Titrimetric procedure and Acid
and Base Titrations Ch 20
Apparatus used in Volumetric analysis
• A Graduated cylinder
• Roughly measures the volume of a liquid
Apparatus used in Volumetric analysis
The volumetric flask
• Is used to hold a definite volume of a liquid.
• Should always be clean so needs to be washed out
with deionised water before you start!
Apparatus used in Volumetric analysis
A pipette
• Is used to deliver an exact
amount of a liquid.
• Should always be clean so
needs to be washed out with
• (i) deionised water
• (ii) solution it is to contain
before you start!
Apparatus used in Volumetric analysis
The burette
• Is used to accurately measure the
volume of a liquid.
• Should be washed out with
• (i) deionised water
• (ii) the solution that it is to contain
before use.
Apparatus used in Volumetric analysis
• Used for mixing liquids together – designed to
prevent the splashing of liquids.
Acid – Base Titrations
• Not every substance can be prepared directly in a
standard solution!
•Example:
Hydrochloric acid is too volatile
to make a standard solution
directly – as soon as you open a
bottle some escapes as a gas!
Preparing a standard solution of
Hydrochloric acid
• Hydrochloric acid can be standardised (its
exact concentration can be found out) by
reacting it with a a standard solution of
sodium carbonate:
Balanced equation:
• 2HCl + Na2CO3
2NaCl + H20 +CO2
Finding the unknown concentration:
• You use the following equation:
Molarity of acid
Volume of acid
reacted
Volume of base
reacted
V1 X M1 = V2 x M2
n1
Number of moles of acid
indicated in the balanced
equation
Molarity of base
n2
Number of moles of base indicated
in the balanced equation
Question 234(e)
HCl + NaOH
•
NaCl+ H20
V1 X M1 = V2 x M2
n1
n2
Solution 1 – HCl
Solution 2 - NaOH
V1 = 27.5cm3
V2 = 25cm3
M1 = 0.1 M
M2 = ?
n1 =
1
n2 =
1
Question
234(e)(i)
V1 X M1 = V2 x M2
n1
n2
(27.5)X (0.1) = (25) x (M2)
1
1
(27.5)X (0.1)X (1) = M2
(1) x (25)
0.11 = M2
The concentration of NaOH solution is 0.11 M (moles per litre)
(ii) What is the concentration in grams per litre?
• Given Moles PER LITRE
Find Grams PER LITRE
How many moles in 1 litre of solution?
0.11 x RMM = ?
0.11 x 40 = 4.4g
There are 4.4g of NaOH in one litre.
Answer = 4.4gL-1
Question 235he)
2HCl + Na2CO3
•
2NaCl+ H20 +CO2
V1 X M1 = V2 x M2
n1
n2
Solution 1 – HCl
Solution 2 – Na2CO3
V1 = 30cm3
V2 = 25cm3
M1 = ?
M2 = 0.06M
n1 =
2
n2 =
1
Q235d
How many grams of sodium carbonate is needed to make up a 0.06M
solution?
Given Moles PER LITRE
Find Grams PER LITRE
How many moles in 1 litre of solution?
0.06 moles x RMM = ?
0.06 X 106g = 6.36
There are 6.36g of Na2CO3 in one litre.
Answer = 6.36g would be needed
Question
235(h)(i)
V1 X M1 = V2 x M2
n1
n2
(30)X (M1) = (25) x (0.06)
2
1
M1 = (25) x (0.06) x (2)
(1) x (30)
M1 = 0.1
The concentration of HCl solution is 0.1 M (moles per litre)
(ii) What is the concentration in grams per litre?
• Given Moles PER LITRE
Find Grams PER LITRE
0.1 moles X RMM = ?
(0.1)(36.5) = 3.65
There are 3.65g of HCl in one litre.
Answer = 3.65gL-1
Vinegar Titrations
Check your learning…PEQ
Why is the vinegar diluted?
