Transcript Unit 4 Solutions

Unit 4 Solutions
4.3.1 SOLUBILITY CURVES
4.3.2 SOLUBILITY TABLES
4.3.3 SOLUBILITY CONSTANT
4.3.4 PRECIPITATION
4.3.5 COMMON ION EFFECT
4.3.1 Solubility Curves
 Solubility curves tell us what mass of solute will
dissolve in 100g (or 100mL) of water over a range of
temperatures.
 For most substances, solubility increases as
temperature increases. Can you find any exceptions
on the chart?
 Here's an example of reading the chart. Find the
curve for KClO3.
 At 30°C approximately 10g of KClO3 will dissolve in
100g of water. If the temperature is increased to
80°C, approximately 40g of the substance will
dissolve in 100g of water.
Here are some for you to try.
What mass of solute will dissolve in 100mL of
water at the following temperatures.
2. Determine which of the three substances is most
soluble in water at 15°C.
 1. KNO3at 70°C
 2. NaCl at 100°C
 3. NH4Cl at 90°C
1.
Solution:
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Solubility
1. KNO3at 70°C 140g / 100mL
2. NaCl at 100°C 40g / 100mL
3. NH4Cl at 90°C 72g / 100mL
4. NaCl is the most soluble at 15°C
 The curves indicate the concentration of a
saturated solution – (the maximum amount of
solute that will dissolve at that specific temperature).
 The molar concentration of the substance can be
calculated, example:
 Determine the molarity of a saturated NaCl solution
at 25°C.
Solution
 We can see from the curve that about 38 g of NaCl
dissolves in 100mL at 25°C. We want to convert 38g
/100mL-1 to mol·L-1.
 First let's convert mL to L:
 Next we need to convert from grams to moles, we
will need to use the molar mass of NaCl, which is
58.5 g·mol-1
Solubility Chart
 Values below the curve represent unsaturated
solutions – i.e more solute could be dissolved at that
temperature.
 Values above a curve represent supersaturated
solutions, i.e a solution which holds more solute that
can normally dissolve in that volume.
Some examples:
 What term - saturated, unsaturated, or
supersaturated - best describes:
 a solution that contains 70g of NaNO3 per 100 mL
H2O at 30°C
 a solution that contains 60g of dissolved KCl per 100
mL H2O at 80°C
Solution
 The NaNO3 solution is unsaturated. At 30°C a
saturated solution would be able to dissolve
approximately 95 g of NaNO3 . Since there are only
70g in the solution, 25 more grams of NaNO3 could
dissolve.
 The KCl solution is supersaturated. At 80°C a
saturated KCl solution contains 50 g KCl per 100 mL
H2O. This solution is holding 10 g of excess KCl.
3.2 Solubility Tables & the Solubility of
Ionic Compounds
 Solubility tables allow us to predict whether a certain
ionic compound will dissolve in water at 25°C.
 Reading the table is not difficult. It tells which
positive cations are soluble when combined with
negative anions
Soluble or not???
 NaCl,
 CH3COO-  BaSO4
 Cu(NO3)2
 Pb(CH3COO)2
 PbCl2
Saturated Solutions & Equilibrium
 It is important to remember that saturated systems are
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equilibrium systems.
For example, the equation illustrating a saturated
solution of silver sulfate would be:
Ag2SO4 (s) 2Ag+(aq) + SO42-(aq)
remember that equilibrium means "equal rates" not
"equal amounts".
in a saturated solution at equilibrium you will not be able
to detect any changes in concentration because the solid
dissolves into ions at the same rate that the ions form
back into a solid (crystallization).
Equilibrium does NOT mean that the concentrations of
reactants and products are equal.
 Practice problems 4.3.2
4.3.3 The Solubility Product Constant, Ksp
 Recall the equilibrium constant expression for any
general reaction:
 aA + bB cD + dD
 Keq= [C]c × [D]d
[A]a × [B]b
 We can apply this mathematical relationship to
solutions. We will refer to our equilibrium constant
as Ksp, where "sp" stands for "solubility product"
 For our silver sulfate saturated solution:
 Ag2SO4 (s)
2Ag+(aq) + SO42-(aq) we can write our
solubility product constant expression as
 Ksp = [Ag+]2[SO42-]
[Ag2SO4]
 Remember that the concentrations of solids and
liquids are NOT included in the expression because
their concentrations remain constant.
