Dilution of Solutions Calculations (c_{1}V_{1}=c_{2}V_{2})

Key Concepts

Dilution means to reduce the concentration of a solution.

A solution is produced when a solute dissolves in a solvent.

term

description

example

solute:

substance which dissolves

solid sodium chloride, NaCl(s)

solvent:

substance which enables the solute to dissolve

liquid water, H_{2}O(l)

solution:

prepared by dissolving a solute in a solvent

NaCl_{(aq)} is NaCl(s) dissolved in H_{2}O(l)

Concentration refers to the amount of solute dissolved in a given volume of solution.

amount of solute

volume of solution

concentration

units of measurement

moles mol

litres L

moles per litre (molarity) mol L^{-1} (mol/L or M)

concentration of a stock solution in mol L^{-1} = moles of solute ÷ volume of solution in litres
c_{1} = n_{1} ÷ V_{1} c_{1} = molarity of stock solution (concentration of stock solution in mol L^{-1})
n_{1} = moles of solute dissolved (in mol)
V_{1} = volume of stock solution (in L)

A solution can be diluted by adding more solvent to the stock solution (the starting solution before dilution) in the same vessel.

c_{1}V_{1} = c_{2}V_{2} c_{1} = concentration of stock solution (before dilution) in mol L^{-1} V_{1} = volume of stock solution present before dilution in L
c_{2} = concentration of new dilute solution (after more solvent added) in mol L^{-1} V_{2} = total volume of the new dilute solution in L

Note: V_{2} > V_{1} and c_{2} < c_{1}

find

calculation

determine concentration of dilute solution

c_{2} =

c_{1}V_{1} ÷ V_{2}

determine volume of dilute solution

V_{2} =

c_{1}V_{1} ÷ c_{2}

determine concentration of stock solution before dilution

c_{1} =

c_{2}V_{2} ÷ V_{1}

determine volume of stock solution used (before dilution)

V_{1} =

c_{2}V_{2} ÷ c_{1}

A solution can be diluted by using a pipette to transfer some of the stock solution to a volumetric flask and then adding solvent up to the mark:

c_{1}V_{1} = c_{2}V_{2} c_{1} = concentration of stock solution (before dilution) in mol L^{-1} V_{1} = volume of pipette used to transfer the stock solution in L
c_{2} = concentration of new dilute solution (after water added) in mol L^{-1} V_{2} = volume of volumetric flask used to prepare the dilute solution in L

find

calculation

determine concentration of dilute solution

c_{2} =

c_{1}V_{1} ÷ V_{2}

determine volume of dilute solution

V_{2} =

c_{1}V_{1} ÷ c_{2}

determine concentration of stock solution

c_{1} =

c_{2}V_{2} ÷ V_{1}

determine volume of pipette used

V_{1} =

c_{2}V_{2} ÷ c_{1}

Dilution Technique

Dilution means to reduce the concentration of a solution.

A solution can be diluted by adding solvent to a given volume of stock solution.

In order to dilute a solution we need:

A container of solution of known concentration. This is the stock solution, the solution to be diluted.

A pipette to remove a known volume of this solution from the container and transfer it to a new container such as a volumetric flask.

A clean volumetric flask.

A container of pure solvent, the same solvent used to make the original solution, such as water.

A way of adding the pure solvent to the volumetric flask, for example a glass funnel and a pasteur pipette.

Step

Description

Calculation

1

The solution to be diluted will be in a vessel such as a volumetric flask. This solution is referred to as the stock solution.
In order to prevent contamination of this stock solution, place a small amount of it in a clean, dry, conical flask (erlenmeyer flask), swirl it around and then discard this solution appropriately.
DO NOT use water or any other solvent to rinse the flask because adding more solvent would dilute the solution and we would no longer know what the concentration of this solution is.
Pour some of your stock solution into the rinsed conical flask.
Stopper the conical flask when you are not using it to prevent contamination from the atmosphere.
Label it with the date, and the formula of the solution and its concentration, for example [NaCl(aq)] = 0.10 mol L^{-1} Record the concentration of this undiluted stock solution as c_{1} in mol L^{-1}