• Diluting the vinegar reduces the volume of
vinegar and the volume of sodium hydroxide
solution needed in the experiment
Outline the correct procedure for
bringing the solution in the volumetric
flask precisely to the 250cm3 mark.
•
•
•
•
•
•
•
rinse pipette (burette) with water //
and then with vinegar //
fill with pipette filler / have bottom of meniscus on calibration mark
deliver 25 cm3 to 250 cm3 volumetric flask //
add deionised (distilled, pure) water until level of water near mark //
add dropwise (by dropper / by pipette / by wash bottle) //
bring bottom of meniscus to (on, at) mark / vol. flask at eye-level
(vertical) //
• stopper and invert several times / mix thoroughly / solution
homogeneous (even concentration, same concentration throughout)
ANY FIVE: (5 × 3)
Outline the procedure used in
preparing the burette so that it is
ready for the first titration.
• Rinse the burette with deionised water, and then
with diluted vinegar solution.
• Fill the burette with diluted vinegar solution
above the zero mark.
• Remove the funnel.
• Using the tap at the base of the burette, allow
the acid to flow into a beaker until the level of
liquid is at the zero mark.
• Ensure that there are no air bubbles in the nozzle
of the burette.
Give two other precautions which
should be taken to ensure that the
burette readings are accurate.
• Make sure that the burette is clamped
vertically.
• Read the level of liquid in the burette by
noting the lower level of the meniscus at eye
level.
Why is phenolphthalein used as the
indicator in this titration?
• Because this is a weak acid/strong base
titration,
• phenolphthalein changes colour in the
appropriate pH range.
Why is a rough titration carried out?
• To find the approximate end-point. This
information enables the subsequent titrations
to be carried out more quickly.
Why are three accurate titrations
carried out?
• To reduce experimental error by calculating
the average value.
What colour change happens at the
end point?
• The indicator changes colour from pink to
colourless.
. Vinegar is a solution of ethanoic acid (acetic acid). Some
bottles of vinegar are labelled “White Wine Vinegar”.
(a) What compound in white wine is converted to ethanoic acid
in vinegar?
What type of chemical process converts this compound to
ethanoic acid? (8)
• COMPOUND: ethanol
• PROCESS: oxidation (4)
Describe the procedure for accurately
measuring the 50 cm3 sample of
vinegar and diluting it to 500 cm3. (12)
DESCRIBE:
pipette (burette) vinegar (3) can be shown on diagram
• into volumetric flask (3) can be shown with diagram
provided line on neck present
• add deionised water (3)
• when near mark, add dropwise (using
dropper/pipette/wash bottle) / until
• bottom of meniscus on (at) mark / read bottom of
meniscus (3)
Name the piece of equipment that should be used to measure
the ethanoic acid solution during the titration.
• burette (3)
State the procedure for washing and filling this piece of
equipment in preparation for the titration.
• WASHING & with deionised water / then
solution (ethanoic acid, diluted vinegar) /
• FILLING: use of funnel (pour in at top) /
ensure that the area below (jet, tip, nozzle)
the tap is filled
Name a suitable indicator for this
titration
• INDICATOR: phenolphthalein /
Question 236
• A 20 cm3 sample of vinegar was diluted to 100 cm3 in a
volumetric flask.
• This diluted solution was then titrated against 25 cm3 of
0.11 M NaOH solution.
• The balanced equation for the reaction is:
CH3COOH + NaOH
CH3COONa + H2O
The average titration figure was 15.8 cm3. Calculate the
concentration of ethanoic acid in the original vinegar in
(a) moles per litre
(b) grams per litre
(c) % (w/v)
(a) Find concentration of ethanoic acid in original
vinegar solution in moles per litre
• We will first find the molarity of the made up vinegar
solution, (then we can use this to find the molarity of the
original vinegar solution)
NaOH + CH3COOH
•
CH3COOH + H20
V1 X M1 = V2 x M2
n1
n2
Solution 1 – NaOH
Solution 2 – CH3COOH
V1 = 25cm3
V2 = 15.8cm3
M1 = 0.11
M2 = ?
