# Chloride in Seawater Gravimetric Analysis

## Key Concepts

• Gravimetric analysis is the quantitative isolation of a substance by precipitation and weighing of the precipitate.1
• Seawater contains dissolved chloride ions, Cl-(aq).2
• Gravimetric analysis can be used to determine the amount of chloride present in seawater by precipitation of the chloride as insoluble silver chloride 3, AgCl
Ag+(aq) + Cl-(aq) → AgCl(s)
• The mass of silver chloride precipitated is used to calculate:

(i) moles of AgCl(s) precipitated
moles = mass ÷ molar mass

(ii) moles of chloride ion, Cl-, present
mole ratio AgCl:Cl- is 1:1

(iii) mass of chloride ion, Cl-, present in the seawater.
mass = moles × molar mass

• The amount of chloride dissolved in the seawater, the concentration of chloride ion in seawater, is usually expressed in units of the mass of chloride per volume of seawater:
(i) grams of chloride in 100 mL of seawater: g/100 mL
(that is, a percentage, w/v%)

(ii) grams of chloride per litre of seawater: g L-1

The concentration of chloride ions in seawater can also be given as a concentration in moles per litre:
moles of chloride in 1 L of seawater: mol L-1 (molarity)

No ads = no money for us = no free stuff for you!

## Theory

Natural waters such as seawater, water in rivers and lakes, contain soluble chloride salts.
Because these salts are soluble in water, they release chloride ions, Cl-, into the water.
Natural waters are therefore aqueous solutions.

Silver chloride, AgCl, is insoluble in water, that is, silver chloride does not dissolve to form ions in aqueous solution, instead the ions stay bonded to each other in a crystal lattice forming a solid.

We can add a water soluble silver salt such as silver nitrate, AgNO3(aq), to a water sample containing chloride ions, Cl-(aq), to form an insoluble solid (precipitate) of silver chloride, AgCl(s):

Ag+(aq) + Cl-(aq) → AgCl(s)

This precipitate of silver chloride, AgCl(aq), can be separated from the water by filtration, then dried and weighed.

We can calculate the moles of AgCl(s) using its mass and molar mass:

moles(AgCl(s)) = mass(AgCl(s)) ÷ molar mass(AgCl(s))

We can calculate the moles of Cl- using the mole ratio (stoichiometric ratio) of silver chloride to chloride ions:

AgCl(s) : Cl- is 1:1
moles(Cl-) = moles(AgCl(s))

Since we know the volume of the water sample used for the analysis, we can calculate the concentration of chloride ions in the water in units of moles per litre, mol L-1 (molarity):

concentration(Cl-) = moles(Cl-) ÷ volume of water sample in litres

If required, we can convert the concentration in mol L-1 to a concentration in g L-1:

Since moles(Cl-) = mass(Cl-) ÷ molar mass(Cl-)
So, mass(Cl-) = moles(Cl-) × molar mass(Cl-)
Therefore,
 concentration of Cl- in g L-1 = moles(Cl-) × molar mass(Cl-)1 L

If required, we can then convert this to a concentration in g/100 mL:

since 1L = 1000 mL
100 mL is 100/1000 of a litre, or, 0.1 of a litre
concentration in g/100 mL = 0.1 × concentration in g L-1

Do you know this?

Play the game now!

## Procedure:

Safety Considerations
• Nitric acid is corrosive. Avoid contact.
Any spills should be rinsed with water immediately.
• Silver nitrate solution causes staining of skin and fabric. Avoid contact.
Any spills should be rinsed with water immediately.
• Wear safety googles, enclosed shoes, and lab coats.

