Gravimetric Analysis

1. Steps in a Gravimetric Analysis

2. Relative Supersaturation 

3. Impurities in Precipitates

4. Gravimetric Calculations

5. Precipitation Equilibria

6. Solubility in Pure Water

7. Solubility in Presence of a Common Ion

8. Solubility in Presence of Diverse Ions

In this technique, the analyte is converted to an insoluble form which can then be washed, dried, and weighed in order to determine the concentration of the analyte in the original solution. Gravimetry is applied to samples where a good precipitating agent is available. The precipitate should be quantitative, easily washed and filtered and is of in suitable quantity for accurate weighing. Therefore, gravimetry is regarded as a macro analytical technique. However, it is considered, when appropriately done, one of the most accurate analytical techniques. Also, Gravimetry is one of a few analytical methods that do not require standard solutions as the weight of precipitate is the only important parameter in analyte determination.

Steps in a Gravimetric Analysis

After appropriate dissolution of the sample the following steps should be followed for successful gravimetric procedure:

1. Preparation of the Solution: This may involve several steps including adjustment of the pH of the solution in order for the precipitate to occur quantitatively and get a precipitate of desired properties, removing interferences, adjusting the volume of the sample to suit the amount of precipitating agent to be added.

2. Precipitation: This requires addition of a precipitating agent solution to the sample solution. Upon addition of the first drops of the precipitating agent, supersaturation occurs, then nucleation starts to occur where every few molecules of precipitate aggregate together forming a nucleous. At this point, addition of extra precipitating agent will either form new nuclei or will build up on existing nuclei to give a precipitate. This can be predicted by Von Weimarn ratio where, according to this relation the particle size is inversely proportional to a quantity called the relative supersaturation where

Relative Supersaturation = (Q – S)/S

The Q is the concentration of reactants before precipitation, S is the solubility of precipitate in the medium from which it is being precipitated. Therefore, in order to get particle growth instead of further nucleation we need to make the relative supersaturation ratio as small as possible. The optimum conditions for precipitation which make the supersaturation low are:

a. Precipitation using dilute solutions to decrease Q

b. Slow addition of precipitating agent to keep Q as low as possible

c. Stirring the solution during addition of precipitating agent to avoid concentration sites and keep Q low

d. Increase solubility by precipitation from hot solution

e. Adjust the pH in order to increase S but not a too much increase as we do not want to loose precipitate by dissolution

f. Usually add a little excess of the precipitating agent for quantitative precipitation and check for completeness of the precipitation

3. Digestion of the Precipitate: The precipitate is left hot (below boiling) for 30 min to 1 hour in order for the particles to be digested. Digestion involves dissolution of small particles and reprecipitation on larger ones resulting in particle growth and better precipitate characteristics. This process is called Ostwald ripening. An important advantage of digestion is observed for colloidal precipitates where large amounts of adsorbed ions cover the huge area of the precipitate. Digestion forces the small colloidal particles to agglomerate which decreases their surface area and thus adsorption. You should know that adsorption is a major problem in gravimetry in case of colloidal precipitate since a precipitate tends to adsorb its own ions present in excess, Therefore forming what is called a primary ion layer which attracts ions from solution forming a secondary or counter ion layer. Individual particles repel each other keeping the colloidal properties of the precipitate. Particle coagulation can be forced by either digestion or addition of a high concentration of a diverse ions strong electrolytic solution in order to shield the charges on colloidal particles and force agglomeration. Usually, coagulated particles return to the colloidal state if washed with water, a process called peptization.

4. Washing and Filtering the Precipitate: It is crucial to wash the precipitate very well in order to remove all adsorbed species which will add to weight of precipitate. One should be careful nor to use too much water since part of the precipitate may be lost. Also, in case of colloidal precipitates we should not use water as a washing solution since peptization would occur. In such situations dilute nitric acid, ammonium nitrate, or dilute acetic acid may be used. Usually, it is a good practice to check for the presence of precipitating agent in the filtrate of the final washing solution. The presence of precipitating agent means that extra washing is required. Filtration should be done in appropriate sized Goosh or ignition filter paper.

5. Drying and Ignition: The purpose of drying (heating at about 120-150 ^oC in an oven) or ignition in a muffle furnace at temperatures ranging from 600-1200 ^oC is to get a material with exactly known chemical structure so that the amount of analyte can be accurately determined.

