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### Transcript of Chapter 5 Gravimetric

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TOPIC 5:

GRAVIMETRIC ANALYSIS

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Gravimetric analysis

A + B = C

An excess of B, at unknown [ ] , isadded to A so that A is completelytransformed in C.

C = Obtained by weighing solid product in apure form.

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Introduction Gravimetric analysis is the quantitative

determination of analyte through a processof precipitation/volatilization, isolation ofthe precipitate, and weighing the isolatedproduct.

Analyte + Reagent Precipitate

Precipitating pure

agent can be filtered,

well separated

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Divided into 2:

i. Precipitation gravimetry Analyte is separated from the solution of

the sample as a ppt & is converted to acompound of known composition that canbe weighed.

Eg: Cl- + AgNO3 AgCl + NO3-

ii. Volatilization gravimetry

Analyte is separated from otherconstituents of a gas then serves as ameasure of the analyte [ ].

Eg: NaHCO3 + H2SO4 CO2 + H2O +NaHSO

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CONTINUE

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CONTINUE

The ideal product should be;

very insoluble,

easily filtered, and

very pure and

posses a known and constant composition

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Steps of a Gravimetric Analysis which requirecondition optimization to eliminate systematicerror :

Preparation of the solution PrecipitationDigestion FiltrationWashingDrying or IgnitingWeighing Calculation

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Dissolution

Sample is dried, weighed, dissolved inacid, etc.

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Precipitation

Precipitation Techniques

Add precipitating reagent to sample solution.

Reacts with analyte to form insolublematerial.

Precipitate has known composition or can beconverted to known composition.

2 types of Precipitating agents:

Specific (react only with 1 chemical sp.) Selective (react with a limited number of sp.)

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Desirable properties of analyticalprecipitates:

Low solubility, preventing losses duringfiltration and washing.

Stable final form (unreactive).

Known composition after drying orignition.

CONTINUE

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Ions in colloid Precipitate

Solution

(10-7cm) (10-7 - 10-4cm) (10-4cm)

During precipitation process, supersaturationoccurs, followed by nucleation and precipitation.

First supersaturation (soln. contains more of thedissolved salt than at equilibrium) occurs.

Nucleation particles come together to produce

microscopic nuclei. The the degree of supersaturation, the greater

the rate of nucleation.

Nucleation can be induced by scratching vessel

surface, dusts, etc

CONTINUE

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Nucleation

Formation of nucleus or primary particlesfrom ion, atoms or molecules, whichaggregates to form a stable second phase.

Growth of particles

Precipitation on top of formed nucleus.

The higher the degree of supersaturation,

the higher the rate of nucleation. Ifrate of nucleation < rate growth of

particles, big particles (crystals ) areformed (easy to filter, less impurities)

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Relative supersaturation = Q-S

S

Q : degree supersaturation ( [ ] Of mixedreagents before ppt)

S : solubility of ppts at equilibrium

If(Q-S)/S >, small crystals (surface area) If (Q-S)/S < , larger crystals ( surface area)

Therefore, we want to keep Q and S during pptn.

CONTINUE

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Favourable conditions for pptn

Keep Q by:Precipitate from dilute solution.

Add dilute ppt reagents slowly, withstirring.

Keep S

by:

Precipitate from hot solution.

Precipitate at as pH as possible.

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Digestion

Solution containing ppts are heated before filtering.

Heating the ppts within the mother liquor (or solution fromwhich it precipitated) for a certain period of time.

Done for crystalline ppts (e.g: BaSO4) diameter >10-4cm.

During digestion, small particles tend to dissolve andreprecipitate on larger ones.

Trapped impurities will be dissolved.

Objective: bigger and purer particles that are more

easily filtered from solution.

DT

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Filtering & Washing

Impurities on the surface can be removed bywashing the precipitate after filtering.

Water is not always a good choice. Washing with water will dilute the counter layer

and the primary layer charge causes theparticles to revert to the colloidal state(peptization).

