Lab Report.docx

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ABSTRACT :-

This experiment was carried out first to prepare the sodium hydroxide solution , second to standardise the base against potassium hydrogen phthalate and lastly to analyse the unknown vinegar sample. The sodium hydroxide solution was very important as in this experiment, we prepared 900 L of approximately 0.25 M NaOH by diluting a stock solution of approximately 50 % ( w/w ) NaOH by mass. Besides , in this experiment also need us to standardise the NaOH as a base against potassium hydrogen phthalate , KHP. For this part , it show us about weighing technique called weighing by difference as we can get accurate weight for the KHP. Next was to analyse the unknown vinegar sample which this was the main part of this experiment. For this part , it tech us how to find the density of vinegar sample first as phenolphthalein indicator was used for titration with standardized NaOH solution.

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OBJECTIVE : -

i)

To prepare the sodium hydroxide solution , NaOH.

ii)

To standardise the base against potassium hydrogen phthalate , KHP.

iii)

To analyse the unknown vinegar sample.

INTRODUCTION : -

Diagram 1 As shown in Diagram 1, it is molecular structure of a acetic acid. Vinegar which consist of acetic acid has a molecular formula of CH3COOH.

Vinegar is a liquid that is produced from the fermentation of ethanol into acetic acid as this fermentation process is carried out by bacteria. Vinegar consist of acetic acid and water. In this experiment, the titration technique was been done as to determine the concentration of acetic acid in vinegar by adding base , NaOH to the solution as this is an acid/base reaction. Besides , titration technique is a common laboratory method of quantitative chemical analysis that used to determine the unknown concentration of an identified analyte.

Next, a reagent or also known as titrant is prepared as a standard solution. A known concentration and volume titrant will react with the solution of an analyte to determine the concentration. The accurate concentration can be obtain by standardized it with potassium hydrogen phthalate, KHP. The KHP is used to determine acetic acid contain in an unknown vinegar sample by titration technique as the colour changes occur.

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PROCEDURE :-

A. Preparation of the Sodium Hydroxide Solution

i)

The 50 mL of an clean and empty beaker being weight.

ii)

12 mL of NaOH solution was poured into the 50 mL beaker and being weight again.

iii)

Then, poured 890 mL of distilled water into a clean plastic bottle using a pipette.

iv)

After that , poured the 12 mL of NaOH into the plastic bottle containing 890 mL of distilled water.

v)

The total volume of the solution containing NaOH and distilled water was 900 mL.

vi)

The bottle was closed with the lid and carefully mix vigorously the solution.

B. Standardisation of the base against Potassium hydrogen phthalate , KHP.

i)

For this step, first the potassium hydrogen phthalate, KHP was weight about 1 g of sample. Prepared four times.

ii)

Then, poured the KHP into each of the 250 mL of conical flask.

iii)

Add 35 mL of distilled water into each of the 250 mL conical flask containing the KHP.

iv)

Each of the 250 mL conical flask containing distilled water and KHP were added with 3 drop phenolphthalein indicator.

v)

Next, poured diluted NaOH that been prepared earlier into the 50 mL burette and make sure there was no bubbles at the tip of the burette.

vi)

Begin the titration for each of the conical flask until it changes the colour to pink. But for the first conical flask , do a rough titration.

vii)

Repeat the process for the other sample.

C. Analysis of the unknown vinegar sample

i)

For this part, first 50 mL of clean and empty beaker was being weight.

ii)

Then , 10 mL of unknown vinegar was being pipette into the 50 mL beaker and being weight again.

iii)

The difference of the weight was a weight of the 10 mL unknown vinegar as it can be used to calculate the density of vinegar sample.

iv)

Prepared for four time and poured the 10 mL of unknown vinegar into each of 250 mL conical flask and then added 3 drops of phenolphthalein.

v)

Then added NaOH that being prepared earlier into the 50 mL burette.

vi)

Next, begin the titration as the first one begin with a rough titration.

vii)

Repeat the process for the other sample until there was changes in colour.

