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AqueousEquilibria
Entry Task: Feb 17th Wednesday
Notes on Precipitate and ions
HW: Precipitate and ions ws
MAYHAN
AqueousEquilibria
I can…• Explain the factors that can affect the Ksp of a
substance• Calculate the changes that occurs when these
factors are applied to a solution
MAYHAN
AqueousEquilibria
AqueousEquilibria
Information from the Curve:There are several things you can read from the titration curve itself. Consider this titration curve.
1.What type of titration curve is above? Identify the titrant.
2. Place a dot () on the curve at the equivalence point. The pH at the equivalence point is ____. Choose a good indicator for this titration from Figure 16.07 on page 604 of your textbook.
Strong base into a weak acid. The titrant is the strong base
9
Phenolphthalein
AqueousEquilibria
Information from the Curve:There are several things you can read from the titration curve itself. Consider this titration curve.
3. What volume of base was used to titrate the acid solution? _______ mL 4.Place a box () on the curve
where the pH of the solution = the pKa of the acid.
What is the pH at this point? _____What is the pKa of the acid? _____
What is the Ka of the acid?
25
4.84.8
1.6 x10-5
AqueousEquilibria
Calculations knowing the Acid:
5. Hydrofluoric acid, HF, has a Ka = 7.2 x 10-4. Calculate the pH of 10.0 mL of a 0.050 M solution of HF. Plot this point on the axes.
x2 0.050M
= 7.2 x 10-4 x2= 3.6 x 10-5
6.0 x 10-3 = pH = 2.22
AqueousEquilibria
Calculations knowing the Acid:
6. A 0.020 M solution of NaOH is used for the titration. What volume will be needed to reach the equivalence point?
(10 mls)(0.050M) = (x mls)(0.020M)
= 25.0 mls
AqueousEquilibria
Calculations knowing the Acid:
7. Write the net reaction for the neutralization of a solution of HF with a solution of NaOH. HF + OH- F- + H2O
AqueousEquilibria
Calculations knowing the Acid:
8. Calculate the moles of F- at the equivalence point. What is the total volume? _______ L
The [F-] at the equivalence point is _________
10 mls (HF) + 25.0 mls (NaOH) = 35.0 mls
0.0350
(x)(35.0 ml) = (0.020M)(25.0ml)
0.0143 M
AqueousEquilibria
Calculations knowing the Acid:
9. Calculate the pH of the solution at the equivalence point. Use this information and the answer to question 6 to plot the equivalence point on your graph. Choose a good indicator for this titration from Figure 16.07 on page 604 of your textbook.
(10.0 mls)(0.050M) = 0.50 mmol HF(25.0 mls)(0.020M) = 0.50 mmol NaOH
The acid was neutralized but the c-base is left at the same amount of OH- which is 0.50 mmol.
0.50 mmol/35.0 ml = 0.0143 M
X2
0.0143 M= 1.389 x10-11
Since its more base- change Ka to Kb 1.0 x10-14 / 7.2 x10-4
x2 = 1.99 x10-13
x= 4.46 x10-7 6.35 - 14 = pH = 7.65
AqueousEquilibria
Calculations knowing the Acid:
9. Calculate the pH of the solution at the equivalence point. Use this information and the answer to question 6 to plot the equivalence point on your graph. Choose a good indicator for this titration from Figure 16.07 on page 604 of your textbook.
0.0143 M
Bromthymol blue
AqueousEquilibria
Calculations knowing the Acid:
10. What is the pH halfway to the equivalence point? Plot this point on your graph.
pH is pKa = 3.14 at 12.5 mls
AqueousEquilibria
Calculations knowing the Acid:
11. How many moles of HF are in the original 10.0 mL sample of HF? _______ (0.010 L)(0.050mol/L) = 0.00050 moles of HF
AqueousEquilibria
Calculations knowing the Acid:
12. When only 5.0 mL of 0.020 M NaOH has been added, calculate the moles of HF left and F- produced.
(0.0050 L)(0.020 mol/L) = 0.00010 moles of OH-
HF OH- H2O F-
i -------
c -------
e -------
0.000500.00010
-0-
-0.00010-0.00010 +0.00010
0.00040 0 0.00010
0.00040 moles of HF 0.015Liters
= 0.0267 M of HF
AqueousEquilibria
Calculations knowing the Acid:
12. When only 5.0 mL of 0.020 M NaOH has been added, calculate the moles of HF left and F- produced.
(0.0050 L)(0.020 mol/L) = 0.00010 moles of OH-
HF OH- H2O F-
i -------
c -------
e -------
0.000500.00010
-0-
-0.00010-0.00010 +0.00010
0.00040 0 0.00010
0.00010 moles of F 0.015Liters
= 0.00667 M of F-
AqueousEquilibria
AqueousEquilibria
AqueousEquilibria
AqueousEquilibria
AqueousEquilibria
AqueousEquilibria
Factors Affecting Solubility
• pH If a substance has a
basic anion, it is more soluble in an acidic solution.
