Entry Task: March
th
5
Tuesday
Question:
(a)Calculate the solubility of AgI in water in
moles per liter given its Ksp value of 8.3 x
10-17.
(b)A saturated solution of AgI also containing
NaI is found to have an iodide ion
concentration of 0.020 M. What is the
concentration of silver ions?
You have 10 minutes
Aqueous
Equilibria
Agenda:
• Discuss Solubility, Precipitation and Ions ws
• Major self-check on content so far & discuss it
• HW: Pre-lab Determine Ksp
Aqueous
Equilibria
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17.37 A 1.00L solution is saturated at 25°C with
calcium oxalate, CaC2O4, is evaporated to dryness,
giving a 0.0061 g residue of CaC2O4. Calculate the
solubility-product constant for this salt.
CaC2O4(s)
Ca2+(aq) + C2O42-(aq)
Calculate the molarity
0.0061g/127.99 = 4.77 x10-5M of CaC2O4
Ksp = [4.77 x10-5][4.77 x10-5]
Ksp = 2.3
x10-9
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17.41 Calculate the solubility of Mn(OH)2 in grams per
liter when buffered at
(a) at pH of 7.0;
Mn(OH)2(s)
Mn2+(aq) + 2OH1-(aq)
pH of 7 means there are 1.0 x10-7 OH ions
1.6x10-13= [x][1.0 x10-7]2
1.6x10-13 = [x]
The molarity is 16M
1.0 x10-14
(16)(1 L) = 16 moles * 89 =
1424 or 1.4 x103g/ liter
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17.41 Calculate the solubility of Mn(OH)2 in grams per
liter when buffered at
(b) at pH of 9.5;
Mn(OH)2(s)
Mn2+(aq) + 2OH1-(aq)
pH of 9.5= pOH 4.5 means there are 3.2 x10-5 OH ions
1.6x10-13= [x][3.2 x10-5 ]2
1.6x10-13 = [x]
The molarity is 1.6x10-4M
1.0 x10-9
(1.6x10-4)(1 L) = 1.6x10-4 moles * 89 =
1.4x10-2g/ liter
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17.41 Calculate the solubility of Mn(OH)2 in grams per
liter when buffered at
(c) at pH of 11.8;
Mn(OH)2(s)
Mn2+(aq) + 2OH1-(aq)
pH of 11.8= pOH 2.2 means there are 6.3 x10-3 OH ions
1.6x10-13= [x][6.3 x10-3]2
1.6x10-13 = [x]
The molarity is 4.02x10-9M
3.98 x10-5
(4.02x10-9)(1 L) = 4.02x10-9 moles * 89 =
3.6x10-7g/ liter
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17.49 (a) Will Ca(OH)2 precipitate from solution if the
pH of a 0.050M solution of CaCl2 is adjusted to 8.0?
Ksp= 6.5x10-6= [Ca+]OH-]2
pH of 8 = pOH of 6 = 1.0x10-6M of OH
Q = [0.050]1.0x10-6]2
Q = 5.0 x 10-14
Ksp = 6.5 x 10-6
Ksp is bigger meaning no precipitate
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M
AgNO3 is mixed with 10 mls of 5.0 x10-2M NaSO4
solution?
In 0.100 L of 0.050 M AgNO3 there are
(0.100 L) (0.050M) =
5.0 x 10-3 moles of Ag+1 ions
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M
AgNO3 is mixed with 10 mls of 5.0 x10-2M Na2SO4
solution?
In 0.010 L of 5.0 x10-2M NaNO3 there are
(0.010 L) (5.0 x10-2M) =
5.0 x 10-4 moles of SO4-2 ions
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M
AgNO3 is mixed with 10 mls of 5.0 x10-2M Na2SO4
solution?
• We have to convert the moles in to molarity but use
the combined volume.
5.0 x 10-3 moles of Ag+1 ions
5.0 x 10-4 moles of SO4-2 ions
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M
AgNO3 is mixed with 10 mls of 5.0 x10-2M Na2SO4
solution?
• We have to convert the moles in to molarity but use
the combined volume.
5.0 x10-3 moles/0.110L =
4.55 x10-2 M of Ag+1 ions
5.0 x10-4 moles/0.110L = 4.55 x10-3 M of SO4-2 ions
Substitute the values into the Ksp expression and
Aqueous
solve for Q
Equilibria
17.6- Precipitation and
Separation of Ions
17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M
AgNO3 is mixed with 10 mls of 5.0 x10-2M Na2SO4
solution?
Q = [Ag+]2[SO42]
(4.55 x10-2)2(4.55 x10-3) = 9.4  106
Q= 9.4  106
Ksp= 1.5 x 10-5
Q is smaller than Ksp that means
No precipitate will occur
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17. 53 A solution contains 2.0 x10-4M Ag+ and
1.5 x 10-3 M Pb+2. If NaI is added, will AgI (Ksp=
8.3x10-17) or PbI2 (Ksp= 7.9 x10-9) precipitate
first? Specify the concentration of I- needed to
begin precipitation.
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17. 53 A solution contains 2.0x10-4M Ag+ and 1.5 x 10-3 M Pb+2.
If NaI is added, will AgI (Ksp= 8.3x10-17) or PbI2 (Ksp= 7.9 x10-9)
precipitate first? Specify the concentration of I- needed to begin
precipitation.
Lets look at Ag+ with I-:
Ksp = [Ag+][l-]
8.3 x10-17 = (2.0 x10-4)(x) = l- ions
8.3 x10-17 =
2.0 x10-4
4.2 x 10-13 l- ions
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17. 53 A solution contains 2.0x10-4M Ag+ and 1.5 x 10-3 M Pb+2.
If NaI is added, will AgI (Ksp= 8.3x10-17) or PbI2 (Ksp= 7.9 x10-9)
precipitate first? Specify the concentration of I- needed to begin
precipitation.
Lets look at Pb+2 with I-:
Ksp = [Pb+][l-]2
7.9 x10-9 = (1.5 x 10-3)(x)2 = l- ions
7.9 x10-9 =
1.5 x 10-3
x2=5.3 x 10-6 l- ions
x=2.3 x 10-3 l- ions
Aqueous
Equilibria
17.6- Precipitation and
Separation of Ions
17. 53 A solution contains 2.0x10-4M Ag+ and 1.5 x 10-3 M Pb+2.
If NaI is added, will AgI (Ksp= 8.3x10-17) or PbI2 (Ksp= 7.9 x10-9)
precipitate first? Specify the concentration of I- needed to begin
precipitation.
Which concentration is smaller?
4.2 x 10-13 l- ions with Ag+
2.3 x 10-3 l- ions- ions with Pb+2
This means that it will Agl precipitate as such a small
concentration verses Pbl2.
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria