2HPO4 (the "phosphate" buffer inside body cells)
(c) NH4Cl and NH3
Which buffer would be better able to hold a steady pH on the addition of strong acid,
buffer 1 or buffer 2? Explain.
Buffer 1: a solution containing 0.10 M NH4Cl and 1 M NH3. Buffer 2: a solution
containing 1 M NH4Cl and 0.10 M NH3.
How many grams of sodium acetate, NaC2H3O2, would have to be added to 1 L of 0.15
M acetic acid (pKa=4.74) to make the solution a buffer for pH 5.00?
How many grams of sodium formate, NaCHO2, would have to be dissolved in 1.0 L of
0.12 formic acid (pKa) to make the solution a buffer for pH 3.80?
What ratio of molar concentrations of NH4Cl and NH3 would buffer a solution at pH
9.25?
To study the effect of a weakly acidic medium on the rate of corrosion of a metal, a
chemist prepared a buffer solution by making it 0.11 M NaC2H5O2 and also 0.090 M
HC2H3O2. What is the pH of this solution?
How is the pH of a buffered solution related to the pKa of the weak acid in the buffer
if
(a) the ratio of [anion] to [acid] is 10 to 1?
(b) The ratio of [anion] to ]acid] is 1 to 10?
A solution of acetic acid is pink in the presence of the indicator methyl orange. When
solid sodium acetate is added to the acid solution, the colour changes to yellow. Why?
A Basic Buffer
NH3 + H2O <-------> NH4+1 + OH-1
So
Take the -log of both sides
A generalized version of this basic Henderson-Hasselbach equation is
A chemistry student needs 250 mL of a solution buffered at a pH of 9.00. How many
grams of ammonium chloride have to be added to 250 mL of 0.2 mol/L NH3 to make such a
buffer? (Volume is assumed not to change.)
pH = 9.0
[base] = 0.2
pOH = 5.0
pKb of ammonia = 4.74 (look up from datatable)
pOH = pKb +log [cation]
[base]
5.00 = 4.74 + log [cation]
(0.2)
0.26 = log [cation]
(0.2)
[cation] = 100.26 X 0.2
= 1.8 * 0.2
= 0.36 mol/L of the NH4+1.
But we only need enough for 250 mL so 0.36 mol = __x___
1000 mL 250 mL
x = 0.09 moles of NH4+1 ions needed
Since the NH4+1 comes from NH4Cl then we also need 0.09 moles of NH4Cl.
g = n * mm
= 0.09 moles * 53.5 grams/mole
= 4.8 grams of the salt are required.
Volumetric Analysis
In many acid-base reactions, the equilibrium is displaced almost completely toward
the product side. These reactions may be considered quantitative and can be used as the
basis for the analysis of the amount of acid or base in a given sample. The process is termed
volumetric analysis.
The requirements are:
1) Only a single, specific reaction must take place between the unknown substance
and the known substance used for the analysis.
2) The unknown substance must react completely and rapidly with the added
standard reagent. ie; it must be a quantitative reaction.
3) An indicator or method must be available to signal when all the unknown
substance has reacted with the added standard reagent.
The usual objective is to determine the mass or percentage of a qualitatively
identified component in a sample whose quantitative amount is unknown. If the sample is a
solution, the objective may be to determine its molar concentration.
Acid-Base Titration
An acid-base titration is just a method by which we can perform a volumetric
analysis. The concentration of an acid-base solution may be determined by measuring the
volume of the base of known concentration needed to react completely with a specific
volume of an acid solution.
Definitions
1) Standard solution: A solution of known concentration.
2) Titration: the process of adding the standard solution from a graduated tube in
controlled amounts.
3) Buret: a graduated tube with a dispensor control at the bottom.
4) Stoichiometric Point: the point at which equal molar quantities of reactants are
present.
5) Chemical Indicators: a chemical that will change colour at or very near the ph of
the stoichiometric point.
6) Endpoint: The place in the titration when the indicator changes colour.
7) The indicator used, should in theory coincide with the stoichiometric point of the
reaction.
A 0.660 mol/L NaOH solution is used to determine the molar concentration of
H2SO4 solution. What is the molarity of the acid, 20.0 mL of which is just neutralized by
36.0 mL of the standard base?
This is a standard problem which can be solved using:
MaVa = MbVb where M is the molarity, V is the volume, a is 'of the acid', b is 'of the base'.
Therefore Ma = Mb * Vb = 0.660 mol/L * 36.0 mL = 1.19 mol/L
Va
20.0 mL
Follow-Up Problems
1. What is the molarity of a hydrochloric acid solution, 30.0 mL of which is just
neutralized by 48.0 mL of 0.100 mol/L NaOH?
2. How many mL of 0.100 mol/L HCl are required to neutralize 25.0 mL of 0.100
mol/L Ba(OH)2?
Titration Curves
Below is a graph depicting the changes that occur in the pH of a solution during a
titration of a typical strong acid with a typical strong base
At the start, before any NaOH has been added, the solution is simply 25.00 mL of
0.20 M HCl. Since the HCl is a strong acid and is 100% dissociated, the initial pH is quite
low.
As we add more and more NaOH solution we are of course affecting the
concentration because of volume changes. Between 0 and approximately 24 mL of NaOH
added, a large increase in the pH is not seen. All that is happening is that the added base is
changing some of the acid to the salt, NaCl. NaCl, neither of which ions hydrolyses, so the
concentration of remaining acid gradually becomes more and more dilute. By adding just
enough base to neutralize the acid the pH should change drastically to a pH of 7.00 or
neutrality. The addition of a little more base will cause the resulting solution to become
quite caustic.
The graph above indicated that a rapid change in pH with a small change in volume
of standard solution is desirable if the indicator is to give a sharp colour change at the
endpoint.
Titration of a Weak Acid by a Strong Base
Below is a graph of the titration curve for a typical weak acid with a typical strong
base.
At the equivalence point, the solution contains sodium acetate, which has an anion,
but not a cation, that hydrolyses. The acetate ions however do ionize to give a slightly basic
solution. For this reason, at the equivalence point where all the acid has neutralized all the
base, there is still the acetate ions in solution which ionize the water.
Titration of a Weak Base by a Strong Acid
The graph below shows the titration curve for a typical weak base and a typical
strong acid.
At the equivalence point, when all the acid has neutralized all the base there will still
be the cation NH4+ in solution. This ion will cause the water to ionize and force the resulting
pH of the solution downwards.
Acid-Base Titrations Worksheet
1.
2.
How many millilitres of 0.100 M HCl are required to neutralize 25.0 mL of 0.100 M
Ba(OH)2?
What is the molarity of a hydrochloric acid solution, 30.0 mL of which is just
3.
4.
5.
6.
7.
8.
9.
neutralized by 48.0 mL of 0.100 M NaOH?
Exactly 50.0 mL of HOCl solution of unknown concentration was titrated with 0.100
mol NaOH. An end point was reached when 38.5 mL of the base was added. Calculate
the molar concentration of the HOCl solution.
What can make the titrated solution at the equivalence point in an acid-base titration
have a pH not equal to 7.00. How does this possibility affect the choice of an
indicator?
What is a good indicator for titrating potassium hydroxide with hydrobromic acid?
Explain.
In the titration of an acid with a base, what condition concerning the quantities of
reactants ought to be true at the equivalence point?
When 50 mL of 0.10 M formic acid is titrated with 0.10 M sodium hydroxide, what is
the pH at the equivalence point? (Be sure to take into account the change in volume
during the titration.) What is a good indicator for this titration?
Calculate the pH of the resulting solution after 20.00 mL of 0.20 M NaOH has been
added to 25.00 mL of 0.20 M HC2H3O2.
To obtain the data needed to plot a titration curve for the titration of a strong acid
with a strong base, a chemist used 25.00 mL of 0.1000 M HCl. The molarity of the
base was also 0.1000 M, and this solution was added in small portions to the acid.
Calculate the pH of the resulting solution after each of the following quantities of base
had been added to the original solution (you must take into account the change in
volume). Graph the result.
(a) 0 mL
(d) 24.99 mL
(g) 25.10 mL
(b) 10.00 mL
(e) 25.00 mL
(h) 26.00 mL
(c) 24.90 mL
(f) 25.01 mL
(i) 50.00 mL
10. For each of the following titration, specify the indicator from column II that should
be used. State your reasons, but do not make any quantitative calculations.