n1 = 1
n2 =
1
V1 X M1 = V2 x M2
n1
(25)X (0.11) = (15.8) x (M2)
1
1
(25)X (0.11)X (1) = M2
(1) x (15.8)
0.1741= M2
The concentration of the vinegar solution made up was
0.1741M (moles per litre)
n2
(a) Find concentration of ethanoic acid in original
vinegar solution in moles per litre
• The concentration of the made up vinegar solution was
0.1741M (moles per litre)
•The original vinegar solution was diluted down 5 times
( from 20cm3 to 100cm3)
•So the original vinegar solution was 5 times more concentrated
(O.1741 X 5 = 0.8703M )
Answer (a)
The molarity of ethanoic acid in the original vinegar solution was
0.8703M ( moles per litre)
(b) What is the concentration of ethanoic acid in the original vinegar
solution in grams per litre?
• Given Moles PER LITRE
Find Grams PER LITRE
0.8705 moles x RMM = Mass
(0.8703)(60) = 52.218
Answer b
The concentration of ethanoic acid in the original vinegar
solution was 52.218 g per litre .
(c)Express the concentration of ethanoic acid in the original
vinegar solution in w/v%
Answer:
w/v means how many grams of CH3COOH are in 100cm3 of solution
We know mass in 1000cm3
52.218 /10 = 5.2218
Answer c)
5.2218 w/v%
Want mass in 100cm3
Q237
• A 25 cm3 sample of vinegar was diluted to 250 cm3 in a volumetric
flask.
• This diluted solution was then titrated against 25 cm3 of .09 M
NaOH solution.
• The balanced equation for the reaction is:
CH3COOH + NaOH
CH3COONa + H2O
The average titration figure was 26.55 cm3. Calculate the concentration
of ethanoic acid in the original vinegar in
(a) moles per litre
(b) grams per litre
(c) % (w/v)
(a) Find concentration of ethanoic acid in original
vinegar solution in moles per litre
• We will first find the molarity of the made up vinegar
solution, (then we can use this to find the molarity of the
original vinegar solution)
NaOH + CH3COOH
•
CH3COOH + H20
V1 X M1 = V2 x M2
n1
n2
Solution 1 – NaOH
Solution 2 – CH3COOH
V1 = 25cm3
V2 = 26.55cm3
M1 = 0.09
M2 = ?
n1 = 1
n2 =
1
V1 X M1 = V2 x M2
n1
(25)X (0.09) = (26.55) x (M2)
1
1
(25)X (0.09)X (1) = M2
(1) x (26.55)
0.0847= M2
The concentration of the vinegar solution made up was
0.0847M (moles per litre)
n2
(a) Find concentration of ethanoic acid in original
vinegar solution in moles per litre
• The concentration of the made up vinegar solution was
0.08474576 M (moles per litre)
•The original vinegar solution was diluted down 10 times
( from 25cm3 to 250cm3)
•So the original vinegar solution was 10 times more concentrated
(0.0847M X 10 = 0.8475M )
Answer (a)
The molarity of ethanoic acid in the original vinegar solution was
0.8475M ( moles per litre)
(b) What is the concentration of ethanoic acid in the original vinegar
solution in grams per litre?
• Given Moles PER LITRE
Find Grams PER LITRE
0.8475 moles X rmm = mass
(0.8475)(60) = 50.85
Answer b
The concentration of ethanoic acid in the original vinegar
solution was 50.85g per litre .
(c)Express the concentration of ethanoic acid in the original
vinegar solution in w/v%
Answer:
w/v means how many grams of CH3COOH are in 100cm3 of solution
We know mass in 1000cm3
50.85/10 = 5.085
Answer c)
5.085 w/v%
Want mass in 100cm3
Q238
• To determine the concentration of ethanoic acid, CH3COOH, in a
sample of vinegar, the vinegar was first diluted and then titrated
against 25.0 cm3 portions of a previously standardised 0.10 M
solution of sodium hydroxide, NaOH.
