A. TITLE : ARGENTOMETRIC TITRATION AND ITS APPLICATIONB. PURPOSE : 1. Determining the concentration of AgNO3

2. Determining the percentage of NaCl in table salt C. SUPPORTING THEORY : D. EQUIPMENTS AND MATERIALS

a. Equipments

b. Materials

Name Quantity

NaCl p.a powder 0,0509 g

AgNO3 As needed

Table salt 0,0509 g

Aquadest As needed

K2CrO4 As needed

Name Quantity

Retort stand 2

Holder 2

Erlenmeyer 3

Funnel 1

Beaker glass 1

Volumetric flask 1

Volumetric pipette 1

0,0509 g of NaCl

Pouring in volumetric flask 100 mLDissolving using aquadest Diluting until limit markShaking well

NaCl Solution

Taking 10 mLPouring in Erlenmeyer 250 mLAdding with 25 mL aquadestAdding 3 drops of K2CrO4 indicator

Result

Titrating using AgNO3Stopping until formed red briquette precipitateWriting volume AgNO3 neededRepeating three times

Volume of AgNO3

E. PROCEDURES

a. Determining the concentration of AgNO3

0, 0509 g of table salt

Pouring in volumetric flask 100 mLDissolving using aquadest Diluting until limit markShaking well

Table salt Solution

Taking 10 mL using volumetric pipette Pouring in Erlenmeyer 250 mLAdding with 25 mL aquadestAdding 5 drops of K2CrO4 indicator

Result

Titrating using AgNO3Stopping until formed red briquette precipitateWriting down volume AgNO3 that needed Repeating three times

Volume of AgNO3

b. Determining the percentage of NaCl in table salt

F. EXPERIMENT RESULT

0, 0509 g of NaCl p.a

Put into volumetric flask 100mLDiluted until limit sign and mix well

take 10mL of NaCl solutionPut into ErlenmeyerAdd with 25mL aquadest and 3 drops K2CrO4 as indicator

Analyte

AgNO3 solution

Enter into burette until 0 scale

Titran

doing titrationstop when there is briquette red precipitaterepeat until 3 timescalculated the concentration of AgNO3

Result

NaCl as standard solution

No Procedures Experiment Result

Hypothesis/Reaction

Conclusion

1.

2.

Using burette with accuracy 0,1 mL

NaCl Solution = colourless

K2CrO4= yellow

NaCl solution + K2CrO4 indicator= yellow

Formed briquette precipitate

Volume AgNO3=

V1 = 8,2 mL

V2 = 8,5 mL

V3 = 7,7 mL

Salt = white NaCl

solution = colourless

K2CrO4 = yellow

NaCl solution + K2CrO4 = yellow

Formed

Initial titration:

AgNO3(aq) + NaCl(aq) AgCl(s)

+NaNO3(aq)

In end point with K2CrO4 indicator:2Ag+ + CrO4

2- Ag2CrO4(s)

NaCl normality: 0,0087

Normality of AgNO3: N1 = 0,0106 N2 = 0,0102 N3 = 0,0113 Naverage =

0,0107

Initial titration:

AgNO3(aq) + NaCl(aq) AgCl(s) +NaNO3(aq)

In end point with K2CrO4 indicator:2Ag+ + CrO4

2- Ag2CrO4(s)

Normality of NaCl: N1 =

0,00856

Normality of AgNO3 is 0,0107 N

Percentage NaCl in salt is 99,61%

NaCl solution + Indicator

AgNO3 solution

Enter into burette until 0 scales

doing titrationstop when there is briquette red precipitateMatch with the percentage of the salt on the labelCalculated the error

Result

G. ANALYSIS AND DISCUSSION a. Analysis

1) In first experiment that is to determine the concentration of AgNO3, NaCl pro- analyze powder ,which color is white , is dissolved in aquades then the color of solution is colorless. After adding K2CrO4,which color is yellow, the solution’s color change to yellow. In this case K2CrO4 is as indicator. and this indicator have to poured when we want to do titration, because this indicator can react with our solution and can give different result if we pour it long time before we do this titration. Next, after titrating with AgNO3 the color of solution change to turbid then briquette red precipitation is formed. It is caused because solubility of AgCrO4 (8, 5 x 10-5 mol/L) is greater than solubility of AgCl (10-5 mol/L), so AgCl precipitate first. It means that silver chromate will not form until the silver ion concentration increases to a large enough value to exceed the K sp of silver chromate. We stop the titration when precipitation is formed( Ag2CrO4). The first permanent appearance of the reddish silver chromate precipitate is taken as the end point of the titration. The reaction in this titration can be writen as follows

In the first reaction: AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

In the end point of titration: 2AgNO3(aq)+K2CrO4(aq) Ag2CrO4(s) + 2KNO3(aq)

The volume of AgNO3 needed in this titration are 8,2 ; 8,5 ; 7,7 mL. And it can be known normality of AgNO3 by using formula below:

V1 x N1 = V2 x N2

And we get N AgNO3 is 0,0106 ; 0,0102 ; 0,0113. The result of third titration is have different scale with first and second titration, it makes the value of N3 has different value too with N1 and N2. It is caused by our mistakes. We don’t have good knowledge about the color briquette red precipitation, so we doubt when we have to stop the titration. In first and second titration we do the titration until in 8,2 mL and 8,5 mL and it is showed really briquette red precipitation, but in the last titration we stop the titration in 7,7 mL with unclear briquette red precipitation, based on co- assistant of laboratory instruction. And then we calculate the average of all nor mality of AgNO3 is 0,0107.