Column I
a) NaOH + KHC8H4O4
Column II
neutral red (6.8-8.0)
b) NaOH + HCl
bromocresol green(3.8-5.4)
c) NaHCO3 + HCl
0-cresolphthalein(8.2-9.8)
Polyprotic Acids
Most of the acids like HCl, HNO3, CH3COOH are monoprotic because they only
produce 1 H+ per molecule. Sulphuric acid, H2SO4, and phthalate acid, C6H4(COOH)2 are
diprotic acids. Phosphoric acid, H3PO4, and citric acid are triprotic acids. There are two
commonly encountered acids that have never been isolated as pure species. These are
carbonic acid, H2CO3, and sulphurous acid, H2SO3.
There is a separate ionization constant for the ionization of each hydrogen ion from
a polyprotic acid.
H2CO3(aq) <--------> H+(aq) + HCO3-(aq)
carbonic
bicarbonate
acid
ion
Ka1 = [H+][HCO3-] = 4.5 x 10-7 pKa1 = 6.35
[H2CO3]
The second acid ionization Ka2 for this equilibrium:
HCO3-(aq) <-------> H+(aq) +
bicarbonate
ion
CO32-(aq)
carbonate
ion
Ka2 = [H+][CO32-] = 4.7 x 10-11 pKa2 = 10.33
[HCO3-]
In general for any diprotic acid the value of Ka1 > Ka2 since it is easier to pull a H+
from a neutral molecule. Ka1 is usually 104 to 105 times larger than Ka2.
Because the first acid ionization constant, Ka1, is so much greater than the second,
Ka2, the pH of a dilute solution of a weak diprotic or triprotic acid can be calculated just
from the value of Ka1. The contributions to [H+] from the 2nd or 3rd dissociation are so small
that they can be ignored.
Acid-Base Unit Review Questions
1.
2.
3.
4.
5.
Calculate the pH of a solution in which the [H3O+] equals:
a) 1.0 M
b) 0.1 M
c) 0.01 M
d) 1.0 X 10-7 M
e) 3.0 X 10-7 M
f) 4.0 X 10-10 M
What is the pOH, and the [H3O+1] of a 0.01 M KOH solution?
Calculate the [OH-1] of a solution if the pH=
a) 11.00
b) 4.68
Calculate the [H3O+1] of a solution if the pH=
a) 7.89
b) 3.54
Calculate the pH of a solution if it contains:
a) 0.010 mole/L H3O+1
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
b) 0.0010 mole/L OH-1
c) 0.50 mole/L H3O+1
d) 0.25 mole/L OH-1
e) 0.015 mole/L HNO3, assuming HNO3 is a strong acid.
f) 0.22 mole/L Ba(OH)2, assuming Ba(OH)2 is a strong base.
Determine the pH in each of the following solutions:
a) lemon juice, [H3O+1] = 5 X 10-3 mole/L
b) milk, [H3O+1] = 3 X 10-7 mole/L
c) seawater, [H3O+1] = 3 X 10-9 mole/L
Benzoic acid, C6H5COOH, is an organic acid whose sodium salt, C6H5COONa, has
long been used as a safe food additive to protect beverages and many foods against
harmful yeasts and bacteria. The acid is monoprotic. Write the equation for the Ka
of the acid.
What is the percent ionization of a 0.15 M solution of HF? What is the pH of this
ionization?
Periodic acid, HIO4, is an important oxidizing agent and a moderately strong acid. In
a 0.10 M solution, [H3O+1] = 3.8 X 10-2 mole/L. Calculate the Ka and pKa for periodic
acid.
Para-aminobenzoic acid, PABA, is a powerful sunscreening agent whose salts are
used widely in suntanning and screening lotions. The parent acid which we may
symbolize as H-Paba, is a weak acid with a pKa of 4.92 at 25oC. What will be the
[H3O+1] and pH of a 0.030 M solution of this acid?
Ethylamine, CH3CH2NH2, has a strong pungent odour similar to that of ammonia.
Like ammonia, it is a Bronsted base. A 0.10 M solution has a pH of 11.86. Calculate
the Kb and pKb for the ethylamine, and find the pKa for its conjugate acid,
CH3CH2NH3+1
A sample of blood was found to have a [H3O+1] = 4.6 X 10-8 M. Find the molar
concentration of OH-1, and decide if the sample was acidic, basic, or neutral.
Hydrazine, N2H4, has been used as a rocket fuel. Like ammonia, it is a Bronsted base.
A 0.15 M solution has a pH of 10.70. What is the Kb and pKb for hydrazine and the
pKa of its conjugate acid?
