1. Titrimetric procedure and Acid and Base Titrations Ch 20

2. Apparatus used in Volumetric analysis
A Graduated cylinder
Roughly measures the volume of a liquid

3. The volumetric flask Is used to hold a definite volume of a liquid.
Apparatus used in Volumetric analysis
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!

4. A pipette Is used to deliver an exact amount of a liquid.
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!

5. The burette Is used to accurately measure the volume of a liquid.
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.

6. Apparatus used in Volumetric analysis
Used for mixing liquids together – designed to prevent the splashing of liquids.

7. 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!

8. 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

9. Finding the unknown concentration:
You use the following equation:
Volume of base reacted
Molarity of acid
Volume of acid reacted
Molarity of base
V1 X M1 = V2 x M2
n1 n2
Number of moles of base indicated in the balanced equation
Number of moles of acid indicated in the balanced equation

10. HCl + NaOH NaCl+ H20 V1 X M1 = V2 x M2 Solution 1 – HCl V1 = 27.5cm3
Question 234(e)
HCl + NaOH NaCl+ H20
V1 X M1 = V2 x M2
n n2
Solution 1 – HCl
V1 = 27.5cm3
M1 = 0.1 M
n1 = 1
Solution 2 - NaOH
V2 = 25cm3
M2 = ?
n2 = 1

11. Question 234(e)(i) (27.5)X (0.1)X (1) = M2 (1) x (25) 0.11 = M2
V1 X M1 = V2 x M2
n1 n2
(27.5)X (0.1) = (25) x (M2)
(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)

12. How many moles in 1 litre of solution? 0.11 x RMM = ? 0.11 x 40 = 4.4g
(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

13. 2HCl + Na2CO3 2NaCl+ H20 +CO2 V1 X M1 = V2 x M2 Solution 1 – HCl
Question 235he)
2HCl + Na2CO3 2NaCl+ H20 +CO2
V1 X M1 = V2 x M2
n n2
Solution 1 – HCl
V1 = 30cm3
M1 = ?
n1 = 2
Solution 2 – Na2CO3
V2 = 25cm3
M2 = 0.06M
n2 = 1

14. How many moles in 1 litre of solution? 0.06 moles x RMM = ?
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

15. Question 235(h)(i) M1 = (25) x (0.06) x (2) (1) x (30) M1 = 0.1
V1 X M1 = V2 x M2
n1 n2
(30)X (M1) = (25) x (0.06)
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)

16. There are 3.65g of HCl in one litre. Answer = 3.65gL-1
(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

17. Vinegar Titrations

19. Check your learning…PEQ

20. Why is the vinegar diluted?
Diluting the vinegar reduces the volume of vinegar and the volume of sodium hydroxide solution needed in the experiment

21. 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)

22. 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.

23. 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.

24. 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.

25. 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.

26. Why are three accurate titrations carried out?
To reduce experimental error by calculating the average value.

27. What colour change happens at the end point?
The indicator changes colour from pink to colourless.

28. Vinegar is a solution of ethanoic acid (acetic acid)
. 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)

29. 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)

30. Name the piece of equipment that should be used to measure the ethanoic acid solution during the titration.
burette (3)

31. 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

32. Name a suitable indicator for this titration
INDICATOR: phenolphthalein /

33. CH3COOH + NaOH CH3COONa + H2O
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)

34. (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
n n2
Solution 1 – NaOH
V1 = 25cm3
M1 = 0.11
n1 = 1
Solution 2 – CH3COOH
V2 = 15.8cm3
M2 = ?
n2 = 1

35. V1 X M1 = V2 x M2
n1 n2
(25)X (0.11) = (15.8) x (M2)
(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)

36. (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 = M )
Answer (a)
The molarity of ethanoic acid in the original vinegar solution was M ( moles per litre)

37. (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
moles x RMM = Mass
(0.8703)(60) =
Answer b
The concentration of ethanoic acid in the original vinegar solution was g per litre .