c_{1} = [undiluted solution] mol L^{-1}

2

You will need a clean, dry, pipette with a known volume.
This will be printed on the pipette, for example 10.00 mL, 25.00 mL, etc
Use the pipette filler (bulb) to draw up a small amount of stock solution from the conical flask, remove the pipette from the solution, swirl the solution in the pipette around to rinse it and then discard this solution suitably.
DO NOT use water or any other solvent to rinse the pipette because you would be diluting the solution and we would no longer know what the concentration of this solution is.
Place the rinsed pipette into the stock solution in the conical flask and draw up the solution until it is past the level marked on the pipette.
Remove the pipette from the stock solution.
Suspended the pipette within the confines of a beaker.
Allow the solution to escape from the pipette drop by drop until the bottom of the meniscus (when viewed at eye level) sits just on the level marked on the pipette.
You need to record the volume of this undiluted, stock solution, held in the pipette.
It is a good idea to convert this volume in milliltres (mL) to a volume in litres (L) if your concentration is mol L^{-1} (molarity):
volume in L = volume in mL ÷ 1000
Record the volume of this pipette as V_{1}, for example, V(NaCl)_{1} = 10.00 mL = 10.00/1000 = 0.0100 L
V_{1} = [volume of undiluted solution in pipette] L

V_{1} = V(undiluted solution) L

3

Position the pipette so that the tip of the pipette meets the inside neck of the clean volumetric flask at an angle.
Allow the solution in the pipette to escape and run down the neck of the volumetric flask into the body of the flask.
Pipettes are usually designed to deliver a known volume, this means that the small drop of solution remaining in the tip of the pipette is supposed to be there, so don't try to blow it out.
We now have a known volume of the undiluted stock solution, with a known concentration in this volumetric flask.
Calculate the moles of solute, n(solute), contained in this volumetric flask.
moles of solute (mol) = concentration of solution (mol L^{-1}) x volume of solution delivered from pipette (L)
n(solute) = c_{1} x V_{1} Record the moles of solute, n(solute) in mol.

n(solute) = c_{1} x V_{1} mol

4

Place a glass funnel in the neck of the volumetric flask.
As you pour your solvent, for example water, into the funnel, leave a small air gap between the stem of the funnel and the neck of the volumetric flask so that the solvent flows freely.
When the solvent reaches the base of the neck of the volumetric flask, stop pouring and allow any solvent remaining in the funnel to flow through.
Slowly, and carefully, add more solvent until the bottom of the meniscus is about 1 cm from the level marked on the neck of the volumetric flask when viewed at eye level.
Use a clean pasteur pipette that has been rinsed with the solvent to add more solvent drop-by-drop to the volumetric flask until the bottom of the meniscus sits exactly on the level marked on the neck of the flask when viewed at eye level.
Stopper the volumetric flask.
Record the volume of the diluted solution in this volumetric flask as V_{2} in mL
Convert the volume in millilitres to a volume in litres dividing by 1000.
Record this volume as V_{2} L.

V_{2} = [dilute solution] L

5

Calculate the concentration of the dilute solution, c_{2}:
concentration of dilute solution = moles of solute ÷ volume of dilute solution
molarity of dilute solution (mol L^{-1}) = moles solute (mol) ÷ volume of dilute solution (L)
c_{2} = n(solute) ÷ V_{2} Affix a label to the volumetric flask which states the date the solution was made, its formula and concentration.

The Dilution Equation (dilution formula, or, dilution expression)

We can summarise the process of diluting a stock solution as:

undiluted solution

→ pipette

→ volumetric flask

→ add solvent

→ dilute solution

c(undiluted solution) mol L^{-1}

V(pipette) in mL V(pipette) in mL ÷ 1000 = V(pipette) L

n(solute) = c(undiluted solution) x V(pipette in L) mol

V(dilute solution) in mL
V(dilute solution in mL) ÷ 1000 = V(dilute solution in L)

c(dilue solution) = n(solute) ÷ V(dilute solution in L)

c_{1} in mol L^{-1}

V_{1} in L

n(solute) = c_{1} x V_{1} in mol

V_{2} in L

c_{2} = n(solute) ÷ V_{2} in mol L^{-1} or
c_{2} = (c_{1} x V_{1}) ÷ V_{2} in mol L^{-1}

