Lesson 4 Calculating Molar Solubility From Ksp

Post on 13-Jan-2016

37 views 4 download

description

Lesson 4 Calculating Molar Solubility From Ksp. 1.Calculate the molar solubility @ 25 o C for BaCrO 4 in units of g/L BaCrO 4(s) ⇌ Ba 2+ +CrO 4 2- sss Ksp=[Ba 2+ ][CrO 4 2- ] Ksp= s 2 from page 5 1.2 x 10 -10 = s 2 s= 1.1 x 10 -5 M - PowerPoint PPT Presentation

Transcript of Lesson 4 Calculating Molar Solubility From Ksp

Lesson 4Calculating

Molar Solubility From Ksp

1. Calculate the molar solubility @ 25oC for BaCrO4 in units of

g/L

BaCrO4(s) ⇌ Ba2+ + CrO42-

s s sKsp = [Ba2+][CrO4

2-]

Ksp = s2

from page 5 1.2 x 10-10 = s2

s = 1.1 x 10-5 M

note that solubility units are M!

1.1 x 10-5 mole x 253.3 g = 0.0028 g/L L 1 mole

2. Calculate the molar solubility @ 25oC for Cu(IO3)2 in units

of g/L

Cu(IO3)2 ⇌ Cu2+ + 2IO3-

s s 2s

Ksp = [Cu2+][IO3-]2

Ksp = [s][2s]2

Ksp = 4s3

6.9 x 10-8 = 4s3

s = 2.6 x 10-3 M

note sig figs are 2 like the Ksp!

2.584 x 10-3 moles L

2.584 x 10-3 moles x 413.2 gL 1 mole

2.584 x 10-3 moles x 413.2 g = 1.1 g/LL 1 mole

3. Calculate the molar solubility @ 25oC for Fe(OH)3. Calculate

the mass required to prepare 2.0 L of the above saturated solution.

Fe(OH)3 ⇌ Fe3+ + 3OH-

s s 3s

Ksp = [Fe3+][OH-]3

Ksp = [s][3s]3

2.6 x 10-39 = 27s4

s = 9.906 x 10-11 M

2.0 L

3. Calculate the molar solubility @ 25oC for Fe(OH)3. Calculate

the mass required to prepare 2.0 L of the above solution.

Fe(OH)3 ⇌ Fe3+ + 3OH-

s s 3s

Ksp = [Fe3+][OH-]3

Ksp = [s][3s]3

2.6 x 10-39 = 27s4

s = 9.906 x 10-11 M

2.0 L x 9.906 x 10-11 moles 1 L

3. Calculate the molar solubility @ 25oC for Fe(OH)3. Calculate

the mass required to prepare 2.0 L of the above solution.

Fe(OH)3 ⇌ Fe3+ + 3OH-

s s 3s

Ksp = [Fe3+][OH-]3

Ksp = [s][3s]3

2.6 x 10-39 = 27s4

s = 9.906 x 10-11 M

2.0 L x 9.906 x 10-11 moles x 106.8 g 1 L mole

3. Calculate the molar solubility @ 25oC for Fe(OH)3. Calculate

the mass required to prepare 2.0 L of the above solution.

Fe(OH)3 ⇌ Fe3+ + 3OH-

s s 3s

Ksp = [Fe3+][OH-]3

Ksp = [s][3s]3

2.6 x 10-39 = 27s4

s = 9.906 x 10-11 M

2.0 L x 9.906 x 10-11 moles x 106.8 g = 2.1 x 10-8 g 1 L mole

The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility.  4. Indicate the solid with the greatest solubility.  

PbSO4

 ZnS

 AgCl

 BeS

The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility.  4. Indicate the solid with the greatest solubility.  

PbSO4 start on page 4

 ZnS

 AgCl

BeS

The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility.  4. Indicate the solid with the greatest solubility.  

PbSO4 low start on page 4

 ZnS

 AgCl

BeS

The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility.  4. Indicate the solid with the greatest solubility.  

PbSO4 low start on page 4

 ZnS low

 AgCl

BeS

The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility.  4. Indicate the solid with the greatest solubility.  

PbSO4 low start on page 4

 ZnS low

 AgCl low

BeS

The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility.  4. Indicate the solid with the greatest solubility.  

