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Previous Knowledge – 30S Chem – Solutions,
Unit 1, and Equilibrium
Content – p. 482 - 493
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*CYU
* What is solubility?
* When is a substance soluable? When is it not?
* What is a saturated solution? Non-Saturated solution?
* What factors affect solubility?
* What is a molar solution?
* What is ionization?
*
* Substances chemists believed to be insoluble in water, by
experiment, were shown to be slightly soluble.
* With sensitive instruments, chemists we able to show
that substances that would not dissolve in water actually
do dissolve to a very small extent.
* Dissolving is a dynamic equilibrium between the ions
formed and the substance in its whole or non-ionized
form
*
* As we learned in Acids and Bases, a strong acid or base is
one that ionized 100%. In other words, all ionized.
* The equilibrium favours the formation of product.
* Weak acids and bases do not ionize 100% and only
partially, therefore, putting only a little hydrogen ion or
hydroxide ion into solution.
* The equilibrium favours the formation of reactant.
*
* For salts, when a solution is saturated, there is a point
when the production of ions (dissolving) is in equilibrium
with the ions coming back together to form the salt.
* This is called dynamic equilibrium Animation
* The more ions that form (ie. Very soluable), the greater
the equilibrium constant (Keq is greater than 1)
* The less ions that form (ie. Low solubility), the lower the
equilibrium constant (Keq is less than 1)
*
* Very soluable substance, dissociate 100%
Ex. NaCl
Na+1 +
Cl-1
(100 % ions until it is saturated)
* Slightly soluable substance, does not dissociate to the same extent (ie.
Less ions in solution that NaCl, at equilibrium)
Ex. AgCl
Ag+1
+
Cl-1
* Notice the arrows, which is favoured at equilibrium
* But some does dissolve, just a very small amount goes to ions.
*
* When a sparingly soluble ionic solid is dissolved in water to
form a saturated solution the general equilibrium equation is:
AaBb(s)
aA+ (aq) + bB¯
(aq)
* Where A is a positively charged ion and B is a negatively
charged ion.
* For very soluable substances, we know that 100% will form ions
at equilibrium
*
* At a given temperature, the equilibrium law for this reaction is given
as:
Keq = [A+]a[B-]b
[AaBb]
* However, the term Keq[AaBb] can be replaced by a new constant, Ksp,
called the solubility product.
Ksp = [A+]a[B¯]b
* The solubility product constant is the product of ion concentrations in
a saturated solution. The solubility product constant takes into account
the presence of the solid.
*
* Ksp is the same as Keq and Ka / Kb, except it is the measure of the ion
concentration at equilibrium (Saturation)
Precipitates
* For highly soluable salts, Ksp
will be greater than 1
* We don’t care about these salts, because for the most part, we can
put a large amount into solution, and ions will form.
* The slightly soluable salts are the ones we are concerned about
* For slightly soluable salts, Ksp
will be less than 1
* Why are the slightly soluable salts a concern when we look at nature?
*
*Assignment
* You must come up with one example of how slightly soluable
salts would affect us in our daily lives and how they are used?
* Examples – City of Winnipeg Water, Kidney stones, osteoporosis,
tooth decay and prevention, limestone caves, Lake Winnipeg
*Due next class
*
* Example 1
* Write the dissociation equation and the expression for the
solubility product constant for calcium hydroxide.(Ksp = 1.3 x 10-6)
Solution
Write the chemical formula for calcium hydroxide, Ca(OH)2.
Write the dissociation equation, remembering that the subscript indicates two
hydroxide ions dissociate for every calcium ion:
Ca(OH)2(s)
Ca2+(aq) + 2 OH¯(aq)
When writing the solubility product expression, the molar coefficient in the
dissociation equation becomes the exponent.
Ksp = [Ca2+][OH¯]2
* Example 2
* Write the solubility product expression for lead (II) phosphate
(Pb3(PO4)2).
* Example 3
* What are the concentrations of ions for a saturated solution of
calcium hydroxide? (Ksp = 1.3 x 10-6)
* Example 4
* What is the Ksp value for barium sulfate, if at equilibrium,
barium sulfate has a concentration of 3.87 x 10-5 M?
* Example 5
* If a saturated solution of magnesium fluoride has a Mg+2 ion
concentration of 1.17 x 10
–3
M, what is the Ksp?
*
* Solubility and solubility product are two different terms.
* Solubility is the maximum amount of solute that can dissolve in a
certain amount of solvent at a certain temperature. Solubility has
an infinite number of possible values, depending on temperature
and other solutes present.
* Solubility product is an equilibrium constant and has only one
value for a given solid at a particular temperature.
* Example 6
* The solubility of PbF2 is 0.466 g/L. What is the value of the
solubility product?
Step 1: Determine the number of moles/L of PbF2
Mols = mass/ MM
Mols = 0.466g/245.20g/mol
So, the concentration of PbF2 is 1.90 x 10-3 mol/L.
Step 2: Write the dissociation equation.
PbF2(s)
Pb2+(aq) + 2 F¯(aq)
* Step 3: Calculate ion concentrations.
* The concentration of PbF2 an Pb2+ ions will be equal since their
stoichiometry is 1:1.