Quinine, an important drug in treating malaria, is a weak Bronsted base that we may
represent as Qu. At 25oC its pKb is 5.48. To make it more stable in water, it is put
into solution as its conjugate acid, which we may represent as H-QuCl. What is the
calculated pH of a 0.15 M solution of H-Qu+1?
Butanoic acid (butyric acid), C3H7COOH, is found in small quantities in human
perspiration and is responsible for the foul odor often associated with locker rooms.
The acid is also produced when butter becomes rancid. A 0.0010 mole/L solution of
butanoic acid has a pH of 3.91 at 25oC. Calculate the acid dissociation constant of
butanoic acid.
The Common Ion Effect and Solubility
Le Châtelier's Principle indicates that it should be possible to vary the solubility of a
solid by varying the concentration of the ions at equilibrium. i.e. To decrease the [Ag+] in
a solution of AgBr, we can add excess Br- ions in the form of a soluble salt such as NaBr.
This excess puts a stress on the equilibrium system which adjusts itself by shifting left. As it
consumes Br- ions and forms AgBr(s).
The Ksp is a constant. If one ion gets larger then the other must get smaller.
Ksp = [Ag+][Br-]
Br-]
Ksp = [Ag+][
Each Br- ion that reacts ties up a Ag+ ion which removes it from the solution.
Ksp = [Ag+][Br-] = 4.8 x 10-13
The value of Ksp for KBr is so small that even a little Br- from the NaBr will cause a
ppt. The overall effect is that the solubility of AgBr is decreased. AgBr is less soluble in a
solution of NaBr then it would be in pure water.
Compare the molar solubility of AgBr in pure water and in a 0.10 M NaBr solution.
AgBr(s) <====> Ag+(aq) + Br-(aq)
In pure water
Let 'x' be the [Ag+] and the [Br-] in a saturated solution of AgBr.
Therefore Ksp = [Ag+][Br-]
= (x) (x) = x2
4.8 x 10-13 = x2
= 6.9 x 10-7 mol/L
Therefore in a solution made with pure water the [Ag+] = [Br-] = 6.9 x 10-7 mol/L.
In a 0.10 M NaBr soln.
In a 0.10 mol/L solution, the [Br-] = 0.10 mol/Lalready
The small quantity of Br- furnished from the dissolving of the AgBr is insignificant
since in pure water it was only 6.9 x 10-7 M. Since dissolution is repressed by the presence
of the common Br- ion, it will be even less. Therefore we can forget about it.
Follow-Up Problem: Compare the molar solubility of PbI2 in pure water and in 0.10 NaI
Solubility Unit Problems Worksheet
1.
Calculate the Ksp for each of the salts whose solubility is listed below.
a) CaSO4 = 5.0 x 10-3 mol/L
b) MgF2 = 2.7 x 10-3 mol/L
c) AgC2H3O2 = 1.02 g/100 mL
d) SrF2 = 12.2 mg/100 mL
2.
Calculate
a) the solubility in moles/L of each of three salts and
b) the concentration of the cations in mg/mL in each of the saturated solutions.
i) AgCN Ksp = 2.0 x 10-12
ii) BaSO4 Ksp = 1.5 x 10-9
iii) FeS Ksp = 3.7 x 10-19
iv) Mg(OH)2 Ksp = 9.0 x 10-12
v) Ag2S Ksp = 1.6 x 10-49
vi) CaF2 Ksp = 4.9 x 10-11
3.
Consider these slightly soluble salts:
i) PbS Ksp = 8.4 x 10-28
ii) PbSO4 Ksp = 1.8 x 10-8
iii) Pb(IO3)2 Ksp = 2.6 x 10-13
a) Which is the most soluble?
b) Calculate the solubility in moles/L for PbSO4.
c) How many grams of PbSO4 dissolve in 1 L of solution?
d) How can you decrease the concentration of Pb2+(aq) in a saturated solution of
PbSO4 solution?
e) What is the concentration in moles/L of PbS in a saturated solution of the salt?
4.
For each of these substances, calculate the milligrams of metallic ion that can remain
at equilibrium in a solution having a [OH-] = 1.0 x 10-4 mol/L.
a) Cu(OH)2 Ksp = 1.6 x 10-9
b) Fe(OH)3 Ksp = 6.0 x 10-38
c) Mg(OH)2 Ksp = 6.0 x 10-12
I’m too tired to continue… so that’s all for now…..