• One rough and two accurate titrations were carried out.
• The three titration figures recorded were 22.9, 22.6 and 22.7 cm3,
respectively.
CH3COOH + NaOH
CH3COONa + H2O
Calculate the concentration of the diluted solution of ethanoic acid in
(i) moles per litre, (ii) grams per litre.
State the concentration of ethanoic acid in the original vinegar sample
in grams per litre. Express this concentration in terms of % (w/v).
(d) Find concentration of ethanoic acid in THE DILUTED
vinegar solution in moles per litre
NaOH + CH3COOH
CH3COOH + H20
• V1 X M1 = V2 x M2
n1
n2
Solution 1 – NaOH
Solution 2 – CH3COOH
V1 = 25cm3
V2 = 22.65cm3 (AVERAGE)
M1 = 0.1
M2 = ?
n1 = 1
n2 =
1
V1 X M1 = V2 x M2
n1
n2
(25)X (0.1) = (22.65) x (M2)
1
1
(25)X (0.1)X (1) = M2
(1) x (22.65)
0.1104 = M2
ANSWER: The concentration of the DILUTED solutionwas
0.1104 M (moles per litre)
(b) What is the concentration of ethanoic acid in the DILUTED vinegar
solution in grams per litre?
• Given Moles PER LITRE
Find Grams PER LITRE
Moles x RMM = mass
0.1104 x 60 = 6.624
Answer b
The concentration of ethanoic acid in the original vinegar
solution was 6.624g per litre .
(C) Find concentration of ethanoic acid in original
vinegar solution in grams per litre
• The concentration of the diluted vinegar solution was
6.624g per litre .
•The original vinegar solution was diluted down 10 times
( from 25cm3 to 250cm3)
•So the original vinegar solution was 10 times more concentrated
(6.624g per litre X 10 = 66.24g per litre
Answer (a)
The molarity of ethanoic acid in the original vinegar solution was
66.24g per litre
(c)Express the concentration of ethanoic acid in the original
vinegar solution in w/v%
Answer:
w/v means how many grams of CH3COOH are in 100cm3 of solution
We know mass in 100cm3
66.24g/ 10 = 6.624
Answer c)
6.624 w/v%
Want mass in 100cm3
Q.239
• The concentration of ethanoic acid in vinegar was measured as follows:
• A 50 cm3 sample of vinegar was diluted to 500 cm3 using deionised water.
• The diluted solution was titrated against 25 cm3 portions of a standard
0.12 M sodium hydroxide solution, using a suitable indicator.
The titration reaction is CH3COOH + NaOH → CH3COONa + H2O
• After carrying out a number of accurate titrations of the diluted solution
of ethanoic acid against the 25 cm3 portions of the standard 0.12 M
sodium hydroxide solution, the mean titration figure was found to be 20.5
cm3.
(d) Calculate the concentration of ethanoic acid in the diluted vinegar solution
in moles per litre and hence calculate the concentration of ethanoic acid
in the original sample of vinegar. Express this concentration in terms of %
(w/v).
Find concentration of ethanoic acid in THE
DILUTED vinegar solution in moles per litre
NaOH + CH3COOH
CH3COOH + H20
• V1 X M1 = V2 x M2
n1
n2
Solution 1 – NaOH
Solution 2 – CH3COOH
V1 = 25cm3
V2 = 20.5cm3 (AVERAGE)
M1 = 0.12
M2 = ?