2) In second experiment, we determine the percentage of NaCl in table salt, table salt powder which color is white , is dissolved in aquades then the color of solution is colorless. After adding K2CrO4,which color is yellow, the solution’s color change to yellow. In this case K2CrO4 is as indicator, and this indicator have to poured when we want to do titration, because this indicator can react with our solution and can give different result if we pour it long time before we do this titration. Next, after titrating with AgNO3 the color of solution change to turbid then briquette red precipitation is formed. In that condition we have to stop the titration. The reaction in this titration can be writen as follows:

In the first reaction: AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

In the end point of titration: 2AgNO3(aq)+K2CrO4(aq) Ag2CrO4(s) + 2KNO3(aq)

The volume of AgNO3 needed in this titration are 8 ; 8,3 ; 8 mL. and it can be known normality of salt by using formula below:

V1 x N1 = V2 x N2

In this time,to calculate Normality of NaCl we use normality of AgNO3 that have known before in 1st experiment. It is 0,0107 N. So we get, N1 NaCl is 0,00856; N2

NaCl is 0,01411 and N3 NaCl is 0,00856. And the average of it is 0,01041 N. And then we determine the percentage of NaCl of salt. And we get 98, 38 % ; 102, 07 % ; and also 98, 38 %. In the second titration we get 102, 07 %, it is because of our mistakes. We are not pay good attention when we observe the scale in burette. So, we get that result. Then, we calculate the average of all the NaCl percentage, and we get 99, 61 %. If we compare this result with NaCl percentage that is shown in the wrap of table salt. There is no big differences, in the label of table salt states if NaCl percentage is 99, 54% but from our experiment we get 99, 61 %. The differences is 0, 07 %. This differences is caused by the human resources and the high technology that we used maybe different.

b. Discussion In the first and second experiment, the volume of AgNO3 is not same for three times titration, it happens because of the accurateness of observer eyes to see the color change. So, in this titration experiment, the accurateness of observer is very important. While in the second experiment the percentage gotten is more than 100%. It was caused because in the second titration, we didn’t stop titration. So, volume of AgNO3 was excess.

H. CONCLUSION From the first experiment we can conclude if to know the concentration

of AgNO3, we can standardize it using NaCl p. a as standard solution. The normality of NaCl is 0,0087 and the normality of AgNO3 is 0,0107.

From the second experiment we can determine the percentage of NaCl in table salt using titration between table salt solution with AgNO3 . And the result is 99, 61 %. It has little differences with label of NaCl percentage in our table salt wrap.

I. ANSWER OF QUESTION

J. REFERENCES

ATTACHMENT

In the first experiment:

Normality of NaCl ¿g

BMxV (L )= 0,0509

58,5x 0,1=0,0087

For V1 = 8,2 mL

NAgNO3xVAgNO3 = NNaClxVNaCl

NAgNO3x8,2 = 0,0087x10

NAgNO3 = 0,0106

For V2 = 8,5 mLNAgNO3xVAgNO3 = NNaClxVNaCl

NAgNO3x8,5 = 0,0087x10 NAgNO3 = 0,0102

For V3 = 7,7 mLNAgNO3xVAgNO3 = NNaClxVNaCl

NAgNO3x7,7 = 0,0087x10 NAgNO3 = 0,0113

Naverage of AgNO3 =0,0106+0,0102+0,0113

3=0,0107

In the second experiment:Normality of AgNO3 is 0,0107

For V1 = 8 mLNNaClxVNaCl = NAgNO3xVAgNO3

10xNNaCl = 0,0107x8 NNaCl = 0,00856

For V2 = 8,3 mLNNaClxVNaCl = NAgNO3xVAgNO3

10xNNaCl = 0,0107x8,3 NNaCl = 0,01411

For V3 = 8 mLNNaClxVNaCl = NAgNO3xVAgNO3

10xNNaCl = 0,0107x8 NNaCl = 0,00856

Naverage of NaCl ¿0,00856+0,01411+0,00856

3=0,01041

Calculation Percentage of NaCl in salt

For V1

Percentage of NaCl(%)

¿V AgNO3 (mL) x N AgNO3 xBENaClx

10010

x100 %

massof sample (mg)

¿8 x0,0107 x58,5 x 1000 %

50,9=98,38

For V2

Percentage of NaCl(%)

¿V AgNO3 (mL) x N AgNO3 xBENaClx

10010

x100 %

massof sample (mg)

¿8,3 x0,0107 x58,5 x 1000 %

50,9=102,07

For V3

Percentage of NaCl(%)

¿V AgNO3 (mL) x N AgNO3 xBENaClx

10010

x100 %

massof sample (mg)

¿8 x0,0107 x58,5 x 1000 %

50,9=98,38

Percentage average of NaCl(%) is

¿ 98,38+102,07+98,383

=99,61

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