Titrimetric procedure and Acid and Base Titrations Ch 20

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

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!

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!

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.

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: 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

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 n1 n2 Solution 1 – HCl V1 = 27.5cm3 M1 = 0.1 M n1 = 1 Solution 2 - NaOH V2 = 25cm3 M2 = ? n2 = 1

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) 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)

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

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 n1 n2 Solution 1 – HCl V1 = 30cm3 M1 = ? n1 = 2 Solution 2 – Na2CO3 V2 = 25cm3 M2 = 0.06M n2 = 1

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

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) 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)

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

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) . 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 /

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)

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

V1 X M1 = V2 x M2 n1 n2 (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)

(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 Want mass in 100cm3 52.218 /10 = 5.2218 Answer c) 5.2218 w/v%

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 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 V1 = 25cm3 M1 = 0.09 n1 = 1 Solution 2 – CH3COOH V2 = 26.55cm3 M2 = ? n2 = 1

V1 X M1 = V2 x M2 n1 n2 (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)

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

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 V1 = 25cm3 M1 = 0.1 n1 = 1 Solution 2 – CH3COOH V2 = 22.65cm3 (AVERAGE) M2 = ? 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 Want mass in 100cm3 66.24g/ 10 = 6.624 Answer c) 6.624 w/v%

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).

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 V1 = 25cm3 M1 = 0.12 n1 = 1 Solution 2 – CH3COOH V2 = 20.5cm3 (AVERAGE) M2 = ? 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 w/v% 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 .

(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%

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 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 V1 = 22.8cm3 M1 = 0.11 n1 = 2 Solution 2 – Na2CO3 V2 = 25cm3 M2 = ? 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)

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 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 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 Mass of water/ RMM = Moles of water 0.8894/ RMM = moles of water 0.8894/18 = 0.0494 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 V1 = 26.05cm3 M1 = 0.11 n1 = 2 Solution 2 – Na2CO3 V2 = 25cm3 M2 = ? 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

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 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 V1 = 20cm3 M1 = 0.1 n1 = 2 Solution 2 – Na2CO3 V2 = 25cm3 M2 = ? 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)

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

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 0.01 moles of Na2CO3 : 0.0806 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 = 0.0806 Answer the value of x in the formula Na2CO3.xH2O is 8.06