n1 = 1
n2 =
1
V1 X M1 = V2 x M2
n1
n2
(25)X (0.12) = (22.65) x (M2)
1
1
(25)X (0.12)X (1) = M2
(1) x (20.5)
0.1463 = M2
ANSWER: The concentration of the DILUTED solution was 0.1463 M
(moles per litre)
(a) Find concentration of ethanoic acid in original
vinegar solution in moles per litre
• The concentration of the made up vinegar solution was
0.1463 M (moles per litre)
•The original vinegar solution was diluted down 10 times
( from 50cm3 to 500cm3)
•So the original vinegar solution was 10 times more concentrated
(0.1463 M X 10 = 1.463 M )
Answer (a)
The molarity of ethanoic acid in the original vinegar solution was
1.463M ( moles per litre)
(c)Express the concentration of ethanoic acid in the original
vinegar solution in w/v%
• Given Moles PER LITRE
Find Grams PER LITRE
1.463 M X rmm = mass
(1.463 )(60) = 87.78
The concentration of ethanoic acid in the original vinegar
solution was 87.78 g per litre .
w/v%
(c)Express the concentration of ethanoic acid in the original
vinegar solution in w/v%
Answer:
w/v means how many grams of CH3COOH are in 100cm3 of
solution
We know mass in 1000cm3
100cm3
87.78 / 10 = 8.778g
Answer
8.778 w/v%
Want mass in
To determine the percentage water of crystallisation and the
degree of water of crystallisation, x, in a sample of hydrated
sodium carbonate crystals (Na2CO3.xH2O
Check your learning…PEQ
• What was done to the volumetric flask and its
contents immediately after the solution had
been made up to the mark with deionised
water? Why was it important to do this?
• It was stoppered, and then inverted several
times. To ensure a homogeneous solution.
Identify a primary standard reagent which could
have been used to standardise the hydrochloric
acid solution. (5)
• anhydrous sodium carbonate (Na2CO3)
Sodium carbonate crystals, Na2CO3.xH2O, is not
a primary standard but anhydrous sodium
carbonate, Na2CO3, may be used as a primary
standard. Why is this the case? (6)
• (b) Name a suitable indicator for the titration
and state the colour change observed in the
conical flask at the end point.
• methyl orange
• orange // to pink
Explain why not more than 1 – 2 drops of
indicator should be used. (12)
indicator is a weak acid / indicator is a weak
base (3) so could part in the reaction and affect
the titre result
Describe the correct procedure for rinsing the
burette before filling it with the solution it is to
deliver.
• rinse with deionised (distilled) water //
• rinse with reagent (solution) (2 × 3)
Why is it important to fill the part below the tap
of the burette? (12)
• some of measured volume goes to fill space
below the tap and isnt delivered into the
reaction mixture - distorts titre reading
• (i) Describe the correct procedure for
weighing and making up the solution from
hydrated sodium carbonate crystals. (12)
• (ii) Name a suitable indicator for the titration
and state the colour change at the end point.
(6)
iii) Describe the correct procedure for washing
the pipette and using it to measure the sodium
carbonate solution. (9)
(
Assuming that the burette has been
properly rinsed, state three other
precautions that should be taken when
using it in order to ensure an accurate
measurement. (12)
Q241.
• In an experiment, 3.55 g of hydrated sodium
carbonate, Na2CO3.xH2O, were weighed out,
dissolved in water and the solution was made up
to 500 cm3 in a volumetric flask.
• 25 cm3 of this solution were titrated against 0.11
M hydrochloric acid solution. The average
titration figure was 22.80 cm3.
The balanced equation for the reaction is:
Na2CO3 + 2HCl →2NaCl + H2O + CO2
Calculate:
• (a) the percentage of water of crystallisation in
the compound
• (b) the value of x in the formula Na2CO3.xH2O
Find concentration of titrated sodium carbonate
solution in moles per litre( molarity)
The balanced equation for the reaction is:
Na2CO3 + 2HCl →2NaCl + H2O + CO2
• V1 X M1 = V2 x M2
n1
n2
Solution 1 – HCl
Solution 2 – Na2CO3
V1 = 22.8cm3
V2 = 25cm3
M1 = 0.11
M2 = ?