 Therefore, we will write our solubility product
constant expression for our saturated silver sulfate
solution as:
 Ksp= [Ag+]2[SO42-]
 Please note that it is VERY IMPORTANT to
include the ion charges when writing this equation
Try this example:
 Write the expression for the solubility product
constant, Ksp, for Ca3(PO4)2.
 Step 1: balance equation.
 Ca3(PO4)2 (s)
3 Ca2+(aq) + 2 PO43-(aq)
 Step 2: Write the expression for Ksp:
 Ksp= [Ca2+]3[PO43-]2
 Solubility Product Tables give Ksp values for various
ionic compounds (keep it handy).
 Because temperature affects solubility, values are
given for specific temperatures (usually 25°C).
 Ksp provides us with useful information:
1. A low value of Ksp indicates a substance that has a
low solubility (or insoluble); for ionic compounds
this means that there will be few ions in solution.
 Iron(II) sulfide, FeS, is an example of a low Ksp :
Ksp = 4 ×10-19
 In a saturated solution of FeS there would be few
Fe2+ or S2- ions.
2. A large value of Ksp indicates a soluble substance;
for ionic compounds there will be many ions in
solution.
 An example lead(II) chloride, PbCl2 Ksp = 1.8 ×10-4
 A saturated solution of PbCl2 would have a high
concentration of Pb 2+ and Cl - ions.
1. Calculating Ksp when concentration of a
saturated solution is known
 The concentration of a saturated solution of BaSO4
is 3.90 × 10-5 M. Calculate Ksp for barium sulfate at
25°C
Solution
1.
Write a balanced equation: BaSO4 (s)
(aq)
Ba2+(aq) + SO42-
Next, write the Ksp expression: Ksp= [Ba2+][SO42-]
Find the concentration of the individual ions for our
equation.
 since 1 mole of BaSO4 produces 1 mole of Ba2+ and also 1 mole
of SO42-, then . . .
 [BaSO4] = 3.90 × 10-5M,
 [Ba2+] = 3.90 × 10-5M,
 [SO42-] = 3.90 × 10-5M
4. Substitute values into the Ksp expression and solve
 Ksp = [Ba2+][SO42-] = (3.90 × 10-5)(3.90 × 10-5)
 Ksp = 1.52 × 10-9 Answer
2.
3.
Example
 The concentration of lead ions in a saturated
solution of PbI2 at 25°C is 1.3 × 10-3 M. What is its
Ksp?
Solution
1. Write a balanced equation:
PbI2 (s)
Pb2+(aq) + 2 I-(aq)
2. Write the Ksp expression: Ksp= [Pb2+][I-]2
3. Determine the concentration of the ions.
[PbI2] = 1.30 × 10-3M; [Pb2+] = 1.30 × 10-3M;
[I-] = 2 × 1.30 × 10-3 = 2.60 × 10-3
4. Solve the Ksp expression:
 Ksp = [Pb2+][I-]2
 = (1.30 × 10-3)(2.60 × 10-3)2
 = (1.30 × 10-3)(6.76× 10-6)
 Ksp = 8.79 × 10-9M
Calculating ion concentrations when Ksp is
known.
 Ksp for MgCO3 at 25°C is 2.0 × 10-8. What are the ion
concentrations in a saturated solution at this
temperature?
Solution
1. Balance equation: MgCO3(s) Mg2+(aq) + CO32-(aq)
2. Write the Ksp expression: Ksp= [Mg2+][CO32-]
3. Let the unknown ion concentrations equal x.
 The balanced equation tells us that both Mg2+ and
CO32- will have the same concentration!
4. Substitute values into the equation and solve for the
unknown:
 Ksp = [Mg2+][CO32-]
 2.0 × 10-8 = (x) (x) = x2
 X2 = 2.0 × 10-8
 X = √2.0 × 10-8 = 1.4× 10-4M
 So, x = [Mg2+] = 1.4× 10-4M
and x = [CO32-] = 1.4× 10-4M
4.3.4 Precipitation Reactions
 Sometimes when two aqueous solutions are mixed
together a solid is produced. This solid is called a
precipitate, and the reaction is known as a
precipitation reaction.
 We can use the Solubility Table to predict whether a
precipitate will form.