6. Precipitation from Homogeneous Solution: In order to make Q minimum we can, in some situations, generate the precipitating agent in the precipitation medium rather then adding it. For example, in order to precipitate iron as the hydroxide, we dissolve urea in the sample. Heating of the solution generates hydroxide ions from the hydrolysis of urea. Hydroxide ions are generated at all points in solution and thus there are no sites of concentration. We can also adjust the rate of urea hydrolysis and thus control the hydroxide generation rate. This type of procedure can be very advantageous in case of colloidal precipitates.

Impurities in Precipitates

1. Occlusion: Some constituents of the precipitation medium may be trapped in the crystal structure resulting in positive or negative errors. The trapped materials can be water, analyte ions, precipitating agent ions, or other constituents in the medium. Slow addition of precipitating agent and stirring may avoid occlusion but if it does occur, dissolution of precipitate and repracipitation may have to be done.

2. Inclusion: If the precipitation medium contains ions of the same charge and size as one forming the crystal structure of the precipitate, this extraneous ion can replace an ion from the precipitate in the crystal structure. For example, in the precipitation of NH₄MgPO₄ in presence of K⁺ ammonium leaves the crystal magnesium ammonium phosphate and is replaced by K⁺ since both have the same charge and size. However, the FW fro NH₄⁺ is 18 while that of K⁺ is 39. In this case a positive error occurs as the weight of precipitate will be larger when K⁺ replaces NH₄⁺. In other situations we may get a negative error when the FW of the included species is less than the original replaced species.

3. Surface Adsorption: This always result in positive errors in gravimetric procedures. See previous discussion on colloidal precipitates.

4. Postprecipitation: In cases where there are ions other than analyte ions which form precipitates with the precipitating agent but at much slower rate then analyte, and if the precipitate of the analyte is left for a long time without filtration then the other ions start forming a precipitate over the original precipitate leading to positive error. Examples include precipitation of copper as the sulfide in presence of zinc. Copper sulfide is formed first but if not directly filtered, zinc sulfide starts to precipitate on the top of it. The same is observed in the precipitation of calcium as the oxalate in presence of magnesium.

Gravimetric Calculations

The point here is to find the weight of analyte from the weight of precipitate. We can use the concepts discussed previously in stoichiometric calculations but let us learn something else. Assume Cl₂ is to be precipitated as AgCl, then we can write a stoichiometric factor reading as follows: one mole of Cl₂ gives 2 moles of AgCl. This is in fact what is called the gravimetric factor (GF) where we can substitute for the number of moles by grams to get:

GF for Cl₂ = 1 mol Cl₂/2 mol AgCl = FW Cl₂/2 FW AgCl = x g analyte/y g precipitate

Weight of substance sought = weight of precipitate x GF

One can also consider the problem by looking at the number of mmoles of analyte in terms of the mmoles of the precipitate where for the precipitation of Cl₂ as AgCl, we can write

Cl₂ = 2 AgCl

mmol Cl₂ = 1/2 mmol AgCl

(mg Cl₂/FW Cl₂) = 1/2 (mg AgCl/FW AgCl)

Let us now look at some examples:

Example

Calculate the grams analyte to mg precipitate for the following: P (at wt =30.97) in Ag₃PO₄ (FW = 711.22), Bi₂S₃ (FW 514.15) in BaSO₄ (FW = 233.40)

Solution

P = Ag₃PO₄

mmol p = mmol Ag₃PO₄

Mg P/30.97 = mg Ag₃PO₄/711.22

Mg P/mg Ag₃PO₄ = 30.97/711.22 = 0.04354

Bi₂S₃ = 3 BaSO₄

mmol Bi₂S₃ = 1/3 mmol BaSO₄

mg Bi₂S₃/FW Bi₂S₃ = 1/3 mg BaSO₄/FW BaSO₄

mg Bi₂S₃/514.15 = 1/3 mg BaSO₄/233.40

mg Bi₂S₃/ BaSO₄ = 1/3 (514.15/ 233.40) = 0.73429

Example

Phosphate in a 0.2711 g sample was precipitated giving 1.1682 g of (NH₄)₂PO₄.12 MoO₃ (FW = 1876.5). Find percentage P (at wt = 30.97) and percentage P₂O₅ (FW = 141.95) in the sample.