n1 = 2
n2 =
1
V1 X M1 = V2 x M2
n1
n2
(22.8)X (0.11) = (25) x (M2)
2
1
(22.8)X (0.11)X (1) = M2
(2) x (25)
0.0502 = M2
The concentration of the Na2CO3 solution was 0.0502 M (moles per
litre)
Find how many moles of sodium carbonate were
in the volumetric flask ( 500cm3 of the solution)
• The concentration of the Na2CO3 solution was 0.0502
M (moles per litre)
• 0.0502 /2 = 0.0251 moles
There were 0.0251 moles of Na2CO3 in the
volumetric flask
Find the mass of sodium carbonate in the
volumetric flask ( 500cm3 of the solution)
• There were 0.0251 moles of Na2CO3 in the volumetric flask
• 0.02508 moles x RMM = mass
(0.02508)(106) = 2.6606
There were 2.6606g of sodium carbonate in the volumetric flask
Find the mass of water in the hydrated sodium
carbonate weighed out
There were g of 2.606g pure sodium carbonate in the sample
weighed out.
Mass of hydrated sodium carbonate = mass of sodium carbonate + mass of water
3.55g = 2.6606g
+ mass of water
3.55 - 2.6606g = mass of water
0.8894g = mass of water
0.8894g is the mass of water in the hydrated sodium carbonate
weighed out
Find the percentage of water of
crystallisation in the compound
• % water of crystallisation = mass of water in compound x 100
Total mass of compound
1
=
0.8894g x 100
3.55g
1
= 25.0535%
Answer (i)
the percentage of water of crystallisation in the compound is
25.0535%
Find the value of x in the formula
Na2CO3.xH2O
• We must find the simplest ratio between the amounts of the two substances
• Moles of sodium carbonate: Moles of water
• 0.0251 moles of Na2CO3 : ? Moles of water
Mass of water/ RMM = Moles of water
0.8894/ RMM = moles of water
0.8894/18 = 0.0494
• 0.0251 moles of Na2CO3 : 0.0494 moles of water
• 1 Na2CO3
: 0.0494/ 0.0251 moles of water
• 1 Na2CO3
: 1.9681 moles of water
Answer (ii)
the value of x in the formula Na2CO3.xH2O is 1.9681
Q242.
An experiment was carried out to determine the percentage water of
crystallisation and the degree of water of crystallisation, x, in a sample of
hydrated sodium carbonate crystals Na2CO3.xH2O, An 8.20 g sample of
the crystals was weighed accurately on a clock glass and then made up to
500 cm3 of solution in a volumetric flask. A pipette was used to transfer
25.0 cm3 portions of this solution to a conical flask. A previously
standardised 0.11 M hydrochloric acid (HCl) solution was used to titrate
each sample. A number of accurate titrations were carried out. The
average volume of hydrochloric acid solution required in these titrations
was 26.05 cm3.
• The balanced equation for the reaction is:
Na2CO3 + 2HCl →2NaCl + H2O + CO2
Calculate:
(d) Calculate the concentration of sodium carbonate in (i) moles per litre (ii)
grams per litre
• (e) the percentage of water of crystallisation in the compound
Find concentration of titrated sodium carbonate
solution in moles per litre( molarity)
The balanced equation for the reaction is:
Na2CO3 + 2HCl →2NaCl + H2O + CO2
• V1 X M1 = V2 x M2
n1
n2
Solution 1 – HCl
Solution 2 – Na2CO3
V1 = 26.05cm3
V2 = 25cm3
M1 = 0.11
M2 = ?
n1 = 2
n2 =
1
V1 X M1 = V2 x M2
n1
n2
(26.05)X (0.11) = (25) x (M2)
2
1
(22.8)X (0.11)X (1) = M2
(2) x (25)
0.0573 = M2
Answer d(i)The concentration of the Na2CO3 solution was 0.0573 M
(moles per litre)
Find the concentration of sodium carbonate in the
grams per litre
• There were 0.0573 moles in 1 litre ( that was the molarity)
• 0.0573 moles X rmm = mass
• (0.0573)(106) = 6.0738
Answer (d) (ii)
The concentration of sodium carbonate is 6.0738 grams per litre
Find how many grams of sodium carbonate were
in the volumetric flask ( 500cm3 of the solution)
• The concentration of the sodium carbonate solution
is 6.0738 grams per litre
• 6.0738/2 = 3.0369g
There were 3.0369 grams of Na2CO3 in the
volumetric flask
Find the mass of water in the hydrated sodium
carbonate weighed out
There were 3.0369 g pure sodium carbonate in the sample weighed
out.
