aqueous equilibria entry task: feb 17 th wednesday notes on precipitate and ions hw: precipitate and...

AqueousEquilibria

Entry Task: Feb 17th Wednesday

Notes on Precipitate and ions

HW: Precipitate and ions ws

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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

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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

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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

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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

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7. Write the net reaction for the neutralization of a solution of HF with a solution of NaOH. HF + OH- F- + H2O

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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

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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

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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

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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

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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

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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

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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-

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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.

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• Complex IonsThe formation of

these complex ions increases the solubility of these salts.

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• 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+.

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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

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Chapter 17Additional Aspects of Aqueous Equilibria

Sections 6

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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

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BaSO4(s)Ba2+(aq) + SO42-(aq)

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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

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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.

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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

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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

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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

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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

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(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

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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

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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

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AqueousEquilibria


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

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AqueousEquilibria


(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

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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

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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?

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AqueousEquilibria


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

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AqueousEquilibria


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

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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.

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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

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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

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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.

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AqueousEquilibria

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aqueous equilibria entry task: feb 17 th wednesday notes on precipitate and ions hw: precipitate and...

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