So we now have an equation (expression or formula) to calculate the concentration of a solution after it has been diluted:

c_{2}

=

c_{1} x V_{1} V_{2}

c_{1} = concentration of undiluted solution in mol L^{-1} V_{1} = volume of stock solution added to the volumetric flask = volume of pipette used in L
c_{2} = concentration of the dilute solution in mol L^{-1} V_{2} = volume of volumetric flask holding the dilute solution in L

If we rearrange this equation (expression or formula) by multiplying both sides of the equation by V_{2} we get:

c_{1}V_{1} = c_{2}V_{2}

which simply means that the moles of solute transferred by pipette and placed in the volumetric flask (n_{1}) equals the moles of solute present in the volumetric flask after more solvent was added during the dilution (n_{2}).

moles solute transferred by pipette to volumetric flask

=

moles solute in volumetric flask after solvent added (dilute solution)

c_{1}V_{1}

=

c_{2}V_{2}

n_{1}

=

n_{2}

This equation (formula or expression) is very useful because we can rearrange it find:

c_{1} the concentration of the undiluted stock solution:

c_{1} = c_{2}V_{2} ÷ V_{1}

V_{1} the volume of pipette required to transfer the undiluted solution to the volumetric flask for dilution:

V_{1} = c_{2}V_{2} ÷ c_{1}

V_{2} the volume of volumetric flask required to make the dilute solution:

V_{2} = c_{1}V_{1} ÷ c_{2}

c_{2} the concentration of solution in the volumetric flask after dilution:

c_{2} = c_{1}V_{1} ÷ V_{2}

Examples

Calculate concentration of solution after dilution: c_{2} = (c_{1}V_{1}) ÷ V_{2}

Calculate the new concentration in mol L^{-1} (molarity) if enough water is added to 100.00 mL of 0.25 mol L^{-1} sodium chloride solution to make up 1.5 L.

What is the question asking you to do?
Calculate concentration of NaCl(aq) in mol L^{-1} after dilution
c_{2} = ? mol L^{-1}

What information (data) has been given in the question?
Extract the data from the question :
c_{1} = 0.25 mol L^{-1} V_{1} = 100.00 mL = 100.00 ÷ 1000 = 0.10 L (volume must be in litres)
V_{2} = 1.5 L

What is the relationship between what you know and what you need to find out?
Write the equation (formula or expression):
c_{2} = (c_{1}V_{1}) ÷ V_{2}

Solve the problem:
Substitute the values into the equation and solve :
c_{2} = (0.25 x 0.10) ÷ 1.5 = 0.017 mol L^{-1} [NaCl_{(aq)}] = 0.017 mol L^{-1} (or 0.017 mol/L or 0.017 M)

Calculate the volume of solution after dilution (V_{2} = (c_{1}V_{1}) ÷ c_{2})

Calculate the volume in litres to which 500.00 mL of 0.020 mol L^{-1} coppper sulfate solution must be diluted to make a new solution with a concentration of 0.0010 mol L^{-1}.

What is the question asking you to do?
Calculate the volume of CuSO_{4}(aq) after dilution in L
V_{2} = ? L

What information (data) has been given in the question?
Extract the data from the question :
c_{1} = 0.020 mol L^{-1} V_{1} = 500.00 mL = 500.00 ÷ 1000 = 500.00 x 10^{-3} L = 0.50 L (since there are 1000 mL in 1L)
c_{2} = 0.0010 mol L^{-1}

What is the relationship between what you know and what you need to find out?
Write the equation (formula or expression):
V_{2} = (c_{1}V_{1}) ÷ c_{2}

Solve the problem:
Substitute the values into the equation and solve :
V_{2} = V(CuSO_{4})_{new} = (0.020 x 0.50) ÷ 0.0010 = 10 L

Calculate the concentration in mol L^{-1} of solution before dilution c_{1} = (c_{2}V_{2}) ÷ V_{1}

Calculate the concentration of the undiluted CuSO_{4}(aq) if 10.00 mL of this solution was used to make 100.00 mL of dilute solution with a concentration of 0.20 mol L^{-1}.

What is the question asking you to do?
Calculate the concentration of the undiluted CuSO_{4}(aq) in mol L^{-1} c_{1} = ? mol L^{-1}

What information (data) has been given in the question?
Extract the data from the question :
V_{1} = 10.00 mL = 10.00 ÷ 1000 = 10 x 10^{-3} L = 0.010 L (since there are 1000 mL in 1 L)
c_{2} = 0.20 mol L^{-1} V_{2} = 100.00 mL = 100.00 ÷ 1000 = 100 x 10^{-3} = 0.10 L

What is the relationship between what you know and what you need to find out?
Write the equation (formula or expression):
c_{1} = (c_{2}V_{2}) ÷ V_{1}

Solve the problem:
Substitute the values into the equation and solve :
c_{1} = c(CuSO_{4})_{undiluted} = (0.20 x 0.10) ÷ 0.010 = 2.0 mol L^{-1}

Calculate the volume of the pipette used to dilute a solution (V_{2} = (c_{1}V_{1}) ÷ c_{2})

Calculate the volume of the pipette in millilitres used to transfer a quantity of 0.15 mol L^{-1} CuSO_{4}(aq) to a 250.00 mL volumetric flask made up with enough water to produce a solution with a concentration of 0.012 mol L^{-1}.

What is the question asking you to do?
Calculate the volume of the pipette in millilitres
V_{1} = ? mL

Extract the data from the question :
c_{1} = 0.15 mol L^{-1} c_{2} = 0.012 mol L^{-1} V_{2} = 250.00 mL = 250.00 ÷ 1000 = 0.25 L

What is the relationship between what you know and what you need to find out?
Write the equation (formula or expression):
V_{1} = (c_{2}V_{2}) ÷ c_{1}

Solve the problem:
Substitute the values into the equation and solve :
V_{1} = V(CuSO_{4})_{undiluted} = (0.012 x 0.25) ÷ 0.15 = 0.020 L = 0.020 ÷ 1000 = 20 mL

Problem Solving for Dilution Questions

The Problem: Chris the chemist requires two aqueous solutions of sodium carbonate, Na_{2}CO_{3}(aq), of different concentrations for an analytical procedure.
Chris carefully weighs out 5.00 g of sodium carbonate and transfers this to a 500.00 mL volumetric flask, adding enough water to make the solution up to the mark.
Then Chris uses a pipette to transfer 50.00 mL of this solution to a 250.00 mL volumetric flask, adding enough water to make the solution up to the mark.
Calculate the concentration of the second solution in mol L^{-1}.

What is the question asking you to do?
Calculate the concentration of the second solution in mol L^{-1}.

What chemical principle will you need to apply?
Apply stoichoimetry (chemical calculations)

What information (data) have you been given?

solute is Na_{2}CO_{3}

solvent is water

m(Na_{2}CO_{3}) = mass of solute = 5.00 g

V_{stock solution} = volume of volumetric flask for undiluted (stock) solution = 500.00 mL

V_{pipette} = volume of pipette = 50.00 mL

V_{dilute} = volume of dilute solution in second volumetric flask = 250.00 mL

c_{dilute} = concentation of dilute solution in second volumetric flask = ? mol L^{-1}

PAUSE!

Plan.

Step 1: Calculate moles of Na_{2}CO_{3} used to make stock (undiluted) solution Assume the Na_{2}CO_{3} is 100% pure (no impurities) m(Na_{2}CO_{3}) = 5.00 g Use Periodic Table to find relative atomic masses:
M_{r}(Na)
M_{r}(C)
M_{r}(O)
Calculate molar mass of Na_{2}CO_{3}:
M(Na_{2}CO_{3}) = 2 x M_{r}(Na) + M_{r}(C) + 3 x M_{r}(O)

Step 2: Calculate concentration of the stock (undiluted) solution n(Na_{2}CO_{3}) from above
Assume temperature of laboratory is the same as that required by the volumetric flask (25^{o}C for example) V_{stock solution} = 500.00 mL convert volume in mL to volume in L:
V_{stock in L} = V_{stock in mL}/1000
Assume the water being used as the solvent contains no Na_{2}CO_{3}

c_{stock} = n(Na_{2}CO_{3}) ÷ V_{stock solution in L}

Step 3: Calculate the moles of Na_{2}CO_{3} in the pipette which is transferred to the second volumetric flask for dilution c_{stock} calculated above
Assume temperature of laboratory is the same as that required by the pipette (25^{o}C for example) V_{pipette} = 50.00 mL convert volume of pipette in mL to volume in L
V_{pipette in L} = V_{pipette in mL}/1000

n(Na_{2}CO_{3} pipette) = c_{stock} x V_{pipette in L}

Step 4: Calculate concentration of dilute solution of Na_{2}CO_{3} in second volumetric flask. n(Na_{2}CO_{3} pipette) = n(Na_{2}CO_{3} second volumetric flask) calculated above
Assume temperature of laboratory is the same as that required by the volumetric flask (25^{o}C for example) V_{dilute} = 250.00 mL convert volume in mL to volume in L
V_{dilute in L} = V_{dilute in mL}/1000
Assume the water being used as the solvent contains no Na_{2}CO_{3}

c_{dilute} = n(Na_{2}CO_{3} pipette) ÷ V_{dilute in L}

GO!

