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## Entry Task: March 5 th Tuesday

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### Entry Task: March 5th Tuesday

Question:

Calculate the solubility of AgI in water in moles per liter given its Ksp value of 8.3 x 10-17.

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

Agenda:

- Discuss Solubility, Precipitation and Ions ws
- Major self-check on content so far & discuss it
- HW: Pre-lab Determine Ksp

Ca2+(aq) + C2O42-(aq)

17.6- Precipitation and Separation of Ions17.37 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.

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

Mn2+(aq) + 2OH1-(aq)

17.6- Precipitation and Separation of Ions17.41 Calculate the solubility of Mn(OH)2 in grams per liter when buffered at

(a) at pH of 7.0;

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]

1.0 x10-14

The molarity is 16M

(16)(1 L) = 16 moles * 89 =

- 1424 or 1.4 x103g/ liter

Mn2+(aq) + 2OH1-(aq)

17.6- Precipitation and Separation of Ions17.41 Calculate the solubility of Mn(OH)2 in grams per liter when buffered at

(b) at pH of 9.5;

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]

1.0 x10-9

The molarity is 1.6x10-4M

(1.6x10-4)(1 L) = 1.6x10-4 moles * 89 =

- 1.4x10-2g/ liter

Mn2+(aq) + 2OH1-(aq)

17.6- Precipitation and Separation of Ions17.41 Calculate the solubility of Mn(OH)2 in grams per liter when buffered at

(c) at pH of 11.8;

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]

3.98 x10-5

The molarity is 4.02x10-9M

(4.02x10-9)(1 L) = 4.02x10-9 moles * 89 =

- 3.6x10-7g/ liter

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

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 MAgNO3 there are

(0.100 L) (0.050M) =

5.0 x 10-3 moles of Ag+1 ions

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 Na2SO4solution?

In 0.010 L of 5.0 x10-2MNaNO3 there are

(0.010 L) (5.0 x10-2M) =

5.0 x 10-4 moles of SO4-2 ions

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 Na2SO4solution?

- 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

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 Na2SO4solution?

- We have to convert the moles in to molarity but use the combined volume.

5.0 x10-3 moles/0.110L =

- 4.55 x10-2M of Ag+1 ions

5.0 x10-4 moles/0.110L =

4.55 x10-3M of SO4-2 ions

Substitute the values into the Ksp expression and solve for Q

17.6- Precipitation and Separation of Ions17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M AgNO3 is mixed with 10 mls of 5.0 x10-2M Na2SO4solution?

Q = [Ag+]2[SO42]

- 9.4 106

(4.55 x10-2)2(4.55 x10-3) =

- Q= 9.4 106

Ksp= 1.5 x 10-5

- Q is smaller than Ksp that means

- No precipitate will occur

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.

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

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

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 10-13l- 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.

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