Substances with acidic cations are more soluble in basic solutions.
MAYHAN
AqueousEquilibria
Factors Affecting Solubility
• Complex IonsThe formation of
these complex ions increases the solubility of these salts.
MAYHAN
AqueousEquilibria
Factors Affecting Solubility
• Amphoterism Amphoteric metal
oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
Examples of such cations are Al3+, Zn2+, and Sn2+.
MAYHAN
AqueousEquilibria
I can…• Explain the ion product and how its value will
affect the equilibrium.• Predict whether a precipitate will form when two
ions are mixed in solution
MAYHAN
AqueousEquilibria
Chapter 17Additional Aspects of Aqueous Equilibria
Sections 6
MAYHAN
AqueousEquilibria
17.6- Precipitation and Separation of Ions
Up till now we have placed ionic substances in water so it can dissociate until its saturated and reaches equilibrium.
BaSO4(s) Ba2+(aq) + SO42-(aq)
Equilibrium can also be reached by mixing two solutions containing the ions to create a precipitate
MAYHAN
BaSO4(s)Ba2+(aq) + SO42-(aq)
AqueousEquilibria
17.6- Precipitation and Separation of Ions
By mixing BaCl2 and Na2SO4 together will create a precipitate of BaSO4 if the product of the ion concentrations, Q = [Ba2+][SO4
2-], greater than Ksp.
BaSO4(s)Ba2+(aq) + SO42-(aq)
Q is referred to simply as the ion product
MAYHAN
AqueousEquilibria
Will a Precipitate Form?
• In a solution, If Q = Ksp, the system is at equilibrium and the
solution is saturated. If Q < Ksp, more solid will dissolve until Q = Ksp.
If Q > Ksp, the salt will precipitate until Q = Ksp.
MAYHAN
AqueousEquilibria
17.6- Precipitation and Separation of Ions
SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0 103 M Pb(NO3)2 is added to 0.40 L of 5.0 103 M Na2SO4?
What precipitate COULD be made?_______________
Look up the Ksp in Appendix D:_________________
To determine if PbSO4 will precipitate, we have to calculate the ion product, Q=[Pb+2][SO4
-2] and compare it with Ksp.
PbSO4
6.3 x10-7
MAYHAN
AqueousEquilibria
SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0 103 M Pb(NO3)2 is added to 0.40 L of 5.0 103 M Na2SO4?
In 0.10 L of 8.0 103 M Pb(NO3)2 there are:
(0.10 L) (8.0 103) =
8.0 x 10-4 moles of Pb+2 ions
MAYHAN
AqueousEquilibria
SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0 103 M Pb(NO3)2 is added to 0.40 L of 5.0 103 M Na2SO4?
In 0.40 L of 5.0 103 M Na2SO4 there are:
(0.40 L) (5.0 103) =
2.0 x 10-3 moles of SO4-2 ions
MAYHAN
AqueousEquilibria
SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0 103 M Pb(NO3)2 is added to 0.40 L of 5.0 103 M Na2SO4?
We have to convert the moles in to molarity but use the combined volume.
2.0 x 10-3 moles/0.50L=
8.0 x 10-4 moles/0.50L = 1.6 x10-3 of Pb+2 ions
4.0 x10-3 of SO4-2 ions
Substitute the values into the Ksp expression and solve for Q
MAYHAN
AqueousEquilibria
SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0 103 M Pb(NO3)2 is added to 0.40 L of 5.0 103 M Na2SO4?
(1.6 x10-3)(4.0 x10-3) =
Q = [Pb+2][SO42]
6.4 106
Q= 6.4 106 Ksp= 6.3 x10-7
Q is larger than Ksp that means
A precipitate will occurMAYHAN
AqueousEquilibria
Does a precipitate form when 0.050 L of 2.0 102 M NaF is mixed with 0.010 L of 1.0 102 M Ca(NO3)2
What precipitate COULD be made?___________
Look up the Ksp in Appendix D:_______________
CaF2
3.9 x10-11
MAYHAN
AqueousEquilibria
Does a precipitate form when 0.050 L of 2.0 102 M NaF is mixed with 0.010 L of 1.0 102 M Ca(NO3)2
In 0.050 L of 2.0 102 M NaF there are:
(0.050 L) (2.0 102) =
1.0 x 10-3 moles of F-1 ions
MAYHAN
AqueousEquilibria
Does a precipitate form when 0.050 L of 2.0 102 M NaF is mixed with 0.010 L of 1.0 102 M Ca(NO3)2?
In 0.010 L of 1.0 102 M Ca(NO3)2 there are:
(0.010 L) (1.0 102) =
1.0 x 10-4 moles of Ca+2 ions
MAYHAN
AqueousEquilibria
Does a precipitate form when 0.050 L of 2.0 102 M NaF is mixed with 0.010 L of 1.0 102 M Ca(NO3)2
We have to convert the moles in to molarity but use the combined volume.