Go with the Plan.

Step 1: Calculate moles of Na_{2}CO_{3} used to make stock (undiluted) solution Assume the Na_{2}CO_{3} is 100% pure (no impurities) m(Na_{2}CO_{3}) = 5.00 g Use Periodic Table to find relative atomic masses:
M_{r}(Na) = 22.99 M_{r}(C) = 12.01 M_{r}(O) = 16.00 Calculate molar mass of Na_{2}CO_{3}:
M(Na_{2}CO_{3}) = 2 x M_{r}(Na) + M_{r}(C) + 3 x M_{r}(O)
M(Na_{2}CO_{<3}) = 2 x 22.99 + 12.01 + 3 x 16.00 = 105.99 g mol^{-1}

Step 2: Calculate concentration of the stock (undiluted) solution n(Na_{2}CO_{3}) from above = 0.04717 mol Assume temperature of laboratory is the same as that required by the volumetric flask (25^{o}C for example) V_{stock solution} = 500.00 mL convert volume in mL to volume in L:
V_{stock in L} = V_{stock in mL}/1000 = 500.00/1000 = 0.50 L Assume the water being used as the solvent contains no Na_{2}CO_{3}

Step 3: Calculate the moles of Na_{2}CO_{3} in the pipette which is transferred to the second volumetric flask for dilution c_{stock} calculated above = 0.0943 mol L^{-1} Assume temperature of laboratory is the same as that required by the pipette (25^{o}C for example) V_{pipette} = 50.00 mL convert volume of pipette in mL to volume in L
V_{pipette in L} = V_{pipette in mL}/1000 = 50.00 ÷ 1000 = 0.050 L

n(Na_{2}CO_{3} pipette) = c_{stock} x V_{pipette in L} n(Na_{2}CO_{3}) = 0.0943 x 0.050 = 0.00472 mol

Step 4: Calculate concentration of dilute solution of Na_{2}CO_{3} in second volumetric flask. n(Na_{2}CO_{3} pipette) = n(Na_{2}CO_{3} second volumetric flask) calculated above = 0.00472 mol Assume temperature of laboratory is the same as that required by the volumetric flask (25^{o}C for example) V_{dilute} = 250.00 mL convert volume in mL to volume in L
V_{dilute in L} = V_{dilute in mL}/1000 = 250.00 ÷ 1000 = 0.25 L Assume the water being used as the solvent contains no Na_{2}CO_{3}

Have you answered the question that was asked?
You were asked to find the concentration of the second (dilute) solution, you have done this, so yes, you have answered the question that was asked.

Is your solution to the question reasonable?
A dilute solution will have a lower concentration than the undiluted stock solution.
Stock solution concentration was 0.0943 mol L^{-1} Dilute solution was 0.019 mol L^{-1} 0.019 < 0.0943 so solution is reasonable.

Check your calculations by working backwards:
Calculate concentration of stock solution using your value for the dilute solution:
c_{stock} = c_{dilute}V_{pipette} ÷ V_{stock} c_{stock} = 0.019 x 0.25 ÷ 0.050 = 0.095
n(Na_{2}CO_{3} in stock) = c_{stock} x V_{stock} n(Na_{2}CO_{3}) = 0.095 x 0.50 = 0.0475 mol
m(Na_{2}CO_{3}) = n(Na_{2}CO_{3}) x M(Na_{2}CO_{3})
m(Na_{2}CO_{3}) = 0.0475 x 105.99 = 5 g
Since 5.00 g of Na_{2}CO_{3} was used to make the stock solution, we are confident that our calculations are correct.

STOP!

State the solution.

The concentration of the second solution is 0.019 mol L^{-1}.

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