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## PowerPoint Slideshow about ' Solubility Review' - madaline-osborn

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Review

1. Solubility is a measure of the maximum amount of solid that will dissolve in a volume of water.

Units: g/Lmol/Lg/100mL

Or even: mL/L for CO2(g) in water

Ksp

Solubility Product

Saturated solutions

No Units

Only Changes with Temperature

No ICE!

Step #1 Write a dissociation equation for the low solubility salt

2. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp.

Dissociation net ionic BaCO3(s)⇌ Ba2+ + CO32-

s s s

Ksp = [Ba2+][CO32-]

Ksp = s2

Ksp = (5.1 x 10-5)2

Ksp = 2.6 x 10-9

Adding BaCO3does notincrease [Ba2+] or [CO32-] but it does increase the rate of dissolving and crystallization.

Formation Net Ionic: Ba2+ + CO32- BaCO3(s)

Ba2+

CO32-

BaCO3(s)

2. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp.

BaCO3(s)⇌ Ba2+ + CO32-

s s s

Ksp = [Ba2+][CO32-]

Ksp = s2

Ksp = (5.1 x 10-5)2

Ksp = 2.6 x 10-9

Adding BaCO3does notincrease [Ba2+] or [CO32-] but it does increase the rate of dissolving and crystallization.

Formation Net Ionic:Ba2+ + CO32- BaCO3(s)

Ba2+

CO32-

BaCO3(s)

3. PbCl2(s)⇌ Pb2+ + 2Cl-

Ksp = 4s3

4. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur?

PbCl2(s)⇌ Pb2+ + 2Cl-

200 0.10 M 300 0.20 M 500 500

0.040 M 0.12 M

TIP = [Pb2+][Cl-]2

TIP = [0.040][0.12] 2

= 5.8 x 10-4

Ksp = 1.2 x 10-5 TIP > Ksp ppt forms

5. Changing the Solubility of a Salt

E +Ca(OH)2(s)⇄ Ca2+ + 2OH-

Substance Added Effect on Molar Solubility Effect on Ksp

Ca(NO3)2Decrease none

Na2SO4Increase none

NaOHDecrease none

HCl Increase none

NaCl none none

Ca(OH)2none none

Increase Temperature Increase Increase

Decrease TemperatureDecrease Decrease

6. Mg(OH)2 will have the greatest solubility in:

Mg(OH)2(s) ⇌ Mg2+ + 2OH-

A. NaOH OH- lowers solubility

B. Mg(NO3)2 Mg2+ lowers solubility

C. H2O No effect solubility

D. AgNO3Ag+ increases solubility by reacting with OH-

7. Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

AgCl(s)⇄ Ag+ + Cl-

0.20 M

Ksp = [Ag+][Cl-]

1.8 x 10-10 = [0.20][Cl-]

[Cl-] = 9.0 x 10-10 M

BaCl2(s)⇄ Ba2+ + 2Cl-

4.5 x 10-10 M9.0 x 10-10 M

0.50 L x 4.5 x 10-10mole x 208.3 g = 4.7 x 10-8 g

1 L mole

8. Ionic Solutions:NaCl HCl NH4NO3

Molecular Solutions:C12H22O11

9. What is the answer?

A. NaCl

B. CaCl2

C. AgCl

D. AgBr

10. Ionic Solutions: NaCl HCl NH4NO3

Molecular Solutions: C12H22O11

11. What is the answer?

A. NaCl High

B. CaCl2 High

C. AgCl Low 1.8 x 10-12

D. AgBrLow 5.4 x 10-13

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