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RESULT and CALCULATION :-

A. Preparation of the Sodium Hydroxide Solution Table 1 : Volume NaOH from 50% ( w/w ) , 0.25 M 12 mL

Volume of NaOH taken from the 50% stock solution

Calculations :Preparation of NaOH from 50% ( w/w ) stock solution , 0.25 M , density = 1.525 g/mL 50 𝑔 𝑁𝑎𝑂𝐻

i)

50% ( w/w ) -

ii)

Number of mol NaOH -

iii)

Density =

iv)

65.574 mL

v)

Molarity of NaOH =

vi)

M₁V₁ = M₂V₂

100 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

𝑚𝑎𝑠𝑠 𝑣𝑜𝑙𝑢𝑚𝑒

𝑚𝑎𝑠𝑠 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

V₂ ~ 12 mL

50 𝑔 40 𝑔/𝑚𝑜𝑙

𝑚𝑎𝑠𝑠

Volume = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =

0.0656 L

1.25 𝑚𝑜𝑙 0.0656 𝐿

= 10.06 𝑀

( 0.25 M ) ( 900 mL ) = ( 19.06 M ) ( V₂ ) V₂ = 11.8 mL

=

= 1.25 mol NaOH

100 𝑔 1.525 𝑔𝑚𝐿

= 65.574 𝑚𝐿

Table 2 : Result for weight of 12 mL NaOH Weight of empty 50 mL beaker

35.3545 g

Weight of 50 mL beaker + 12 mL NaOH

51.3804 g

Weight of 12 mL NaOH

16.0259 g

Density of 12 mL NaOH :Density =

𝑚𝑎𝑠𝑠 𝑣𝑜𝑙𝑢𝑚𝑒

=

16.0259 𝑔 12 𝑚𝐿

= 1.3355 g/ mL

B. Standardisation of the base against Potassium hydrogen phthalate , KHP. Table 3 : Result of standardisation of the base against KHP Rough

1

2

3

Weight of KHP

1.001 g

1.005 g

1.005 g

1.009 g

Final reading of NaOH

33.1 mL

33.0 mL

33.0 mL

33.2 mL

Initial reading of NaOH

0

0

0

0

33.10 mL

33.00 mL

33.00 mL

33.20 mL

33.06

32.84

32.84

32.90

Volume of NaOH ( mL ) used Ratio volume of NaOH / weight of KHP

Calculations :Ratio volume NaOH / weight of KHP –

Rough

=

1st

=

𝟑𝟑.𝟏𝟎 𝒎𝑳 𝟏.𝟎𝟎𝟏 𝒈 𝟑𝟑.𝟎𝟎 𝒎𝑳 𝟏.𝟎𝟎𝟓 𝒈

= 𝟑𝟑. 𝟎𝟔

2nd

=

= 𝟑𝟐. 𝟖𝟒

3rd

=

𝟑𝟑.𝟎𝟎𝒎𝑳 𝟏.𝟎𝟎𝟓 𝒈 𝟑𝟐.𝟐𝟎 𝒎𝑳 𝟏.𝟎𝟎𝟗 𝒈

= 𝟑𝟐. 𝟖𝟒 = 𝟑𝟐. 𝟗𝟎

C. Analysis of the unknown vinegar sample Table 4 : Result for the Unknown vinegar sample Rough

1

2

3

10

10

10

10

Final reading of standard NaOH

52.0

53.0

51.0

51.0

Initial reading of standard NaOH

0

0

0

0

52.0

53.0

51.0

51.0

Volume of unknown vinegar ( mL )

Volume of standard NaOH ( mL ) used

Table 5 : Calculation density of unknown vinegar Weight of empty 50 mL beaker

29.843 g

Weight of 50 mL beaker + 10 mL unknown vinegar

39.827 g

Weight of 10 mL unknown vinegar

9.984 g

Calculations :Density of unknown vinegar =

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𝑚𝑎𝑠𝑠 𝑣𝑜𝑙𝑢𝑚𝑒

=

9.984 𝑔 10 𝑚𝐿

= 0.9984 𝑔/𝑚𝐿

DISCUSSION :In this analysis of an unknown vinegar sample experiment , the purpose were to prepare

the sodium hydroxide solution , to standardise the base against potassium hydrogen phthalate , KHP and to analyse the unknown vinegar sample. While doing this experiment , we have followed all the step based on the procedure given. In this experiment in order to analysis the unknown vinegar sample , we divided into three part which were for part A was to prepare the sodium hydroxide solution , as for part B was to standardise the base against potassium hydrogen phthalate and for part C was where we analyse the unknown vinegar. For part A which was to prepared the sodium hydroxide solution. We need to prepare 900 mL of NaOH from a stock solution of 50% ( w/w ) as its molarity was 0.25 M and its density