38. (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 Want mass in 100cm3
/10 =
Answer c)
w/v%

39. CH3COOH + NaOH CH3COONa + H2O
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 cm3. Calculate the concentration of ethanoic acid in the original vinegar in
(a) moles per litre
(b) grams per litre
(c) % (w/v)

40. (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
n n2
Solution 1 – NaOH
V1 = 25cm3
M1 = 0.09
n1 = 1
Solution 2 – CH3COOH
V2 = 26.55cm3
M2 = ?
n2 = 1

41. V1 X M1 = V2 x M2
n1 n2
(25)X (0.09) = (26.55) x (M2)
(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)

42. (a) Find concentration of ethanoic acid in original vinegar solution in moles per litre
The concentration of the made up vinegar solution was
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 = M )
Answer (a)
The molarity of ethanoic acid in the original vinegar solution was M ( moles per litre)

43. (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
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 .

44. (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 Want mass in 100cm3
50.85/10 = 5.085
Answer c)
5.085 w/v%

45. 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).

46. (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
n n2
Solution 1 – NaOH
V1 = 25cm3
M1 = 0.1
n1 = 1
Solution 2 – CH3COOH
V2 = 22.65cm3 (AVERAGE)
M2 = ?
n2 = 1

47. V1 X M1 = V2 x M2
n1 n2
(25)X (0.1) = (22.65) x (M2)
(25)X (0.1)X (1) = M2
(1) x (22.65)
= M2
ANSWER: The concentration of the DILUTED solutionwas
M (moles per litre)

48. (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
x 60 = 6.624
Answer b
The concentration of ethanoic acid in the original vinegar solution was 6.624g per litre .

49. (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

50. (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 Want mass in 100cm3
66.24g/ 10 = 6.624
Answer c)
6.624 w/v%

51. The titration reaction is CH3COOH + NaOH → CH3COONa + H2O
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).

52. Find concentration of ethanoic acid in THE DILUTED vinegar solution in moles per litre
NaOH + CH3COOH CH3COOH + H20
V1 X M1 = V2 x M2
n n2
Solution 1 – NaOH
V1 = 25cm3
M1 = 0.12
n1 = 1
Solution 2 – CH3COOH
V2 = 20.5cm3 (AVERAGE)
M2 = ?
n2 = 1

53. V1 X M1 = V2 x M2
n1 n2
(25)X (0.12) = (22.65) x (M2)
(25)X (0.12)X (1) = M2
(1) x (20.5)
= M2
ANSWER: The concentration of the DILUTED solution was M (moles per litre)

54. (a) Find concentration of ethanoic acid in original vinegar solution in moles per litre
The concentration of the made up vinegar solution was
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
( M X 10 = M )
Answer (a)
The molarity of ethanoic acid in the original vinegar solution was 1.463M ( moles per litre)

55. (c)Express the concentration of ethanoic acid in the original vinegar solution in w/v%
Given Moles PER LITRE Find Grams PER LITRE w/v%
1.463 M X rmm = mass
(1.463 )(60) = 87.78
The concentration of ethanoic acid in the original vinegar solution was g per litre .

56. (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 Want mass in 100cm3
87.78 / 10 = 8.778g
Answer
8.778 w/v%

57. 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

58. Check your learning…PEQ

59. 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.

60. Identify a primary standard reagent which could have been used to standardise the hydrochloric acid solution. (5)
anhydrous sodium carbonate (Na2CO3)

61. Sodium carbonate crystals, 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)

62. (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

63. 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

64. 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)

65. 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

66. (i) Describe the correct procedure for weighing and making up the solution from hydrated sodium carbonate crystals. (12)

67. (ii) Name a suitable indicator for the titration and state the colour change at the end point. (6)

68. (iii) Describe the correct procedure for washing the pipette and using it to measure the sodium carbonate solution. (9)

69. 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)

70. 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 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

71. 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
n n2
Solution 1 – HCl
V1 = 22.8cm3
M1 = 0.11
n1 = 2
Solution 2 – Na2CO3
V2 = 25cm3
M2 = ?
n2 = 1

