Chapter 16 Worksheet 3 (ws16.3)
Quantitative Analysis of Solubility Equilibria
Solubility of Ionic Compunds
In chapter 4, some solubility guidelines for ionic compounds were given in Table 4.2. A similar table
from another textbook is given below. These compounds are classified as either soluble or insoluble.
In reality many ionic compounds classified as insoluble are slightly soluble. We will now develop the
tools to handle solubility quantitatively using equilibrium constants for dissolution reactions. Such an
equilibrium constant is called Ksp where sp stands for solubility product. Some Ksp values are listed in
table 16.2 in your textbook and reproduced at the end of this worksheet.
We are interested in calculating the solubility of slightly soluble salts such as silver carbonate.
Ksp is called the “solubility product” but it is NOT the solubility of the compound. It is the equilibrium
constant for the dissolving (dissolution) reaction. For example:
Ag2CO3 (s) = 2Ag+(aq) + CO32−(aq) Ksp = [Ag+]2[ CO32−] = 8.1 x 10−12
When the reaction is at equilibrium, the solution is saturated.
The solubility of an ionic compound is its concentration in a saturated solution (its equilibrium
concentration). This is not quite right since after an ionic compound dissolves, 100% of it dissociates.
In the reaction above, the solubility of silver carbonate is actually the equilibrium concentration of the
carbonate ion or ½ of the silver ion concentration.
Will a precipitate form?
5. In chapter 4, you studied precipitation reactions in which two solutions of soluble salts were mixed
to produce at least one insoluble salt.
a. Use table 5.1 to predict whether or not a precipitate will form when a solution oflead (II) nitrate
is mixed with a solution of sodium chloride.
II I ~( u~ t~ /I
ft,cl2, t<:.
~
i;sHtd e~.)1 pl ~fe .
(+
b. Write the molecular, total ionic, and net ionic equations for the reaction. . f)\OL"-lO~)2 (<:l.<1) +;ltJq(t(Ctq) --t P\o(\2 (5) l.}JQ~ (Qq '\
MoIecuIar. I _ . / Total ionic: PI.,'l+ J+ 2
'\ {. 2<:"-(<:1<1)""""
*
~"1
~CQq)~ ~-\-~<{
Pl?"2.+COq '
Net ionic:
2Cl-(o.C{)
of
---+
_J... _
~dz.(S)t ~4-(o.q)1 ~ G'Ci<Z
pl:,C(z.(")\
(re~iS~t G~ d6c:d\.Jl,",~ ~~ct-(<l'-'
f(~G::
U:<;p> )
c. In this type of problem one ofthe reactants is often a sodium salt and the other reactant is often
a nitrate. Why? (Consult table 5.1).
ScdlU""",, t<JV)
",,~{'C"-\~
Q-
(0'-'" C<::JW\()GU....o'S
CJ'~.
q(W<;"\:f)
'S.~./u~t~ ~ q
6. In the previous problem, it was assumed that a precipitate would form because PbCh is classified as
"insoluble". Although the solubility ofPbCh is very small, it is not zero. In general, a reaction
between two soluble salts produces a precipitate only when the amount of product formed exceeds the
solubility of the compound. To predict whether or not a precipitate will form, simply compare Q to
Ksp:
If Q < Ksp, no precipitate will form
If Q > Ksp, a precipitate will form
If Q = Ksp, the solution is saturated (no precipitate will form)
a. Look up Ksp for PbCh in table 16.2 and calculate its solubility. «s{' -;:. 2.'-\ X l6
-q
f'o7;)'\c
=t
( '1 ¥-\
.
2
2C t - C:>"~)
G
C
X
+- 2')( ><
~)( (X ').:: ~)(
~
-Lf
= 2.~ >(1 G
____ -2 -...
J.C\. ~\<J fV'\\
4
)
b. If S.OO mL of 0.300 M lead (II) nitrate is mixed with 10.0 mL ofO.lS0 M sodium chloride, a
precipitate ofPbChforms as predicted above. However, ifS.OO mL of 0.0300 M lead (II)
nitrate is mixed with 10.0 mL of 0.0 ISO M sodium chloride, no precipitate forms. Explain these
observations by calculating Q for each reaction and compare Q to Ksp PbCh. The calculations
for the more concentrated solutions are done for you. Do the same calculations for the less
concentrated solutions.
Calculations for the more concentrated solutions
Since two solutions are mixed, you must calculate the initial concentration of each salt using the
dilution equation: M JV 1 M2V 2 (M J and V I are the molarity and volume before dilution. M2 and V 2
are the molarity and volume after dilution.)