PbSO4 low start on page 4

 ZnS low

 AgCl low

BeS high

5. Indicate the solid with the least solubility.  

PbSO4

 ZnS

 AgCl

 BeS

5. Indicate the solid with the least solubility.  

PbSO4 low start on page 4

 ZnS low

 AgCl low

 BeS high

5. Indicate the solid with the least solubility.  

PbSO4 low start on page 4

 ZnS low use page 5 if required

 AgCl low

 BeS high

5. Indicate the solid with the least solubility.  

PbSO4 low 1.8 x 10-8 start on page 4

 ZnS low use page 5 if required

 AgCl low

 BeS high

5. Indicate the solid with the least solubility.  

PbSO4 low 1.8 x 10-8 start on page 4

 ZnS low 2.0 x 10-25 use page 5 if required

 AgCl low

 BeS high

5. Indicate the solid with the least solubility.  

PbSO4 low 1.8 x 10-8 start on page 4

 ZnS low 2.0 x 10-25 use page 5 if required

 AgCl low 1.8 x 10-10

 BeS high

5. Indicate the solid with the least solubility.  

PbSO4 low 1.8 x 10-8 start on page 4

 ZnS low 2.0 x 10-25 use page 5 if required

 AgCl low 1.8 x 10-10

 BeS high

6. How many of the following salts could produce a solution with a concentration more than 0.10 M?

  Greater than 0.10 M means high solubility!

FeSO4

 ZnSO4  

Al2(SO4)3

 Na2SO4

6. How many of the following salts could produce a solution with a concentration more than 0.10 M?

  Greater than 0.10 M means high solubility!

FeSO4 high

 ZnSO4

 Al2(SO4)3

 Na2SO4

6. How many of the following salts could produce a solution with a concentration more than 0.10 M?

  Greater than 0.10 M means high solubility!

FeSO4 high

 ZnSO4 high

 Al2(SO4)3

 Na2SO4

6. How many of the following salts could produce a solution with a concentration more than 0.10 M?

  Greater than 0.10 M means high solubility!

FeSO4 high

 ZnSO4 high

 Al2(SO4)3 high

 Na2SO4

6. How many of the following salts could produce a solution with a concentration more than 0.10 M?

  Greater than 0.10 M means high solubility!

FeSO4 high

 ZnSO4 high

 Al2(SO4)3 high

 Na2SO4 high

6. How many of the following salts could produce a solution with a concentration more than 0.10 M?

  Greater than 0.10 M means high solubility!

FeSO4 high

 ZnSO4 high

 Al2(SO4)3 high

 Na2SO4 high four!

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] =

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles 1 L

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + 1Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

s = 0.1091 M

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

s = 0.1091 M

B. Ksp

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

s = 0.1091 M

B. Ksp Ba(OH)2 ⇌ Ba2+ + 2OH-

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

s = 0.1091 M

B. Ksp Ba(OH)2 ⇌ Ba2+ + 2OH-

s s 2s

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

s = 0.1091 M

B. Ksp Ba(OH)2 ⇌ Ba2+ + 2OH-

s s 2sKsp = [Ba2+][OH-]2

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

s = 0.1091 M

B. Ksp Ba(OH)2 ⇌ Ba2+ + 2OH-

s s 2sKsp = [Ba2+][OH-]2 = [s][2s]2

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

s = 0.1091 M

B. Ksp Ba(OH)2 ⇌ Ba2+ + 2OH-

s s 2sKsp = [Ba2+][OH-]2 = [s][2s]2 = 4s3

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

s = 0.1091 M

B. Ksp Ba(OH)2 ⇌ Ba2+ + 2OH-

s s 2sKsp = [Ba2+][OH-]2 = [s][2s]2 = 4s3

= 4(0.1091)3

7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by

adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

A. Titration 2HCl + Ba(OH)2

0.02910 L 0.04000 L0.300 M ? M

[Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2

1 L 2 moles HCl 0.0400 L

s = 0.1091 M

B. Ksp Ba(OH)2 ⇌ Ba2+ + 2OH-

s s 2sKsp = [Ba2+][OH-]2 = [s][2s]2 = 4s3

= 4(0.1091)3 = 5.20 x 10-3