What weight of sulfur (FW = 32.064) ore which should be taken so that the weight of BaSO
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In the same manner g AlCl 3 = g Al ( FW AlCl 3 /at wt Al) g FeCl 3 + g AlCl 3 = 5.95 PowerPoint PPT Presentation


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What weight of sulfur (FW = 32.064) ore which should be taken so that the weight of BaSO 4 (FW = 233.40) precipitate will be equal to half of the percentage sulfur in the sample. Solution S D BaSO 4 mmol S = mmol BaSO 4 mg S/at wt S = mg BaSO 4 /FW BaSO 4

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In the same manner g AlCl 3 = g Al ( FW AlCl 3 /at wt Al) g FeCl 3 + g AlCl 3 = 5.95

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In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

What weight of sulfur (FW = 32.064) ore which should be taken so that the weight of BaSO4 (FW = 233.40) precipitate will be equal to half of the percentage sulfur in the sample.

Solution

S D BaSO4

mmol S = mmol BaSO4

mg S/at wt S = mg BaSO4/FW BaSO4

mg S = at wt S x ( mg BaSO4 / FW BaSO4)

mg S = 32.064 x (1/2 %S/233.40)

mg S = 0.068689 %S

%S = (mg S/mg sample) x 100

%S = 0.068689 %S/mg sample) x 100

mg sample = 0.068689 %S x 100 /%S = 6.869 mg


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

A mixture containing only FeCl3 (FW = 162.2) and AlCl3 (FW = 133.34) weighs 5.95 g. The chlorides are converted to hydroxides and ignited to Fe2O3 (FW = 159.7) and Al2O3 (FW = 101.96). The oxide mixture weighs 2.62 g. Calculate the percentage Fe (at wt = 55.85) and Al (at wt = 26.98) in the sample.

Solution

Fe D FeCl3

1 mol Fe = 1 mol FeCl3

g Fe/at wt Fe = g FeCl3/ FW FeCl3

Rearrangement gives

g FeCl3 = g Fe (FW FeCl3/at wt Fe)


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

In the same manner

g AlCl3= g Al ( FW AlCl3/at wt Al)

g FeCl3 + g AlCl3 = 5.95

g Fe (FW FeCl3/at wt Fe) + g Al ( FW AlCl3/at wt Al) = 5.95

assume g Fe = x, g Al = y then:

x (FW FeCl3/at wt Fe) + y ( FW AlCl3/at wt Al) = 5.95

x (162.2/55.85) + y (133.34/26.98) = 5.95

2.90 x + 4.94 y = 5.95 (1)


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

The same treatment with the oxides gives

2 Fe D Fe2O3

mol Fe = 2 mol Fe2O3

g Fe/at wt Fe = 2 (gFe2O3/FW Fe2O3)

g Fe2O3 = 1/2 g Fe (FW Fe2O3/at wt Fe)

In the same manner

g Al2O3 = 1/2 g Al (FW Al2O3/at wt Al)

g Fe2O3 + g Al2O3 = 2.62

1/2 g Fe (FW Fe2O3/at wt Fe) + 1/2 g Al (FW Al2O3/at wt Al) = 2.62

1/2 x (159.7/55.85) + 1/2 y (101.96/26.98) = 2.62

1.43 x + 1.89 y = 2.62 (2)


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

2.90 x + 4.94 y = 5.95 (1)

1.43 x + 1.89 y = 2.62 (2)

from (1) and (2) we get

x = 1.07

y = 0.58

% Fe = (1.07/5.95) x 100 = 18.0%

% Al = (0.58/5.95) x 100 = 9.8%


Problem

Problem

Consider a 1.0000 g sample containing 75% potassium sulfate (FW 174.25) and 25% MSO4. The sample is dissolved and the sulfate is precipated as BaSO4 (FW 233.39). If the BaSO4 ppt weighs 1.4900g, what is the atomic weight of M2+ in MSO4?

K2SO4g BaSO4

MSO4g BaSO4


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

mmol BaSO4 = mmol K2SO4 + mmol MSO4

Rearranging and solving:


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

A mixture of mercurous chloride (FW 472.09) and mercurous bromide (FW 560.99) weighs 2.00 g. The mixture is quantitatively reduced to mercury metal (At wt 200.59) which weighs 1.50 g. Calculate the % mercurous chloride and mercurous bromide in the original mixture.