Solution

First we set the mol relationship between analyte and precipitate

P = (NH₄)₂PO₄.12 MoO₃

mmol P = mmol (NH₄)₂PO₄.12 MoO₃

mg P/at wt P = mg (NH₄)₂PO₄.12 MoO₃/ FW (NH₄)₂PO₄.12 MoO₃

mg P = at wt P x (mg (NH₄)₂PO₄.12 MoO₃/ FW (NH₄)₂PO₄.12 MoO₃)

mg P = 30.97 (1.1682x10³ / 1876.5) = 19.280 mg

% P = (19.280/271.1) x 100 = 7.111%

The same procedure is applied for finding the percentage of P₂O₅

P₂O₅ = 2 (NH₄)₂PO₄.12 MoO₃

mmol P₂O₅ = 1/2 (NH₄)₂PO₄.12 MoO₃

mg P₂O₅/FW P₂O₅ = 1/2 (mg (NH₄)₂PO₄.12 MoO₃/ FW (NH₄)₂PO₄.12 MoO₃)

mg P₂O₅ = 1/2 x FW P₂O₅ x (mg (NH₄)₂PO₄.12 MoO₃/ FW (NH₄)₂PO₄.12 MoO₃)

mg P₂O₅ = 1/2 x 141.95 (1.1682x10³/1876.5) = 44.185 mg

% P₂O₅ = (44.185/271.1) x 100 = 16.30%

Example

Manganese in a 1.52 g sample was precipitated as Mn₃O₄ (FW = 228.8) weighing 0.126 g. Find percentage Mn₂O₃ (FW = 157.9) and Mn (at wt = 54.94) in the sample.

Solution

3 Mn₂O₃ = 2 Mn₃O₄

mmol Mn₂O₃ = 3/2 mmol Mn₃O₄

mg Mn₂O₃ / FW Mn₂O₃ = 3/2 (mg Mn₃O₄/FW Mn₃O₄)

mg Mn₂O₃ = 3/2 FW Mn₂O₃ (mg Mn₃O₄/FW Mn₃O₄)

mg Mn₂O₃ = 3/2 x 157.9 ( 126/228.8) = 130 mg

% Mn₂O₃ = (130/1520) x 100 = 8.58%

The same idea is applied for the determination of Mn in the sample

3 Mn = Mn₃O₄

mmol Mn = 3 mmol Mn₃O₄

mg Mn / at wt Mn = 3 (mg Mn₃O₄/FW Mn₃O₄)

mg Mn = 3 at wt Mn (mg Mn₃O₄/FW Mn₃O₄)

mg Mn₂O₃ = 3 x 54.94 ( 126/228.8) = 90.8 mg

% Mn = (90.8/1520) x 100 = 5.97%

Example

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 = 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

By substitution we have

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

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

Example

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 2.26 g. Calculate the percentage Fe (at wt = 55.85) and Al (at wt = 26.98) in the sample.

Solution

Fe = 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 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)

The same treatment with the oxides gives

2 Fe =Fe₂O₃

mol Fe = 2 mol Fe₂O₃

g Fe/at wt Fe = 2 (g Fe₂O₃/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 = 2.26

1/2 g Fe (FW Fe₂O₃/at wt Fe) + 1/2 g Al (FW Al₂O₃/at wt Al) = 2.26

1/2 x (159.7/55.85) + 1/2 y (101.96/26.98) = 2.26

1.43 x + 1.89 y = 2.26 (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%

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) = 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, k_sp where

K_sp = [Ag⁺][Cl^-]

It should be clear 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 K_sp: No precipitate is formed and we have clear solution

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

3. When the product exceeds the k_sp : 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. The same equilibrium concepts discussed earlier control equilibrium govern the behavior of solutions containing sparingly soluble substances. Three situations will be studied here

a. Solubility in pure water

b. Solubility in presence of a common ion

c. Solubility in presence of diverse ions

Solubility in Pure Water

The calculations involved in this type of equilibrium is straightforward and the following examples show such a calculation.