1. Filter about 100 mL of seawater through a filter paper into a clean, dry conical flask (erlenmeyer flask).
The seawater is filtered to remove any insoluble solids before the analysis begins.
If the seawater contained solids that were not filtered off, these would add to the final weight of the precipitate and the calculated concentration of chloride ions in seawater would be too high.
2. Pipette 50.00 mL of the filtered seawater into a 100.0 mL volumetric flask and make up to the mark with deionised water.
The concentration of chloride ions in seawater is typically around 0.5 mol L-1.
We dilute the seawater sample because the particle size of the precipitate will be greater if the concentration is lower.
3. Pipette 20.00 mL of the diluted seawater into a clean, dry 100 mL beaker.
Using a smaller sample of the diluted seawater means we will use a smaller amount of the expensive silver nitrate to precipitate out all the chloride.
4. Add about 1 mL of 6 mol L-1 nitric acid to this sample of seawater.
Acidifying the solution with dilute nitric acid prevents the precipitation of some other silver salts, such as silver carbonate and silver phosphate, which are insoluble in neutral solution but soluble in acidic solution.4
Note that the choice of acid is important!
Hydrochloric acid, HCl(aq), is unsuitable because it would increase the concentration of chloride ions, Cl-(aq), in solution.
Sulfuric acid, H2SO4(aq), is unsuitable because it would add sulfate ions, SO42-(aq), and when silver nitrate is added the silver sulfate would precipitate out along with the silver chloride.
5. While stirring, use a glass dropper to add 0.2 mol L-1 silver nitrate solution slowly to the acidified seawater sample.
Continue adding silver nitrate solution until no more white precipitate forms.
The addition of silver ions slowly is important because this will help increase the size of the silver chloride particles.
The first AgCl(s) particles formed will act as nuclei which grow as further AgCl(s) precipitates.
6. While stirring, heat the mixture until it nearly boils.
Continue heating and stirring at this temperature until the precipitate coagulates and the supernatant liquid is clear (2-3 minutes).
Heating the precipitate helps it to coagulate, form larger particles.
Larger particles are less likely to go through the filter paper and be lost.
7. Remove the beaker from the heat, cover with a watch glass (clock glass), and allow the precipitate to settle.
Then test for complete precipitation by adding a few more drops of silver nitrate. If no more precipitate forms, cover the beaker, wrap it all in foil, and place in a dark cupboard until the next day.
Silver chloride is light sensitive!
In the presence of light, silver chloride decomposes into silver and chlorine with the minute silver particles being dispersed through the silver chloride making it look purplish in colour.
Therefore, it is best to keep your solutions out of direct sunlight, and store them in dark places if you are not going to use them straight away.
8. Weigh a filter paper and record its mass.
9. Filter the settled mixture through the filter paper.
Decant the clear, supernatant liquid through the filter paper without disturbing the settled precipitate.
Transfer the precipitate from the beaker to the filter paper with the aid of a jet of dilute nitric acid from a wash bottle.
Use a rubber-tipped glass rod (a "policeman") to remove any precipitate adhering to the sides of the beaker and wash that onto the filter paper also using the dilute nitric acid.
Do not use deionised water to wash the precipitate as this will tend to break the particles up into smaller particles which will be harder to filter.
10. Place the filter paper containing the precipitate on a watch glass.
Place this in an oven to dry overnight.
This will remove any water which would increase the mass of the precipitate.
Any volatile nitric acid remaining on the precipitate will also be lost during the drying process.
11. Remove the filter paper and precipitate from the oven and place it in a dessicator to cool.
It is essential to use a dessicator to prevent water from the atmosphere entering the system and adding to the mass of the filter paper and precipitate.
12. Weigh the filter paper with the precipitate and record its mass.
13. Repeat the process of drying in the oven, cooling in the dessicator, and weighing, until a constant mass is achieved.

Do you understand this?

Take the test now!