given was 1.525 g/mL . So, in order to determine volume that we should took from the stock solution in order to prepare 900 mL , we can determine through the calculation. First was by calculate number of mol of NaOH which the number of mol for NaOH was 1.25 mol. Next, as the density of the 50 % ( w/w ) was 1.525 g/mL, we can determine the volume as the formula for density was mass over volume. So, from the calculation the volume we got was 0.0656 L. After that , we used molarity formula to find the concentration of the solution which the concentration we got was 19.06 M. So that , to determine the volume of NaOH in order to prepare 900 mL NaOH from the 50% ( w/w ) stock solution, we used dilution formula which was M₁V₁ = M₂V₂ , and the volume we got to prepare the 900 mL solution of NaOH was 12 mL. From here , we calculate the density of the 12 mL of NaOH as the weight of the 12 mL NaOH we can get by the difference of the 50 mL empty beaker and 50 mL beaker with 12 mL of NaOH. The weight of the 12 mL of NaOH was 16.0259 g. So then, we calculate the density of 12 mL of NaOH solution and the density was 1.3355 g/mL. The 12 mL of NaOH solution was diluted with the distilled water in the clean plastic bottle. For the next part which was for part B was to standardise the base against potassium hydrogen phthalate , KHP. Standardization was a process to determine the exact concentration ( molarity ) of a solution. As in this experiment, titration technique was used as it was one type of analytical procedure often used in standardization. In a titration , an exact volume of one substances was reacted with a known amount of another substances. During the titration, the point at which the reaction was complete in a titration was referred as endpoint. A chemical substance known as indicator was used to indicate the endpoint as there will be the change in colour. The type of indicator that was used in this experiment was phenolphthalein as it was organic compound and colourless in acidic solution but pink colour in base solution. The endpoint can be detect by the changes in the colour due to the indicator , as we can see from the figure below :-

Figure 2 : Part B

From the figure above , shown the appearance of pink colour shown the endpoint of the solution. So , we can determine the volume of the NaOH used to reach the endpoint. The volume of the NaOH used as for rough was 33.1 mL , for 1st reading was 33.0 mL , for the 2nd reading was 33.0 mL and for the 3rd reading was 33.2 mL. Also , the weight of the KHP used for this part also was noted as the weight for rought was 1.001 g , as for 1st reading was 1.005 g , as for 2nd reading was 1.005 g and also for the 3rd reading was 1.009 g. Next , for part C was to analyse the unknown vinegar. For this part , the density of the unknown vinegar was determine by using the formula density which was mass over volume. The volume of unknown vinegar used in this experiment was 10 mL. As for the mass of the unknown vinegar being determine by the difference weight of 50 mL empty beaker with the weight of the 50 mL beaker contain 10 mL of unknown vinegar. From there , the weight of the 10 mL of the unknown vinegar was 9.984 g. So that, we calculate the density of the unknown vinegar and we got the density of the unknown vinegar was 0.9984 g/mL. The result for the titration process of the unknown vinegar can be seen below :-

Figure 3 : Part C

Lastly, every experiment there might be an error occur for an example observational error which due to the human on how to read the reading as the eye must be perpendicular to the instrument. Next might be source error which due to the person might record a wrong value , misread scale , forget a digit when reading a scale or recording a measurement. This type of error was an outright mistake.

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CONCLUSION :As a conclusion , 12 mL of sodium hydroxide solution was prepared and the density

of the unknown vinegar was 0.9984 g/mL.

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REFERENCES :-

i)

Abdullah Munir , Hamzah & Md Yunus ( 2018 ). Analytical Chemistry – laboratory manual.

ii)

Eddy, D. (n.d.). Retrieved March 18, 2019, from http://www.chem.latech.edu/~deddy/chem104/104Standard.htm

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QUESTION and ANSWER :-

i)

Explain how weighing by difference eliminates systematic errors.

Because the mass is determined by the difference between two readings, a systematic error in the absolute mass on the balance will be removed by subtracting the final weight from the initial weight

ii)

Why does it not matter how much water you add when dissolving the acid ( KHP ) or when carrying out the titration.

Because the calculations are based on the volume of acid you put in the flask before you dilute it, not after. You have the same moles of acid, regardless of how much water is also present.