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) = Ag⁺ + Cl^-

Before Equil	Solid	0	0
Equation	AgCl(s)	Ag^+	Cl^-
At Equilibrium		s	s

K_sp = [Ag⁺][Cl^-]

1.0x10^-10 = s x s = s²

s = 1.0 x 10^-5 M

[Ag⁺] = [Cl^-] = 1.0 x 10^-5 M

Example

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

Solution

Before Equil	Solid	0	0
Equation	PbSO4(s)	Pb^2+	SO4^2-
At Equilibrium		s	s

K_sp = [Pb²⁺][SO₄^2-]

1.6 x 10^-8 = s x s = s²

s = 1.3x10^-4 M

Example

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

Solution

The molar solubility is equal to the concentration of lead or half the concentration of iodide

Before Equil	Solid	0	0
Equation	PbI2(s)	Pb^2+	2I^-
At Equilibrium		s	2s

K_sp = [Pb²⁺][I^-]²

7.1x10^-9 = s x (2s)²

7.1x10^-9 = 4s³

s = 1.2x10^-3 M

If we compare between 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 k_sp. 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.

Example

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

Solution

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

K_sp = [Ag⁺][Cl^-]

1.0x10^-10 = [Ag⁺] x 1.0 x 10^-3

[Ag⁺] = 1.0x10^-7 M

Example

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

Solution

Fe(OH)₃ = Fe³⁺ + 3 OH^-

K_sp = [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

Of course, we expect a decrease in solubility according to Le Chatelier’s principle, where a common ion will shift the equilibrium to left (reactants side). Look at the following two examples:

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.

Solution

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

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

[Na⁺] = 0.1/20 = 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⁺

Before Equil	Solid	0.05	0
Equation	AgCl(s)	Ag^+	Cl^-
At Equilibrium		0.05 + s	s

K_sp = [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 decreased in presence of the common ion.

Example

25 mL of 0.100 M AgNO₃ are mixed with 35 mL of 0.050 M K₂CrO₄. Find the concentration of each ion in solution at equilibrium if k_sp of Ag₂CrO₄ = 1.1x10^-13.

Solution

2 Ag²⁺ + CrO₄^2- = Ag₂CrO_4(s)

mmol Ag⁺ = 0.100 x 25 = 2.5

mmol CrO₄^2- = 0.050 x 35 = 1.75

From the stoichiometry we know that two moles of silver react with one mole of chromate, therefore

mmol CrO₄^2- excess = 1.75 – 1.25 = 0.50

[CrO₄^2-] excess = mmol/mL = 0.05/60 = 0.0083 M

[NO₃^-] = mmol/mL = 0.100 x 25/60 = 0.0417 M

[K⁺] = mmol/mL = 2 x 0.050 x 35/60 = 0.0583 M

The silver was consumed in the reaction and the only way to calculate its concentration is through the solubility product:

Before Equil	Solid	0	0.0083
Equation	Ag2CrO4(s)	2Ag^+	CrO4^2-
At Equilibrium		2s	0.0083 + s

K_sp = [Ag⁺]² [CrO₄^2-]

1.1x10^-12 = [Ag⁺]² x (0.0083 + s)

However, s is very small since the equilibrium constant is very small. Therefore, assume 0.0083>>s

1.1x10^-12 = (2s)² x 0.0083

s = 5.7₆x10^-6 M

Relative error = (5.7₆x10^-6/0.0083) x 100 = 0.07%

[Ag⁺] = 2s = 2x5.7₆x10^-6 = 1.1₅x10^-5 M

[CrO₄^2-] = 0.0083 + s = 0.0083 + 5.7₆x10^-6 = 0.0083 M

Solubility in Presence of Diverse Ions

As expected from previous information, diverse ions have a screening effect on dissociated ions which leads to extra dissociation. Solubility will show a clear increase in presence of diverse ions as the solubility product will increase. Look at the following example:

Example

Find the solubility of AgCl (k_sp = 1.0 x 10^-10) in 0.1 M NaNO₃. The activity coefficients for silver and chloride are 0.75 and 0.76, respectively.

Solution

AgCl_(s) = Ag⁺ + Cl^-

We can no longer use the thermodynamic equilibrium constant ( i.e. in absence of diverse ions ) and we have to consider the concentration equilibrium constant or use activities instead of concentration if we use K_th:

K_sp = a_Ag⁺ a_Cl^-

K_sp = [Ag⁺] f_Ag⁺ [Cl^-] f_Cl^-

1.0x10^-10 = s x 0.75 x s x 0.76

s = 1.3x10^-5 M

We have calculated the solubility of AgCl in pure water to be 1.0x10^-5 M, if we compare the this value to that obtained in presence of diverse ions we see

% increase in solubility = {(1.3x10^-5 – 1.0x10^-5)/1.0x10^-5}x 100 = 30%

Therefore, once again we have an evidence for an increase in dissociation or a shift of equilibrium to right in presence of diverse ions.