1.0 x 10-3 moles/0.060L=
1.0 x 10-4 moles/0.060L = 1.67 x10-3 of Ca+2 ions
1.67 x10-2 of F-1 ions
Substitute the values into the Ksp expression and solve for Q
MAYHAN
AqueousEquilibria
Does a precipitate form when 0.050 L of 2.0 102 M NaF is mixed with 0.010 L of 1.0 102 M Ca(NO3)2
(1.67 x10-3)(1.67 x10-2)2 =
Q = [Ca+2][F1]2
4.7 107
Q= 4.7 107 Ksp= 3.9 x10-11
Q is larger than Ksp that means
A precipitate will occur
MAYHAN
AqueousEquilibria
Consider a solution containing Ag+ ions and Cu+2 ions then HCl is added, AgCl precipitate will form and the CuCl2 is soluble and remain as ions in solution.
• Separation of ions in an aquesous solution by using a reagent that forms a precipatate with one or few of the ions is called selective precipatation
MAYHAN
AqueousEquilibria
A solution contains 1.0 102 M Ag+ and 2.0 102 M Pb2+. When Cl is added, both AgCl (Ksp = 1.8 1010) and PbCl2 (Ksp = 1.7 105 M) can precipitate. What concentration of Cl is necessary to begin the precipitation of each salt? Which salt precipitates first?
We know that both Ag+ and Pb+2 would form a precipitate with Cl- but which will form first?
MAYHAN
AqueousEquilibria
A solution contains 1.0 102 M Ag+ and 2.0 102 M Pb2+. When Cl is added, both AgCl (Ksp = 1.8 1010) and PbCl2 (Ksp = 1.7 105 M) can precipitate. What concentration of Cl is necessary to begin the precipitation of each salt? Which salt precipitates first?
Lets look at Ag+ with CI-: Ksp = [Ag+][Cl-]
1.8 x10-10 = (1.0 x10-2)(x) = Cl- ions
1.8 x10-10 = 1.0 x10-2
1.8 x 10-8 Cl- ions
MAYHAN
AqueousEquilibria
A solution contains 1.0 102 M Ag+ and 2.0 102 M Pb2+. When Cl is added, both AgCl (Ksp = 1.8 1010) and PbCl2 (Ksp = 1.7 105 M) can precipitate. What concentration of Cl is necessary to begin the precipitation of each salt? Which salt precipitates first?
Lets look at Pb+2 with CI-: Ksp = [Pb+][Cl-]2
1.7 x10-5 = (2.0 x 10-2)(x)2 = Cl- ions
1.7 x10-5 = 2.0 x 10-2
x2=8.5 x 10-4 Cl- ions
x=2.9 x 10-2 Cl- ionsMAYHAN
AqueousEquilibria
A solution contains 1.0 102 M Ag+ and 2.0 102 M Pb2+. When Cl is added, both AgCl (Ksp = 1.8 1010) and PbCl2 (Ksp = 1.7 105 M) can precipitate. What concentration of Cl is necessary to begin the precipitation of each salt? Which salt precipitates first?
Which concentration is smaller?
2.9 x 10-2 Cl- ions with Pb+2
1.8 x 10-8 Cl- ions with Ag+
This means that it will AgCl precipitate as such a small concentration verses PbCl2.
MAYHAN
AqueousEquilibria
A solution consists of 0.050 M Mg2+ and Cu2+. Which ion precipitates first as OH is added?
What concentration of OH is necessary to begin the precipitation of each cation? [Ksp = 1.8 1011 for Mg(OH)2, and Ksp = 4.8 1020 for Cu(OH)2.]
Lets look at Mg+2 with OH-: Ksp = [Mg+2][OH-]2
1.8 x 10-11 = (0.050)(x)2 = OH- ions
1.8 x 10-11 = 0.050
x2= 3.6 x 10-10
x= 1.9 x 10-5 OH- ions
AqueousEquilibria
A solution consists of 0.050 M Mg2+ and Cu2+. Which ion precipitates first as OH is added?
What concentration of OH is necessary to begin the precipitation of each cation? [Ksp = 1.8 1011 for Mg(OH)2, and Ksp = 4.8 1020 for Cu(OH)2.]
Lets look at Cu+2 with OH-: Ksp = [Cu+2][OH-]2
4.8 x 10-20 = (0.050)(x)2 = OH- ions
4.8 x 10-20 = 0.050
x2= 9.6 x 10-19
x= 9.8 x 10-10 OH- ions
AqueousEquilibria
A solution consists of 0.050 M Mg2+ and Cu2+. Which ion precipitates first as OH is added?
What concentration of OH is necessary to begin the precipitation of each cation? [Ksp = 1.8 1011 for Mg(OH)2, and Ksp = 4.8 1020 for Cu(OH)2.
9.8 x 10-10 OH- ions with Cu+2
1.9 x 10-5 OH- ions with Mg+2
Which concentration is smaller?
This means that it will Cu(OH)2precipitate as such a small concentration verses Mg(OH)2.
MAYHAN