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SOLUBILITY EQUILIBRIA

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Assume x 6.5 X 10-3. 1.8 X 10-10 = (6.5 X 10-3) (x) x = 1.8 X 10-10 divided by ... a) Assume the final volume is 20.0 mL. b) The Ksp for AgCl is 1.8 X 10-10 ... – PowerPoint PPT presentation

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Title: SOLUBILITY EQUILIBRIA


1
Chapter 18
SOLUBILITY EQUILIBRIA
2
  • THE SOLUBILITY PRODUCT CONSTANT

A) The solubility product constant (Ksp) is the
equilibrium constant for the solubility of a
slightly soluble ionic compound.
1) MA(s) ?M(aq) A-(aq) Ksp M A-
3
2) Example AgCl(s) ?Ag(aq) Cl-(aq) Ksp
Ag Cl-
What is the Ksp for AgCl if 1.88 X 10-3 g of AgCl
dissolves to make 1.0 L of solution? Ksp Ag
Cl-
4
Ksp _____________________________
5
3) MA2(s) ? M2(aq) 2A-(aq) Ksp M2
A-2 SrF2(s) ? Sr2(aq) 2F-(aq)
Ksp Sr2 F-2 What is the Ksp of SrF2 if the
solubility of SrF2 is 5.8 X 10-4 M? Ksp Sr2
F-2
Ksp ____________________________
6
At 25oC, the Ksp if Ag2CrO4 is 1.1 X 10-12. What
is the solubility of Ag2CrO4 in grams per L?
Ag2CrO4(s) ?2 Ag(aq) CrO42-(aq) Ksp Ag2
CrO42-
We have to make a chart like we did in the last
chapter.
7
Ksp 2 x2 x Ksp 1.1 X 10-12 4x2 x 4x3
8
x 6.5 X 10-5 mol/L That is not the answer yet.
We have to change the moles to grams by
multiplying by the formula weight.
6.5 X 10-5 mol/L X 331.8 g/mol __________
9
II. COMMON ION EFFECT AND SOLUBILITY
A) At a given temperature, the solubility of a
given compound is decreased by the common ion
effect. Ksp remains the same whether or not other
substances are present in the solution.
10
B) Example AgCl(s) ?Ag(aq) Cl-(aq) 1) If
we add silver ions from another source such as
AgNO3, we will shift the equilibrium to the left
and more AgCl will precipitate from solution
until the ion product is equal to the Ksp.
11
2) Calculate the solubility of AgCl in g/L in a
6.5 X10-3 M silver nitrate solution. The Ksp for
AgCl is 1.8 X 10 10. AgNO3 ?Ag(aq) NO-3(aq)
AgCl(s) ?Ag(aq) Cl-(aq)
12
1.8 X 10-10 (6.5 X 10-3 x) (x)
Assume x ltlt 6.5 X 10-3
1.8 X 10-10 (6.5 X 10-3) (x)
x 1.8 X 10-10 divided by 6.5 X 10-3
x 2.8 X 10-8 (assumption is ok.)
13
The solution must be 2.8 x 10-8 M in Cl- and all
these ions come from AgCl. So only 2.8 X 10-8
moles of AgCl can dissolve in the 1 L of 6.5 X
10-3 M AgNO3 solution.
2.8 X 10-8 mol/L X 143.4 g/mol 4.0 X 10-6g/L
In pure water, the solubility of AgCl is 1.9 X
10-3 g/L.
14
The Ksp of CaCO3 is 3.8 X 10-9. Estimate its
solubility in 0.10 M Na2CO3. Na2CO3 ? 2 Na
CO32-
CaCO3(s) ?Ca2(aq) CO32-(aq)
15
Ksp 3.8 X 10-9 Ca2 CO32-
3.8 X 10-9 (x) (0.10 x) assume x ltlt 0.10 Ksp
3.8 X 10-9 (x) (0.10) 3.8 X 10-8 x
Assumption is ok. What does this answer mean?
16
III. PREDICTING PRECIPITATION REACTIONS
QUANTITATIVELY
A) Earlier we based precipitation reactions on
the basis of solubility rules and metathesis
reactions.
B) Now we will base our prediction on the Ksp and
the knowledge from the first semester.
C) Will a precipitate form if 10.0 mL of 0.010 M
AgNO3 solution and 10.0 mL of 1.0 X 10-4 M NaCl
are mixed?
17
1) Some information you need to know
a) Assume the final volume is 20.0 mL.
b) The Ksp for AgCl is 1.8 X 10-10
2)What does this number have to do with the
problem?
AgCl is the probable precipitate, since the
reaction which might occur is
AgNO3(aq) NaCl(aq)__? AgCl(s) Na(aq)
NO3-(aq)
18
3) You must recall that mixing the solutions
dilutes the species and changes the concentration
of each species.
e) In order to predict whether or not a
precipitate will form you must calculate Qc , the
ion product, for the reaction and compare it with
Ksp.
19
If Qc is less than Ksp then no precipitate
occurs.
If Qc is greater than Ksp then a precipitate is
expected to form.
If Qc is equal to Kc then a saturated solution
forms, solid is in equilibrium with the ions.
20
For the reaction of Ag (aq) Cl- (aq) ?
AgCl(s)
we need to obtain the concentrations of Ag and
Cl- to calculate Qc.
Qc Ag Cl- (5.0 X 10-3) (5.0 X 10-5)
2.5 X 10-7
21
2.5 X 10-7 is greater than 1.8 X 10-10 therefore
______________________ .
IV. pH AND SOLUBILITY
A) The solubilities of many substances depend on
the pH of the solution.
Example Mg(OH)2(s) ? Mg2(aq)
2 OH-(aq)
1) Adding OH- to the reaction mixture decreases
the solubility and shifts the reaction to the
left.
22
If the pH 12.00, then the pOH 2.00, the OH-
1.00 x 10 -2.
Ksp Mg2 OH-2 (1.4 X 10-4) (2 X 1.4 X
10-4)2 1.1 X 10-11 1.1 X 10-11 Mg2 1.00 x
10-22
Mg2 1.1 X 10-11 divided by 1.0 X 10-4 1.1
X 10-7
To what does this number refer?
23
Of what is this another example ?
24
The End
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