## Sample Results and Calculations

On completion of the gravimetric analysis above to determine the concentration of chloride ions in seawater, a student recorded the following results:

item mass /g
filter paper 0.005
(1) filter paper + precipitate 0.836
(2) filter paper + precipitate 0.769
(3) filter paper + precipitate 0.769

Calculations:

1. Calculate mass of AgCl(s) precipitated.
Use ONLY the constant mass, 0.769 g, in your calculations (the first recorded mass must be a "wet" sample, that is, not all the water has been evaporated off during the drying process)
m(AgCl(s)) = mass(filter paper + precipitate) - mass(filter paper)
m(AgCl(s)) = 0.769- 0.005 = 0.764 g
2. Calculate moles of AgCl(s) precipitated:
n(AgCl(s)) = mass(AgCl(s)) ÷ molar mass(AgCl(s))
n(AgCl(s)) = 0.769 ÷ (107.9 + 35.45)
n(AgCl(s)) = 0.769 ÷ 143.35
n(AgCl(s)) = 5.364 × 10-3 mol
3. Calculate moles of Cl- present in the 20.0 mL of diluted seawater (see mole ratio):
1 mole of AgCl contains 1 mole of Cl-
Therefore 5.364 × 10-3 moles AgCl contains 5.364 × 10-3 moles of Cl-
4. Calculate concentration, in mol L-1, of chloride ions in the diluted seawater sample:

 molarity(Cl-) = moles(Cl-)   volume(Cl-)

moles(Cl-) = 5.364 × 10-3 mol
volume(Cl-) = 20.0 mL = 20.0/1000 = 0.0200 L

 molarity(Cl-) = 5.364 × 10-3 0.0200 = 0.2682 mol L-1

5. Calculate concentration, ci in mol L-1, of chloride ions in original, undiluted seawater:
 ciVi = cfVf ci = cfVfVf

Vi = initial volume = 50.00 mL = 50.00/1000 = 0.0500
cf = final concentration = 0.2682 mol L-1
Vf = 100.0 mL = 100.0/1000 = 0.1000 L
 ci = 0.2682 ×0.10000.0500 ci = 0.5364 mol L-1
6. Convert concentration in mol L-1 to a concentration in grams of solute per 100 mL of solution (w/v%):
c(Cl-) = 0.5364 mol L-1
That is, 1 L (1000 mL) of seawater contains 0.5364 moles of Cl-
100 mL is 100/1000 = 1/10 = 0.1 of a litre
moles Cl- in 100 mL = 0.1 L × 0.5364 mol L- = 0.05364 mol in 100 mL
moles(Cl-) = mass(Cl-) ÷ molar mass(Cl-)
mass(Cl-) = moles(Cl-) × molar mass(Cl-) = 0.05364 × 35.45 g mol-1 = 1.902 g
w/v% (Cl-) = 1.902 g/100 mL
7. Convert concentration in g/100 mL (w/v%) to a concentration in grams per litre, g L-1
w/v% (Cl-) = 1.902 g/100 mL
That is, 100 mL of solution contains 1.902 g of Cl-.
1mL of solution contains 1.902/100 = 0.01902 g of Cl-
1000 mL of solution contains 1000 × 0.01902 = 19.02 g of Cl-
1000 mL = 1 L
1 L of solution contains 19.02 g of Cl-
concentration of Cl- = 19.02 g L-1

## Sources of Error:

• Incomplete precipitation results in a value for the amount of chloride in seawater that is too low.
• Particles smaller than the pore size of the filtering medium can pass through it resulting in a value for the amount of chloride in seawater that is too low.

• Incomplete drying of the sample results in a value for the amount of chloride in seawater that is too high.
• Other ions, especially sulfate ions, in the sample may also be precipitated resulting in a value for the amount of chloride in seawater that is too high.

Can you apply this?

Join AUS-e-TUTE!

Take the test now!

## Problem Solving using StoPGoPS method

Question:
Bo the Biologist has a problem: the plant-life in Lake Loch is dying. Bo thinks the salt concentration in the lake has increased and this is killing the plants.
Bo takes a 1 L sample of lake water to Chris the Chemist to analyse.
Chris the Chemist filters the litre of lake water into a clean, dry flask.
Then Chris pipettes 25.00 mL of the filtered lake water into a beaker and adds a drop of concentrated nitric acid to acidify the solution.
AgNO3(aq) was then added to the water drop-by-drop until no further precipitation of AgCl(s) occurred.
This mixture was then heated to coagulate the solids, then cooled, filtered, dried and weighed.