Example
 ZnBr2 (aq) + 2 AgNO3 (aq) → Zn(NO3)2 + 2 AgBr
 But if we look at our solubility table, we see that Ag+
and Br- forms an insoluble compound and form a
solid.
 On the other hand, we see in our solubility table that
Zn2+ and NO3- form a soluble compound. Thus these
two ions will remain in solution
 We now need to return to our double displacement
reaction and add the physical states to the two
products:
 ZnBr2 (aq) + 2 AgNO3 (aq) → Zn(NO3)2 (aq) + 2 AgBr(s)
 Let’s write our solutions out in long form, remember -
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anything that is (aq) is really present as separate ions.
This, however, ONLY applies to aqueous states.
Thus we can rewrite our equation as:
Zn2+(aq) + 2 Br-(aq) + 2 Ag+(aq) + 2 NO3-(aq) →
Zn2+(aq) + 2 NO3-(aq)+ 2 AgBr(s)
Notice that some ions are the same on both sides of the
equation, indicating they did not change during the
reaction.
Ions that are present in a reaction but do not participate
are called spectator ions.
 We can remove spectator ions from equations to
highlight what changes occurred during a reaction.
 Reactions that the spectator ions have been removed
are called net ionic equations.
 Therefore, for the overall equation:
Zn2+(aq) + 2 Br-(aq) + 2 Ag+(aq) + 2 NO3-(aq) → Zn2+(aq) +
2 NO3-(aq)+ 2 AgBr(s)
 The net ionic equation is:
 2 Ag+(aq) + 2 Br-(aq) → 2 AgBr(s)
 Keep in mind that a reaction does not always take place.
 Let's look at what happens when we mix solutions of
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Mg(CH3COO)2 and (NH4)2SO4.
Write a balanced equation and predict if the products.
Then predict the physical states of the products
(reactants are aq.)
Mg(CH3COO)2 (aq) + (NH4)2SO4 (aq)→ MgSO4 + 2
NH4CH3COO
We find that both MgSO4 and NH4CH3COO are soluble
compounds, meaning they do not form precipitates.
Thus, there is NO REACTION occurring, all
participants simply remain in solution!
 Lab 4.3.4
Selective Precipitation
 Knowledge of precipitation reactions also allows us to use
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selective precipitation to separate mixtures of ions.
For example . . .
We have a solution that contains the following cations:
Ag+, Cu2+, and Mg2+ ; each of these cations is associated with
nitrate ions, NO3-.
Compounds containing nitrate are ALWAYS soluble. So the
nitrate ions will remain spectator ions, so we will not need to
consider them
We wish to isolate each ion by causing each one to precipitate
out of solution. Once a solid forms, we can filter the solid
precipitate out, leaving the other ions in solution.
We are given the following solutions to use:
Na2S, NaCl, and NaOH.
 Sodium ions, Na+, always form soluble compounds.
So Na+ will remain as a spectator ion . So we are
given solutions of the following anions:
 S2-, Cl-, and OH We add these one at a time to our cation mixture. We
need to determine the order so that we have ONE
AND ONLY ONE precipitate occur each addition,
removing one of the cations from the solution. Then
continue until all cations are isolated.
 Prepare a chart, with the cations we need to separate
along one axis and anions along the other axis:
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Ag+
Cu2+
Mg2+
S2-
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Cl-
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OH-
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 Use a table of solubility to determine which
combinations produce soluble compounds and which
ones form a precipitate (ppt - precipitate).
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Ag+
Cu2+
Mg2+
S2-
ppt
ppt
sol
Cl-
ppt
sol
sol
OH-
ppt
ppt
ppt
 We need to determine the correct order for adding the negative ions
 Only by adding Cl- first can we separate out one of the ions, Ag+.
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Our first precipitate would be AgCl(s). Note that both S2- and OHalso form a precipitate with Ag+, but it has been removed using Cl- Now add S2- to again form a single precipitate - CuS(s), thus
removing Cu2+
This will leave only Mg2+ still in solution. Sometimes this last cation
is allowed to remain in solution, if add OH- we get our final
precipitate, Mg(OH)2 (s)
To summarize:
First add NaCl to remove Ag+ , forming the precipitate AgCl(s)
Second, add Na2S to remove Cu2+, forming CuS(s)
Last add NaOH to remove Mg2+, forming Mg(OH)2 (s)
 Practice problems 4.3.4
 Assignment 4.3.4