[Pb2+]
=
[Pb(N03h] = MjVj = (0.300 M)(S.OO mL)
IS.0mL
V2
(0.1 SO [cr] = [NaCl]
,~
0.100 M
0.100M
IS.0mL
f.J~~: w~ q( ~d'j
14. C\.""'7~'l
t:: ...c v
Tt..,., QX:c.~Qc) S
't~ '?JOt.J.ol/lhj Cqkvla-W
In r~r-r q ~
PbCh (s) ~ Pb2+ (aq) + 2Cr (aq)
(Remember: Ksp is the equilibrium constant for the dissolving reaction, not the precipitation reaction!) Q = (0.100)(0.100i = 1.00 x 10-3 Q > Ksp so the dissolving reaction will proceed in the reverse direction and a precipitate will form. Calcuations for the less concentrated solutions r
~ Ill",> ~"5 ~+
Cpl,
[
~~1 .::
a. 0
~
lOel M
i
+y
Q.KG~ ~ :;oluL. t ,
,
Cl-) -::; 0,0,00 M
dQ_::
?J (0. (j teo J -=-
-(
!.oo ')( Ie>
Q<K
Sa
~ r ~G( Q(tC.h::- ~I
((
~r~ ~
5
7. Will a precipitate fonn if90.0 mL of 0.200 M tin (II) nitrate is added to 10.0 mL of 0.200 M
sodium iodide? (K,,>p ror- i",V) (odlde. f~ (.0 X:{O~4
J
SII\ (}-jo 3'1- (q~l) + J. 10".L (r~C'1 ') ~
•
~q ]003 (0q 1 + S~::r: 2 ( ~ I
~ 'Sdlth/~ S'"
:[1. (5')
~
Cc,.l,-U(CI\r-~ Q
Q:;
:;VI::z.-+~C\ ~ -+ 2:r -(qct )
~
COtfVl.-pd..A?_
i"<l
I<c,( -;: 1.0 ,><lO-~
IC.sf J..
[Sl'12.T
16 [ 1-30 ~ ~ (o.2.00 jYl) ~
6\0 0
M
tWIA.L
Q <. \C,:> f No
r~(JI f' ~c>).- ~
/pi
II ~V"IN\ .
)
Common-Ion Effect: Solubility of ionic compounds
Note: The "common-ion effect" is nothing new! It is simply a name for what happens when you
add a product to a reaction at equilibrium. (It shifts to the left.) In this case the added product
happens to be an ion!
5.0 x 10-7) in water.
8. a. Calculate the molar solubility ofPbh (Ksp
rb'l't:-(~O ') -l- II­ (<:;9')
dissolution reaction:
-
I
C
o
0
+x
E
( ><. ') (z\(."";'::;
s"
Solubility
t{)("5::
S, ox t C.. .
7
)(=I~
b~CJ X l6- ~
M
<;'\..lfY'll~ S
7
b. Calculate the molar solubility of PbI, (K"
E
VJ =
~5
------
l,Z.C-;:'XlO-5 M
C\. 7')0lr"~
( X ) (C.2D)
2
-'7
S"'..O'KtG
s==- X ~ t L1.5Xt6 M
.
~
O:.2.. G-t-ZX
x: '"-
(X J(G:Z.O-'+"Lx'
\~ JOU
v
42K
t-x.
C
Solubility
JA :r
o
I
I
Zd
P\o 1: 1... (s,) -;;: 'Pb'2..+(a<i) t z ':r:: -GCl~ ')
dissolution reaction:
'\
5.0 x 10. ) inE
o
J. .
(Mucl--
l~S5 '":tJlJ,l~~~
<3.. 2.0 +2.'< --:: 0:2.,0 :
--,
:: 0,040 X' .:::.. S' ,0,<t6
I...nr
('SQJ."..Q QV\S UJ~
I )
v
)
7
The effect of pH on the solubility of ionic compounds
9. For many compounds, such as Ca(OH)2 or CaC0 3, solubility will vary with pH ofthe solution. This
is because the anions, OR and C0 32-, are bases. Using LeChatelier's principle, predict how the
solubility of Ca(OH)2 will vary with each of the following changes.
Ca(OH)2 (s) ~ Ca2+ (aq) + 2 OH- (aq)
a. Lowering the pH of the solution by adding a strong acid, HCl.
'" +
CGI'I"'5UIM.Q')
C"'\C(j~\
to:.
CM -)
~~GTICJ'" '5l.l~ts
~ ~~tJJ..E2..
tv\ CX(dlC
---=)
~lu{.tGV) ~V\
I
V"1
pO~ ~~r,
b. Raising the pH of the solution by adding a strong base, NaOH.
[0
\-.t- J tv--c:~qVS) ~~ cJ IO\--, ~L}."is ~
Ca.(oi1)z
{<":>
J~~S ~{JrJ~
lV\
100.51<. '5old--16'1
tk~
Iv-.
pU~ LJJ.~r"\
7. Circle the compounds that are more soluble in acidic solution than in pure water.
GI(OH?3)NaCI~aF0CuBr, NaCI03~ NaN03~
o
H- )
~)
So <:>dO. ~~
ct-,
B("-)
C l Oz.-
) f'YJz.-
Q II
o,<..d ~,n-S fQ",~t-IC'"
CIG~-) ~
CGu-. ~~<::1\+~
~ rl<
ON<
~Ctd'S Q~ "'Sfro
V'C"+
J
roOSl <: QIII IGI/lS,
- 4 (ecore:
~(J-/e: ) bc13Z..~ ~CQlUS~ ~\ ('
atlaS. 8