The results of the experiment are shown below:

item mass / g
filter paper 0.003
filter paper + precipitate 0.127

Calculate the concentration of chloride in the lake water in mol L-1.

Worked Solution:

 STOP STOP! State the Question. What is the question asking you to do? Calculate the concentration of Cl- in mol L-1 [Cl-(aq)] = ? mol L-1 PAUSE PAUSE to Prepare a Game Plan (1) What information (data) have you been given in the question? volume of water = 25.00 mL mass(filter paper) = 0.003 g mass(filter paper + precipitate) = 0.127 g precipitate is AgCl(s) (2) What is the relationship between what you know and what you need to find out? Ag+(aq) + Cl-(aq) → AgCl(s) moles(Cl-) = moles(AgCl) moles(AgCl) = mass(AgCl) ÷ molar mass(AgCl)     mass(AgCl) = mass(filter paper + precipitate) - mass(filter paper)     molar mass(AgCl) = molar mass(Ag) + molar mass(Cl) [Cl-] = moles(Cl-) ÷ volume of water in L     volume of water = 25.00 mL GO GO with the Game Plan mass(AgCl(s)) = mass(filter paper + precipitate) - mass(filter paper) mass(AgCl(s)) = 0.127 - 0.003 = 0.124 g moles(AgCl(s)) = mass (g) ÷ molar mass (g mol-1) molar mass(AgCl) = molar mass(Ag) + molar mass(Cl) = 107.9 + 35.45 = 143.35 g mol-1 moles(AgCl) = 0.124 ÷ 143.35 = 8.650 × 10-4 mol Ag+(aq) + Cl-(aq) → AgCl(s) moles(Cl-) = moles(AgCl(s)) = 8.650 × 10-4 mol [Cl-(aq)] = moles(Cl-) ÷ volume in L moles(Cl-) = 8.650 × 10-4 mol volume of water = 25.00 mL = 25.00/1000 = 0.02500 L (or 25.00 × 10-3 L) [Cl-(aq)] = 8.650 × 10-4 ÷ 0.02500 = 0.0346 mol L-1 PAUSE PAUSE to Ponder Plausibility Is your answer plausible? Work backwards: assume the water had a chloride ion concentration of 0.0346 mol L-1, how much AgCl(s) would be produced from 25.00 mL of water? Calculate moles of Cl- in 25.00 mL of 0.0346 mol L-1 solution moles(Cl-) = concentration(Cl-) × volume in L = 0.0346 × 25.00/1000 = 8.65 × 10-4 mol moles(Cl-) = moles(AgCl) = 8.65 × 10-4 mol mass(AgCl) = moles(AgCl) × molar mass(AgCl) = 8.65 × 10-4 × (107.9 + 35.45) = 0.124 g mass(AgCl + filter paper) = 0.124 + 0.003 = 0.127 g The question gave us the same value, 0.127 g, so we are confident our answer is correct. STOP STOP! State the Solution The concentration of chloride ions in this lake water is 0.0346 mol L-1

Can you apply this?

Join AUS-e-TUTE!

Take the test now!

1. The separation of the element or the compound containing it may done a number of different ways, precipitation is one of the most common. Other methods include volatilisation and electro-analytical methods.

2. Seawater contains many dissolved ions! The main ones are:
chloride ions, Cl-
sodium ions, Na+
sulfate ions, SO42-
magnesium ions, Mg2+
calcium ions, Ca2+
potassium ions, K+
bromide ions, Br-

3. The solubility of silver chloride is 0.0002 g/100 g at 25°C. Because so very little of the silver chloride dissolves in water it is said to be insoluble in water at 25°C.

4. Unfortunately silver sulfate, Ag2SO4, will still precipitate out. It is best to remove the sulfate ions first before you begin the analysis for chloride ions.
The best way to remove sulfate ions is by anion-exchange.