GRAVIMETRIC ESTIMATION OF CHLORIDE

IONS

Aneeqa Haider, Ariel Tsang, Carrie Fan, Fabiha Nuzhat

Introduction Chloride ion

○ Cl + e− Cl−

Chloride○ Results from the

combination of Cl2 with a metal (e.g. NaCl)

Reactants: NaCl, AgNO3 Chloride Ion

Introduction Gravimetric analysis

○ The set of procedures to determine the quantity of a substance present in the mass of a solid

Seven steps1. Drying and measuring the masses of samples to

be analysed2. Dissolving the sample in distilled water.3. Precipitating the substance by adding a reagent.4. Separating the precipitate from the solution by

filtration.5. Washing precipitate free of impurities.6. Drying precipitate to obtain mass.7. Determine the amount of the original ion based

on the known mass and composition of the precipitate.

Introduction Objective: Determine the amount

of chloride ions present in a given solution of NaCl using AgNO3 as a reagent.

Problem: What is the amount of chloride ions present in 0.2 g of Sodium Chloride (NaCl)?

Hypothesis: The number of chloride ions present in 0.2 g of NaCl is approximately 2.06 x 10²¹.

Materials Erlenmeyer

Flask (2)

Beaker (1)

Funnel (1)

Ashless Filter Paper (1)

Paper Clips (4)

Balance

0.2 g of Sodium Chloride -NaCl

3 g of Silver Nitrate -

AgNO3 (aq)

Distilled Water

Dropper

Test Tubes (2)

Bunsen Burner

Crucible and lid

Crucible tongs

Retort Stand

Ring Clamp

Clay Triangle

Safety Goggles

Spatula

Stirring Rod

Graduated Cylinder

Procedure

1. Formation of the precipitate

2. Filtration of the solution containing the precipitate

3. Measurement of the mass of AgCl by drying the filter paper

4. Measurement of the mass of AgCl by burning the ashless filter paper

Formation of the Precipitate

0.2 g of NaCl was dissolved in Distilled Water in Erlenmeyer Flask

3.4 g of AgNO3 (aq) was poured into Erlenmeyer Flask containing NaCl (aq).

Solution was put into rest until all the precipitate formed.

Filtration of the Solution containing the Precipitate

Filter paper, funnel and Erlenmeyer flask were set up

The solution containing the precipitate was poured through the filter paper

Washed periodically with Distilled water

Measurement of the Mass of AgCl by drying the filter paper

Filter paper and precipitate were completely dried

Mass of the precipitate:○ Mass of the filter paper with precipitate -Mass of the filter paper

Measurement of the Mass of AgCl by burning the ashless filter paper

Retort stand, ring clamp, clay triangle, and Bunsen burner were set up

Filter paper was carefully folded with the precipitate inside, and placed in the crucible

Crucible was heated until no more filter paper was left

Mass of Precipitate:○ Mass of crucible, lid & precipitate –

Mass of crucible & lid

Safety Precautions General Safety

Precautions

Safety precautions specific for this experiment:○ Avoiding contact with Silver Chloride (AgCl)○ Safety precautions while using the Bunsen

burner

Observations Formation of the Precipitate

Objects Mass (g) Calculation

Filter paper 1.04 g

Filter paper with NaCl 1.24 g

NaCl 0.2 g (1.24 g - 1.0 g)

Graduated cylinder 22.48 g

Graduated cylinder with AgNO3

27.08 g

AgNO3 3.4 g (27.48 g - 22.48 g)

Observations

Measurement of the mass of AgCl by drying the filter paper

Objects Mass (g) Calculation

Filter paper 1.04 g

Filter paper with AgCl

1.43 g

AgCl 0.39 g (1.43 g - 1.04 g)

Observations Measurement of the mass of AgCl

by drying the filter paper

Objects Mass (g) Calculation

Crucible and lid 32.13 g

Crucible, lid and contents (AgCl)

32.64 g

AgCl 0.51 g (32.64 g - 32.13 g)

Calculations – Known Information Mass used of Sodium Chloride (NaCl):

○ 0.2 g Molar mass of NaCl:

○ 35.45 g/mol Percentage composition by mass of

Silver Chloride (AgCl):○ Silver (Ag) = 75%○ Chloride (Cl) = 25%

Calculations – By Drying Filter Paper

Mass of filter paper: ○ 1.04g

Mass of filter paper + AgCl: ○ 1.43g

AgCl:○ (1.43g – 1.04g) = 0.39g

Mass of chloride ions present: ○ 0.25 x 0.39g = 0.0975g

Number of moles of chloride ions:= Mass of Cl

Molar Mass of Cl = 0.0975g

35.45g = 0.00275 mol

Calculations – By Drying Filter Paper

Mole = Avogadro's number ○ 6.022 x 1023

Number of chloride ions: = (# of moles) x (Avogadro's number) = (0.00275) x (6.022 x 1023) = 1.656 x 1021 chloride ions present in 0.2 g of NaCl by drying filter paper

Calculations – By Drying Filter Paper

Mass of crucible + lid + filter paper:○ 32.13g

Mass of crucible + lid + filter paper + AgCl ○ 32.64g

AgCl ○ (32.64g – 32.13g) = 0.51g

Mass of chloride ions present: ○ 0.25 x 0.51g = 0.1275g

Calculations – By Burning Ashless Paper

Number of moles of chloride ions: = Mass of Cl

Molar Mass of Cl = 0.1275g

35.45g = 0.00360 mol

Calculations – By Burning Ashless Paper

Mole = Avogadro's number: ○ 6.022 x 1023

Number of chloride ions:○ (# of moles) x (Avogadro's number)○ (0.00360) x (6.022 x 1023)○ 2.167 x 1021 chloride ions present in 0.2 g of

NaCl by burning filter paper

Calculations – By Burning Ashless Paper

How did the Law of Conservation of Mass help predict the amount of Cl ions in AgCl?

Law of Conservation of Mass○ Mass of the reactants = Mass of the

products

Mass of Cl ions in NaCl (reactant) = Mass of Cl ion is AgCl (product)

What type of chemical reaction is taking place in this experiment?

Double displacement reaction:

AB + CD → AD + CB 

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

Why does NaCl dissolve in water? Hydration

provides greater stability than lattice energy

Hydration shell

Water – dipole moment

Why doesn’t AgCl dissolve in water?

More stable as a solid precipitate than separate ions

Hydration energy provided is less than lattice energy released when compound forms

What are some properties of AgCl? White crystalline solid Light sensitive

○ Purple black Change colour when AgCl Ag + Cl Dry powder, doesn’t draw moisture

from the air Very low solubility MP: 455°C BP: 1550°C

Why doesn’t an excess amount of AgNO3 affect the chemical reaction in the experiment? AgNO3 is the excess reagent Amount of chloride ions = product In this double displacement reaction all of the

NaCl must be used up LR = NaCl : limits amount of AgNO3 used

limits amount of products An excess of AgNO3 will not react since all the

NaCl is used up already

Why was it necessary to wash off all the impurities from the AgCl precipitate? Otherwise, mass of the impurities

would be included in the mass of the AgCl precipitate

Causes an inaccurate measurement of mass of AgCl

Causes inaccurate determination of the number of Cl ions

Why were the contents of the crucible slightly gray in colour after heating? Contents: AgCl and ashless filter

paper Ashless filter paper turned into CO2

Remaining content: AgCl AgCl is a white coloured powder at

SATP Upon heating, AgCl undergoes

decomposition to yield Ag and Cl

Percentage Yield and Error Percentage Yield

= Actual Yield x 100 Theoretical Yield= 80%

Percentage Error= (Theoretical Yield – Actual Yield) x 100

Theoretical Yield= 20%

Conclusion By drying filter paper:

○ Number of chloride ions in 0.2g of NaCl is 1.656 x 1021

By burning ashless filter paper:○ Number of chloride ions in 0.2g of NaCl

is 2.167 x 1021 Amount of ions present in NaCl =

amount of ions present in AgCl○ Law of Conservation of Mass

Sources of Error The reading on the electrical balance

was observed to change constantly○ Due to slight air currents○ Contents being weighted were extremely light○ Measured multiple times

Small amounts of AgCl were stuck in the flask after attempts to remove it○ Caused alterations in final mass○ Inaccurate percentage yield

Sources of Error Ashless filter paper not burned

away completely○ final mass greater than expected

Crucible was placed on counter to allow for cooling○ picked up unwanted particles on the

counter

Suggestions Use of better quality ashless filter

paper that will completely burn away without leaving any unwanted residue

Use of clean crucible and lid Minimum transfer of the samples

from container to container

THE END.

Thank you for listening to our presentation.