So we wash with an electrolyte that can be

volatilized on heating (HNO3).

Test for completeness of washing ~ check thepresence of precipitation agent in filtrate.Impurities and excess ions are washed off the

ppt in this step.

CONTINUE

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Type of filter will depend on the particle size ofppts.

filter paperGooch/glass sintered crucible

Colloidal ppts

E.g: AgCl.Mainly small particles. Cannot be filtered

using normal filtration as they are not pptfrom soln.

Bigger particles are not formed

Agglomeration of colloidal particles Is encouraged for formation of matter that

can be easily filtered, can be ppt fromsolutions.

Formation of larger particles

CONTINUE

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Drying/ Igniting & Weighing

Many ppt contain varying amounts of H2O.

Adsorbed from the air (i.e. hygroscopic).

Precipitates are dried for accurate, stable massmeasurements.

Heating removes the solvent & any volatile sp.carried down with ppt.

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Drying/ Igniting & Weighing

Drying T depends on ppt :

AgCl : > 110 oC Al2O3 : > 1000

oC

Ignition at higher temperature is required only if aprecipitate must be converted to a more suitable

form for weighing.

For example, Fe(HCO2)3.nH2O is ignited at 850oC

for 1 hour to give Fe2O3.

The compound finally weighed.

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Gravimetry Calculation

Generally express in units ofgram or %.

In gravimetry analysis, weight of the analyte isgiven by:

GF is gravimetry factor that is given by

a = mol of analyte

b = mol of ppt

xGF =

a

b

FW [analyte]

FW [precipitate]

weight of analyte (g) = weight of ppt (g) x GF

a & b must beequivalent

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CONTINUE

Examples of gravimetry factor (GF)

Analyte ppt GF

CaO CaCO3

Fe3O4 Fe2O3

Mg Mg2P2O7

1 Mg2P2O7

1 CaO

1 CaCO3

2 Fe3O43Fe2O3

2 Mg

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A general equation for calculating the % of

analyte (A) from gravimetric data;

Mass analyte= Mass ppt. x Gravimetric Factor (GF)

% Analyte = mass of ppt x GF x 100mass of sample

CONTINUE

% Analyte = mass of analyte (g) x 100 %

mass of sample (g)

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Example

An ore is analyzed for the manganesecontent by converting the manganese toMn3O4 & weighing it.

If a 1.52g sample yields Mn3O4 weighing0.126g, what would be the % Mn2O3 inthe sample? The % Mn?

(JMR: Mn= 54.94, O =16)

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Solution

%Mn2O3 =0.126g Mn3O4 x

3 Mn2O3

2 Mn3O4x 100%

1.52g sample

= 8.58 %

% Mn = 0.126g Mn3O4 x3 Mn

1 Mn3O4x 100%

1.52g sample

= 5.97 %

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A piece of impure marble (CaCO3) isanalyzed. It is found that 3.000 g ofthe marble react with Ba(NO3)2solution to produce 1.25 grams ofBaCO3. What % of the marble waspure CaCO3?

(Fw : CaCO3 = 100, BaCO3 = 197.3)

Ans: 21.1%

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Solution

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CaCO3 + Ba(NO3)2 BaCO3 + Ca(NO3)2

GF = 1 mole CaCO3 = 100 g CaCO31 mole BaCO3 197. 3 g BaCO3

= 0.507

% CaCO3 = 0.507 x 1.25 x 100% = 21.1%3.00 g

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Exercise

Treatment of a 0.4000 g sample ofimpure KCl with an excess of AgNO3

resulted in the formation of 0.7332 gof AgCl. Calculate the percentage ofKCl in the sample.

(JAR: Ag = 107.9, Cl= 35.5, K= 39.1, N= 14)

Ans: 95.35%

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A sample of ore weighing 1.2504 g containssulphur is treated with HNO3 and KClO3 toconvert sulphur to BaSO4. After treating with

HCL, chlorate and nitrate are removed