Solubility Equilibria And The Solubility Product Constant (Chemistry Worksheet) Page 3
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1
2
3THE COMMON ION EFFECT
2+
2-
Ag
CrO
⇌ 2 Ag
+ CrO
2
4(s)
(aq)
4
(aq)
If AgNO
is added the addition of silver ions will shift the equilbrium to the left in favour of
3(aq)
precipitation. The
is about reducing the solubility (forming a
precipitate) by the addition fo a common ion in solution.
Example Problem 5: Calculating Effect of Common Ions
The molar solubility of solid calcium fluoride CaF
in a 0.025 mol/L solution of NaF
at 25 C.
2(s)
(aq)
Determine the maximum amount of solid CaF
that will dissolve into the NaF
.
2
(aq)
1-
Step 1: Identify the Common Ion – Common Ion is Fluoride Ion, F
Step 2: Determine the Concentration of the Common Ion in NaF
(Solubility Chart suggests this is a Soluble Compound)
+
-
-
-
Na
NaF
+ F
, 1:1 Ratio between NaF and F
, therefore [F
] = 0.025 mol/L
(s)
(aq)
(aq)
Step 3: Determine the concentration of the common ion once the solid has dissolved.
According to the solubility chart, NaF has a low solubility and therefore will have an equilibrium
in solution.
2+
-
-
CaF
⇌ Ca
+ 2F
[F
] = 0.025 mol/L - The fluoride ions will shift this equilibrium to the
2(s)
(aq)
(aq)
left so that less CaF
dissolves.
2
Step 4: Create an ICE table
2+
-
CaF
Ca
+
2F
⇌
2(s)
(aq)
(aq)
I
-
0
0.025
C
-
+x
+2x
E
-
+x
0.025+2x
Step 5: Solve for x using the K
equation. Look up the K
value. Determine whether or not to
sp
sp
follow the 100 Rule.
2+
-
2
K
= [Ca
][F
]
sp
(aq)
(aq)
x will give us the amount of calcium ions that dissolve in this equilibrium system and there is a
2+
1:1 ratio of Ca
ions and CaF
. Therefore x will give us the molar solubility.
2
HOMEWORK: Read Pages 460-470 and do Review Questions 1-3, 5, 7, 9-11 on Page 471
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