Hg2Cl2g 2Hg

Hg2Br2 g 2Hg


Answer

Answer

mmol Hg = 2 mmol Hg2Cl2 + 2 mmol Hg2Br2

Rearranging and solving:


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

A 10.0 mL solution containing Cl- was treated with excess AgNO3 to precipitate 0.4368 g of AgCl. What was the concentration of Cl- in the unknown? (AgCl = 143.321 g/mol)

mmoles of Cl- = mmoles of AgCl

Concentration of Cl-


Precipitation equilibria

Precipitation Equilibria

Inorganic solids which have limited water solubility show an equilibrium in solution represented by the so called solubility product. For example, AgCl slightly dissolve in water giving Cl- and Ag+ where

AgCl (s)D Ag+ + Cl-

K = [Ag+][Cl-]/[AgCl(s)]

However, the concentration of a solid is constant and the equilibrium constant can include the concentration of the solid and thus is referred to, in this case, as the solubility product, ksp where

Ksp = [Ag+][Cl-]


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

It should be clear that the product of the ions raised to appropriate power as the number of moles will fit in one of three cases:

 1. When the product is less than Ksp: No precipitate is formed and we have a clear solution

2. When the product is equal to ksp: We have a saturated solution

3. When the product exceeds the ksp : A precipitate will form

It should also be clear that at equilibrium of the solid with its ions, the concentration of each ion is constant and the precipitation of ions in solution does occur but at the same rate as the solubility of precipitate in solution. Therefore, the concentration of the ions remains constant at equilibrium.


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

The molar solubility depends on the stoichiometry of the salt.

A 1:1 salt is less soluble than a nonsymmetric salt with the same Ksp


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

Three situations will be studied: 

a. Solubility in pure water

b. Solubility in presence of a common ion

c. Solubility in presence of diverse ions

Two other situations will be discussed in later chapters

  • Solubility in acidic solutions

  • solubility in presence of a complexing agent


Solubility in pure water

Solubility in Pure Water

Example

Calculate the concentration of Ag+ and Cl- in pure water containing solid AgCl if the solubility product is 1.0x10-10.

Solution

First, we set the stoichiometric equation and assume that the molar solubility of AgCl is s.

AgCl(s)D Ag+ + Cl-


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

Ksp = [Ag+][Cl-]

1.0x10-10 = s x s = s2

s = 1.0 x 10-5 M

[Ag+] = [Cl-] = 1.0 x 10-5 M


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

Example

Calculate the molar solubility of PbSO4 in pure water if the solubility product is 1.6 x 10-8

Solution

Ksp = [Pb2+][SO42-]

1.6 x 10-8 = s x s = s2

s = 1.3x10-4 M


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

Calculate the molar solubility of PbI2 in pure water if the solubility product is 7.1 x 10-9.

Solution

Ksp = [Pb2+][I-]2

7.1x10-9 = s x (2s)2

7.1x10-9 = 4s3

s = 1.2x10-3 M


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

If we compare the molar solubilities of PbSO4 and PbI2 we find that the solubility of lead iodide is larger than that of lead sulfate although lead iodide has smaller ksp. You should calculate solubilities rather than comparing solubility products to check which substance gives a higher solubility. In presence of a precipitating agent, the substance with the least solubility will be precipitated first.


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

Example

What must be the concentration of Ag+ to just start precipitation of AgCl in a 1.0x10-3 M NaCl solution.

Solution

AgCl just starts to precipitate when the ion product just exceeds ksp

Ksp = [Ag+][Cl-]

1.0x10-10 = [Ag+] x 1.0 x 10-3

[Ag+] = 1.0x10-7 M


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

Example

What pH is required to just start precipitation of Fe(III) hydroxide from a 0.1 M FeCl3 solution. Ksp = 4x10-38.

Solution

Fe(OH)3D Fe3+ + 3 OH-

Ksp = [Fe3+][OH-]3

4x10-3 = 0.1 x [OH-]3

[OH-] = 7x10-13 M

pH = 14 – pOH

pH = 14 – 12.2 = 1.8


Solubility in presence of a common ion

Solubility in Presence of a Common Ion

Example

10 mL of 0.2 M AgNO3 is added to 10 mL of 0.1 M NaCl. Find the concentration of all ions in solution and the solubility of AgCl formed.

First we find mmol AgNO3 and mmol NaCl then determine the excess concentration

mmol Ag+ = 0.2 x 10 = 2

mmol Cl- = 0.1 x 10 = 1

mmol AgCl formed = 1 mmol

mmol Ag+ excess = 2 – 1 = 1

[Ag+]excess = mmol/mL = 1/20 = 0.05 M


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

We can find the concentrations of NO3- = 0.2*10/20 = 0.1 M

[Na+] = 0.1*10/20 = 0.05 M

Chloride ions react to form AgCl, therefore the only source for Cl- is the solubility of AgCl; but now in presence of 0.05 M excess Ag+


In the same manner g alcl 3 g al fw alcl 3 at wt al g fecl 3 g alcl 3 5 95

Ksp = [Ag+][Cl-]

1.0x10-10 = (0.05 + s) (s)

Since the solubility product is very small, we can assume 0.05>>s

1.0x10-10 = 0.05 (s)

s = 2.0x10-9 M

Relative error = ( 2.0x10-9 / 0.05) x 100 = 4x10-6% which is extremely small, therefore:

[Cl-] = 2.0x10-9 M

[Ag+] = 0.05 + 2.0x10-9 = 0.05 M

Look at how the solubility is decreased in presence of the common ion.


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