* So, [Pb2+] = 1.90 x 10-3 mol/L
* From the stoichiometry,
* [F¯] = 2 x [Pb2+] = 3.80 x 10-3 mol/L
* Step 4: Write the solubility product expression and calculate.
* Ksp = [Pb2+][F¯]2
* Ksp = (1.90 x 10-3)(3.80 x 10-3)2
* Ksp = 2.74 x 10-8
* Therefore, the solubility product of PbF2 is 2.74 x 10-8.
* Example 7
* The Ksp of magnesium hydroxide is 8.9 x 10-12. What will be the
equilibrium concentrations of the dissolved ions in a saturated
solution of Mg(OH)2?
* Example 8
* What is the hydroxide concentration in a saturated solution of
iron (III) hydroxide?
* (Ksp = 4.00 x 10-38)
Answer = 5.89 x 10-10 M
* Example 9
* What mass of silver (I) sulfate could be dissolved in 500 mL of
water? (Ksp = 1.20 x 10-5)
* Answer = 2.18 g
*
* Insoluble compounds are generally described as those that
precipitate upon mixing equal volumes of solutions which are
0.10 mol/L in the respective ions.
* A precipitate is the solid that forms when two solutions are
mixed. A precipitate will usually cause a mixture to appear
cloudy.
* In previous chemistry courses you have learned the general rules
for determining combinations of ions that have a low solubility.
*
* Example
* Which of the following ions will precipitate with a solution of
sodium chloride?
* Copper (II) ion
Silver ions
Lead (II) ion
Sulfide ions
* Use the solubility rules to figure out the answer.
*
* Not all reactions produce a precipitate. If the volume of solution is
large enough, and the amount of solute is small enough no
precipitate will form.
* We will use the reaction quotient, Q, to predict whether a
precipitate forms.
* The form of the reaction quotient is the same as the Ksp, just as it
was with Keq and Ka/b values.
* By comparing the value of the reaction quotient with the solubility
product, we can determine if an aqueous solution is saturated or
unsaturated.
*
*If,
*Q = Ksp, the solution is just saturated and no precipitate
forms.
*Q>Ksp, the solution is saturated and a precipitate forms.
*Q<Ksp, the solution is unsaturated and no precipitate forms.
*We determine the Q value, from comparing our
experimental concentrations to the Ksp value for the
saturated solution
*
* Example
* The Ksp of lead (II) chloride is 1.6 x 10-5. If 0.57 g of lead (II)
chloride are added to 1500 mL of water, is the solution saturated?
Assume no volume change.
*
* Things to consider for determining precipitates:
* The initial solutions must be soluable, or you do not have a solution
* The chemical reaction is a double displacement reaction
* The concentrations of the ions are what the stoichiometry states by the
dissociation of the compound
* When solutions are mixed, the volume changes, and so does the
concentration of the ion
* The net ionic equation will tell you what the Ksp expression should be
* Find the Q and compare to the literature Ksp value
*
* Example
* If 20.0 mL of a 0.0010 mol/L silver nitrate solution is mixed with 20.0 mL
of a 3.0 x 10-5 mol/L potassium bromide solution, does silver bromide
(Ksp = 5.0 x 10-13) precipitate? Assume the volumes are additive.
* Example 2
* 100 ml of a 0.01 M potassium hydroxide solution is combined with 50 ml
of a 0.1 M copper (II) nitrate solution. Will a precipitate form?
* (Ksp for the precipitate = 1.60 x 10 –19)
*
* When an ionic compound dissolves in pure water, the initial
concentration of each ion is zero.
* However, if an ionic compound dissolves in a solution that has an
ion in common with the compound, this is not the case.
* Even though the starting concentrations may not be zero, the
product of the ions must still equal the solubility product
constant.
* For example, how would the solubility of silver chloride in pure
water change if we try dissolving it in tap water?
*
Let's write out the dissociation equation:
AgCl(s)
Ag+(aq) + Cl¯(aq)
Tap water often has chlorine added to kill bacteria. The chlorine exists as
chloride ions, so when we dissolve silver chloride in tap water, chloride ions
are present.
According to Le Chatelier's Principle, Adding more chloride ions to a
saturated solution would cause the equilibrium to shift to the left to use up
the excess product. This would result in more solid formed, and a decreased
solubility.
*
* Le Chatelier's Principle predicts that the solubility of an ionic solid
in a solution containing a common ion decreases its solubility.
* Let's see if this is supported by the calculations.
* Example
* Determine the solubility of silver chloride in pure water and in a
solution of 0.10 mol/L sodium chloride.
The Ksp of AgCl is 1.7 x 10-10.
*
* Step 1
* Find the solubility of the ions in pure water at saturation
* Step 2
* Set up an ICE Chart for the addition of the NaCl solution.
What is the
common ion? What is its concentration?
* Step 3
* Substitute the values from your ICE chart into your Ksp equation.
* Step 4
* Solve for x. Since x is small, disregard as it will have little impact on
the final answer.
* Step 5
* Find the value of solubility of the NaCl, using the chloride ion value
*
* The Ksp of lead (II) chloride, PbCl2, is 1.6 x 10-5. What is the
solubility of lead (II) chloride in a 0.10 mol/L solution of
magnesium chloride, MgCl2?
* MgCl2 is a highly soluable salt, with 100% ionization.
Therefore,
the solubility of PbCl2 will be less, due to the common ion in
solution.