72. V1 X M1 = V2 x M2
n1 n2
(22.8)X (0.11) = (25) x (M2)
(22.8)X (0.11)X (1) = M2
(2) x (25)
= M2
The concentration of the Na2CO3 solution was M (moles per litre)

73. There were 0.0251 moles of Na2CO3 in the volumetric flask
Find how many moles of sodium carbonate were in the volumetric flask ( 500cm3 of the solution)
The concentration of the Na2CO3 solution was M (moles per litre)
/2 = moles
There were moles of Na2CO3 in the volumetric flask

74. Find the mass of sodium carbonate in the volumetric flask ( 500cm3 of the solution)
There were moles of Na2CO3 in the volumetric flask
moles x RMM = mass
( )(106) =
There were g of sodium carbonate in the volumetric flask

75. 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 = g + mass of water
g = mass of water
0.8894g = mass of water
0.8894g is the mass of water in the hydrated sodium carbonate weighed out

76. Find the percentage of water of crystallisation in the compound
% water of crystallisation = mass of water in compound x 100
Total mass of compound
= g x 100
3.55g
= %
Answer (i)
the percentage of water of crystallisation in the compound is %

77. 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
moles of Na2CO3 : ? Moles of water
moles of Na2CO3 : moles of water
1 Na2CO3 : / moles of water
1 Na2CO3 : moles of water
Mass of water/ RMM = Moles of water
0.8894/ RMM = moles of water
0.8894/18 =
Answer (ii)
the value of x in the formula Na2CO3.xH2O is

78. 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 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

79. 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
n n2
Solution 1 – HCl
V1 = 26.05cm3
M1 = 0.11
n1 = 2
Solution 2 – Na2CO3
V2 = 25cm3
M2 = ?
n2 = 1

80. V1 X M1 = V2 x M2
n1 n2
(26.05)X (0.11) = (25) x (M2)
(22.8)X (0.11)X (1) = M2
(2) x (25)
= M2
Answer d(i)The concentration of the Na2CO3 solution was M (moles per litre)

81. Find the concentration of sodium carbonate in the grams per litre
There were moles in 1 litre ( that was the molarity)
moles X rmm = mass
(0.0573)(106) =
Answer (d) (ii)
The concentration of sodium carbonate is grams per litre

82. There were 3.0369 grams of Na2CO3 in the volumetric flask
Find how many grams of sodium carbonate were in the volumetric flask ( 500cm3 of the solution)
The concentration of the sodium carbonate solution is grams per litre
6.0738/2 = g
There were grams of Na2CO3 in the volumetric flask

83. Find the mass of water in the hydrated sodium carbonate weighed out
There were g pure sodium carbonate in the sample weighed out.
Mass of hydrated sodium carbonate = mass of sodium carbonate + mass of water
8.20g = g + mass of water
g = mass of water
5.1631g = mass of water
5.1631g is the mass of water in the hydrated sodium carbonate weighed out

84. Find the percentage of water of crystallisation in the compound
% water of crystallisation = mass of water in compound x 100
Total mass of compound
= x 100
8.20g
= %
Answer (e)
the percentage of water of crystallisation in the compound is %

85. 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.

86. 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
n n2
Solution 1 – HCl
V1 = 20cm3
M1 = 0.1
n1 = 2
Solution 2 – Na2CO3
V2 = 25cm3
M2 = ?
n2 = 1

87. V1 X M1 = V2 x M2
n1 n2
(26.05)X (0.11) = (25) x (M2)
(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)

88. There were 0.01 moles of Na2CO3 in the volumetric flask SO
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

89. 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

90. 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
g = mass of water
1.45g = mass of water
1.45 g is the mass of water in the hydrated sodium carbonate weighed out

91. 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
0.01 moles of Na2CO3 : moles of water
1 moles of Na2CO3 : 8.06 moles of water
Mass of water/ RMM = Moles of water
1.45/ RMM = moles of water
1.45/18 =
Answer
the value of x in the formula Na2CO3.xH2O is 8.06