Title: Solubility product constant, Ksp: the equilibrium constant expression for the dissolving of a slightly soluble solid.
1The Solubility Product Constant, Ksp
- Many important ionic compounds are only slightly
soluble in water (we used to call them
insoluble Chapter 4). - An equation can represent the equilibrium between
the compound and the ions present in a saturated
aqueous solution
- Solubility product constant, Ksp the equilibrium
constant expression for the dissolving of a
slightly soluble solid.
Ksp Ba2 SO42
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3- Example 16.1
- Write a solubility product constant expression
for equilibrium in a saturated aqueous solution
of the slightly soluble salts (a) iron(III)
phosphate, FePO4, and (b) chromium(III)
hydroxide, Cr(OH)3.
4Ksp and Molar Solubility
- Ksp is an equilibrium constant
- Molar solubility is the number of moles of
compound that will dissolve per liter of
solution. - Molar solubility is related to the value of Ksp,
but molar solubility and Ksp are not the same
thing. - In fact, smaller Ksp doesnt always mean lower
molar solubility. - Solubility depends on both Ksp and the form of
the equilibrium constant expression.
5- Example 16.2
- At 20 C, a saturated aqueous solution of silver
carbonate contains 32 mg of Ag2CO3 per liter of
solution. Calculate Ksp for Ag2CO3 at 20 C. The
balanced equation is - Ag2CO3(s) 2 Ag(aq) CO32(aq)
Ksp ? - Example 16.3
- From the Ksp value for silver sulfate, calculate
its molar solubility at 25 C. - Ag2SO4(s) 2 Ag(aq) SO42(aq)
- Ksp 1.4 x 105 at 25 C
6- Example 16.4 A Conceptual Example
- Without doing detailed calculations, but using
data from Table 16.1, establish the order of
increasing solubility of these silver halides in
water AgCl, AgBr, AgI.
7The Common Ion Effectin Solubility Equilibria
- The common ion effect affects solubility
equilibria as it does other aqueous equilibria. - The solubility of a slightly soluble ionic
compound is lowered when a second solute that
furnishes a common ion is added to the solution.
8Common Ion Effect Illustrated
When Na2SO4(aq) is added to the saturated
solution of Ag2SO4
Ag attains a new, lower equilibrium
concentration as Ag reacts with SO42 to produce
Ag2SO4.
Calculate the molar solubility of Ag2SO4 in 1.00
M Na2SO4(aq).
9Will Precipitation Occur? Is It Complete?
- Qip is the ion product reaction quotient and is
based on initial conditions of the reaction.
Qip and Qc new look, same great taste!
- Qip can then be compared to Ksp.
- Precipitation should occur if Qip gt Ksp.
- Precipitation cannot occur if Qip lt Ksp.
- A solution is just saturated if Qip Ksp.
- In applying the precipitation criteria, the
effect of dilution when solutions are mixed must
be considered.
10- Example 16.6
- If 1.00 mg of Na2CrO4 is added to 225 mL of
0.00015 M AgNO3, will a precipitate form? - Ag2CrO4(s) 2 Ag(aq) CrO42(aq)
- Ksp 1.1 x 1012
- Example 16.8
- If 0.100 L of 0.0015 M MgCl2 and 0.200 L of 0.025
M NaF are mixed, should a precipitate of MgF2
form? - MgF2(s) Mg2(aq) 2 F(aq) Ksp
3.7 x 108
11- Simple explanation problem
- Pictured here is the result of adding a few drops
of concentrated KI(aq) to a dilute solution of
Pb(NO3)2. What is the solid that first appears?
Explain why it then disappears.
12To Determine Whether Precipitation Is Complete
- A slightly soluble solid does not precipitate
totally from solution - but we generally consider precipitation to be
complete if about 99.9 of the target ion is
precipitated (0.1 or less left in solution). - Three conditions generally favor completeness of
precipitation
- A very small value of Ksp.
- A high initial concentration of the target ion.
- A concentration of common ion that greatly
exceeds that of the target ion.
13- Example 16.9
- To a solution with Ca2 0.0050 M, we add
sufficient solid ammonium oxalate, (NH4)2C2O4(s),
to make the initial C2O42 0.0051 M. Will
precipitation of Ca2 as CaC2O4(s) be complete?
CaC2O4(s) Ca2(aq) C2O42(aq)
Ksp 2.7 x 109
14Effect of pH on Solubilityfly in the Ointment
- If the anion of a precipitate is that of a weak
acid, the precipitate will dissolve somewhat when
the pH is lowered
Added H reacts with, and removes, F
LeChâteliers principle says more F forms.
- If, however, the anion of the precipitate is that
of a strong acid, lowering the pH will have no
effect on the precipitate.
H does not consume Cl acid does not affect
the equilibrium.
15- Example 16.11
- What is the molar solubility of Mg(OH)2(s) in a
buffer solution having OH 1.0 x 105 M, that
is, pH 9.00?
Mg(OH)2(s) Mg2(aq) 2 OH(aq) Ksp
1.8 x 1011
Another one of those explain problems Without
doing detailed calculations, determine in which
of the following solutions Mg(OH)2(s) is most
soluble (a) 1.00 M NH3 (b) 1.00 M NH3 /1.00 M
NH4 (c) 1.00 M NH4Cl.
16Neutralization Reactions
- At the equivalence point in an acidbase
titration, the acid and base have been brought
together in precise stoichiometric proportions. - The endpoint is the point in the titration at
which the indicator changes color. - Ideally, the indicator is selected so that the
endpoint and the equivalence point are very close
together. - The endpoint and the equivalence point for a
neutralization titration can be best matched by
plotting a titration curve, a graph of pH versus
volume of titrant.
17Titration Curve, Strong Acid with Strong Base
Bromphenol blue, bromthymol blue, and
phenolphthalein all change color at very nearly
20.0 mL
At about what volume would we see a color change
if we used methyl violet as the indicator?
18- Example 15.20
- Calculate the pH at the following points in the
titration of 20.00 mL of 0.500 M HCl with 0.500 M
NaOH - H3O Cl Na OH ? Na Cl
2 H2O - (a) before the addition of any NaOH
- (b) after the addition of 10.00 mL of 0.500 M
NaOH - (c) after the addition of 20.00 mL of 0.500 M
NaOH - (d) after the addition of 20.20 mL of 0.500 M
NaOH
19Titration Curve, Weak Acid with Strong Base
The equivalence-point pH is NOT 7.00 here. Why
not??
Bromphenol blue was ok for the strong acid/strong
base titration, but it changes color far too
early to be useful here.
20Sample problem for WA/SB titration
- The titration of 100.0 mL of 0.016 M HOCl (Ka
3.5 x 10-8 with 0.0400 M NaOH. How many mL of
0.040 M NaOH are required to reach the
equivalence point? - Calc the pH after addition of 10.0 mL NaOH
- Halfway to equivalence point
- At equivalence point.