Mass of hydrated sodium carbonate = mass of sodium carbonate + mass of water
8.20g = 3.0369g
+ mass of water
8.20 - 3.0369g = mass of water
5.1631g = mass of water
5.1631g is the mass of water in the hydrated sodium carbonate
weighed out
Find the percentage of water of
crystallisation in the compound
• % water of crystallisation = mass of water in compound x 100
Total mass of compound
1
=
5.1631 x 100
8.20g
1
= 62.9646%
Answer (e)
the percentage of water of crystallisation in the compound is
62.9646%
Q243
• A sample of 2.51 g of hydrated sodium carbonate (washing soda)
crystals, Na2CO3.xH2O, was dissolved in deionised water and the
solution made up to 250 cm3 in a volumetric flask. The molarity of
this solution was found by titrating 25.0 cm3 portions of this
solution against a 0.10 M solution of hydrochloric acid.
• The mean titration figure was found to be 20.0 cm3.
• The equation for the titration reaction is Na2CO3 + 2HCl → 2NaCl +
H2O + CO2
• (e) Calculate the concentration of the sodium carbonate in the
washing soda solution in moles per litre?
• (f) Calculate the value of x, the degree of hydration, of the crystals.
Find concentration of titrated sodium carbonate
solution in moles per litre( molarity)
The balanced equation for the reaction is:
Na2CO3 + 2HCl →2NaCl + H2O + CO2
• V1 X M1 = V2 x M2
n1
n2
Solution 1 – HCl
Solution 2 – Na2CO3
V1 = 20cm3
V2 = 25cm3
M1 = 0.1
M2 = ?
n1 = 2
n2 =
1
V1 X M1 = V2 x M2
n1
n2
(26.05)X (0.11) = (25) x (M2)
2
1
(20)X (0.1)X (1) = M2
(2) x (25)
0.04 = M2
Answer e) The concentration of the Na2CO3 solution was 0.04 M
(moles per litre)
Find how many moles of sodium carbonate were
in the volumetric flask ( 250cm3 of the solution)
• The concentration of the sodium carbonate solution
is 0.04 moles per litre
• 0.04 /4 = 0.01moles
There were 0.01 moles of Na2CO3 in the
volumetric flask SO
There were 0.01 moles of Na2CO3 weighed
out
Find the mass of sodium carbonate in the weighed
out sample
• There were 0.01 moles in weighed out amount
• 0.01 moles x RMM = mass
(0.01)(106) = 1.06g
There were 1.06g of sodium carbonate in weighed out sample
Find the mass of water in the hydrated sodium
carbonate weighed out
There were 1.06 g pure sodium carbonate in the sample weighed
out.
Mass of hydrated sodium carbonate = mass of sodium carbonate + mass of water
2.51g = 1.06g
+ mass of water
2.51 - 1.06g = mass of water
1.45g = mass of water
1.45 g is the mass of water in the hydrated sodium carbonate
weighed out
Find the value of x in the formula
Na2CO3.xH2O
• We must find the simplest ratio between the amounts
of the two substances
• Moles of sodium carbonate: Moles of water
• 0.01 moles of Na2CO3 : ? Moles of water
Mass of water/ RMM = Moles of water
1.45/ RMM = moles of water
1.45/18 = 0.0806
• 0.01 moles of Na2CO3 : 0.0806 moles of water
• 1 moles of Na2CO3
: 8.06 moles of water
Answer
the value of x in the formula Na2CO3.xH2O is 8.06