8/19/2019 01 VA Det Mr of H2X Ans

1/3

1

Catholic Junior College

H2 Chemistry (9729)

Experiment 1:  Acid-Base Titration

FA 1  is a solution containing 28.00 g dm –3 of a weak dibasic acid, H2X.

FA 3  is 0.100 mol dm –3 sodium hydroxide, NaOH.

You are required to determine the  M r  of the weak dibasic acid, H2X.

(a) Dilution of FA 1

1. Fill a burette with FA 1.

2. Place between 22.00 cm3 and 24.00 cm3 of FA 1 into a 250 cm3 graduated flask and make up tothe mark with deionised water.

3. Shake the flask thoroughly to obtain a homogeneous solution. Label this solution FA 2. 

Record your burette readings in the following table.

Final burette reading / cm3 

Initial burette reading / cm3 

Volume of FA 1 used / cm3  volume of FA 1 = ......23.00 cm3...... [1]

(b) Titration of FA 2 against FA 3

(i)  1. Fill a second burette with FA 3.

2. Using a pipette, transfer 25.0 cm3 of FA 2 into a conical flask and titrate it with FA 3 using

thymolphthalein as indicator.

3. Record your titration results in the table provided. Make certain that your recorded resultsshow the precision of your working.

4. Repeat the titration until consistent results are obtained.

Results:

1 2 3 4

Final burette reading / cm3 

Initial burette reading / cm3 

Volume of FA 3 used / cm3 

[5]

(ii) From your titrations, obtain a suitable volume of FA 3 to be used in your calculations. Show

clearly how you obtained this volume and place a tick () under the readings used.

volume of FA 3 = ………..........… [1]

23.00

0.00 

23.00 

23.20 46.35

0.00 23.20 

23.20  23.15

average titre = ..

  = 23.18 cm3 

 Name: ................................................... (1T ) Date: .............................

15 

23.18 cm3 

8/19/2019 01 VA Det Mr of H2X Ans

2/3

2

(c) (i) Calculate the amount of NaOH in the volume of FA 3 which reacts with 25.0 cm3 of FA 2.

amount of NaOH in …………. cm3 of FA 3 = ……..............…… [1]

(ii)  Calculate the amount of H2X in 25.0 cm3 of FA 2. 

amount of H2X in 25.0 cm3 of FA 2 = …….......………… [1]

(iii) Calculate the amount of H2X in 1.00 dm3 of FA 1.

amount of H2X in 1.00 dm3 of FA 1 = …........……… [2]

(iv)  Determine the  M r  of H2X.

 M r  of H2X = …………....... [1]

amt of NaOH in 23.18 cm3 of FA 3 = 0.100 × .

 

= 2.318 × 10–3 mol

H2X + 2NaOH Na2X + 2H2O

amt of H2X in 25 cm

3

 of FA 2 = 21

× amt of NaOH=

21 × 2.318 × 10–3 mol

= 1.16 × 10–3 mol

amt of H2X in 250 cm3 of FA 2 = amt of H2X in 23.00 cm

3 of FA 1

=

.  × 1.16 × 10

–3

 mol= 0.0116 mol

amt of H2X in 1.00 dm3 of FA 1 =

.  × 0.0116

= 0.504 mol

mass of 0.504 mol of H2X = 28.00 g

mass of 1 mol of H2X = ( 

.

  × 28.00) g

= 55.6 g

 M r of H2X = 55.6

2.32 × 10–3 mol23.18

1.16 × 10–3 mol

0.504 mol

55.6

8/19/2019 01 VA Det Mr of H2X Ans

3/3

3

(d) When glassware is manufactured there will always be a maximum error. This is usually marked on

the glassware.

  A 250 cm3

 graduated flask has a maximum error of 0.2 cm3

.  A 25 cm3 pipette has a maximum error of 0.06 cm3.

  A 50 cm3 burette has a maximum error of 0.05 cm3 in each measurement.

(i)  A student carried out this experiment and obtained a mean titre value of 24.20 cm3.

Calculate the total percentage error from the apparatus in this experiment. Assume that volume

of FA 1 used is 24.00 cm3.

total % error from apparatus = …………....... [1]

(ii)  From theoretical calculation, the volume of FA 3 required in this student's experiment should

have been 24.50 cm3. Calculate the percentage error in the student's result.

% error in student's result = …………....... [1]

(iii) Based on your answers to (d)(i) and (d)(ii), is the student's result accurate? Explain your answer.

..………………………..................................................................................................................

…………………………………………………………………………………..………………..

..……....................................................................................................................................… [1]

[Total: 15]

  No, student's result is not accurate 

  because experimental error (1.22 %) is greater than apparatus error (1.15 %)

total % error = ( (.).

  + .

  + ..

  + (.).

  ) × 100

= 0.417 + 0.0800 + 0.240 + 0.413

= 1.15 % 

% error in student's result = ...

  × 100

= 1.22 % 

1.15 %

1.22 %