Chemistry Lab Report on standardization of acid and bases Essay

Purpose: To fix standardize solution of Na hydrated oxide and to find the concentration of unknown sulphuric acid solution. Data and Calculations: This experiment is divided into two parts ( Separate A and Part B ) . In the first portion of experiment. the standardize solution of Na hydrated oxide is prepared by titrating it with base Potassium H phthalate ( KHP ) . The index Phenolphthalein is used to find that whether titration is complete or non.

Part A: Standardization of a Sodium Hydroxide solution NaOH Sample Code = O Trial 1 Mass of KHP transferred = 0. 42 g Volume of Distilled H2O = 25 mL Volume of NaOH used = 22. 50 mL Molar mass of KHP = 204. 22 g/mol No. of moles of KHP = Mass of KHP used / Molar mass = 0. 42 g / 204. 22 g/mol = 0. 0021 moles Concentration of NaOH = No. of moles / Volume = [ 0. 0021 mol / { ( 22. 50 + 25 ) / 1000 } L ] * 100 = 4. 4 M Trial 2 Mass of KHP transferred = 0. 4139 g Volume of Distilled H2O = 25 mL Volume of NaOH used = 22. 80 mL Molar mass of KHP = 204. 22 g/mol No. of moles of KHP = Mass of KHP used / Molar mass = 0. 4139 g / 204. 22 g/mol = 0. 0020267 moles Concentration of NaOH = No. of moles / Volume = [ 0. 0020267 mol / { ( 22. 80 + 25 ) / 1000 } L ] * 100 = 4. 24 M Trial 3 Mass of KHP transferred = 0. 4239 g Volume of Distilled H2O = 25 mL Volume of NaOH used = 23. 10 mL Molar mass of KHP = 204. 22 g/mol No. of moles of KHP = Mass of KHP used / Molar mass = 0. 4239 g / 204. 22 g/mol = 0. 0020757 moles Concentration of NaOH = No. of moles / Volume = [ 0. 0020757 mol / { ( 23. 10 + 25 ) / 1000 } L ] * 100 = 4. 32 M Trial 4 Mass of KHP transferred = 0. 4311 g Volume of Distilled H2O = 25 mL Volume of NaOH used = 22. 60 mL Molar mass of KHP = 204. 22 g/mol No. of moles of KHP = Mass of KHP used / Molar mass = 0. 4311 g / 204. 22 g/mol = 0. 0021109 moles Concentration of NaOH = No. of moles / Volume = [ 0. 0021109 mol / { ( 22. 60 + 25 ) / 1000 } L ] * 100 = 4. 43 M Table: Trail 1 Mass weighing bottle + KHP ( g ) Mass empty weighing bottle ( g ) Mass of KHP transferred ( g ) Initial volume of burette. Vi ( milliliter ) Final Volume of burette. Vf ( milliliter ) Volume of NaOH used ( milliliter ) Trial 2 Trial 3 Trial 4 11. 561 11. 6217 11. 6113 11. 6329 11. 1461 11. 2078 11. 1874 11. 2018 0. 4200 0. 4139 0. 4239 0. 4311 4. 30 6. 30 10. 1 33. 20 26. 80 29. 10 33. 20 55. 80 22. 50 22. 80 23. 10 22. 60 Concentration of NaOH ( moles/L ) 4. 4 4. 24 4. 32 Average concentration of NaOH = [ 4. 4 M + 4. 24 M + 4. 32 M + 4. 43 M ] / 4 = 4. 35 M 1. % Difference between

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Trial 1 and Trail 2 = [ 4. 24 M / 4. 4 M ] * 100 % = 96. 3 % Difference = ( 100 – 96. 3 ) % = 3. 7 % 2. % Difference between Trial 2 and Trail 3 = [ 4. 24 M / 4. 32 M ] * 100 % = 98. 1 % Difference = ( 100 – 98. 1 ) % = 1. 9 % 3. % Difference between Trial 3 and Trail 4 = [ 4. 32 M / 4. 43 M ] * 100 % = 97. 5 % Difference = ( 100 – 97. 5 ) % = 2. 5 % 4. 43 Observations: KHP is white colour crystals and has definite form. NaOH is clear and crystalline solution with no colour. In the first test. after adding 90 beads of NaOH solution at that place was repeatedly appearance and disappearing of light pink colour. When the whole solution of KHP and H2O get titrated so. the colour of solution becomes light pink and it stays for good. The same colour alterations happen with the following three tests. Concentration of NaOH was about similar for every tests. Part B: Concentration of Sulfuric Acid solution H2SO4 Sample Code = 34

Trial 1: Volume diluted acerb = 25 milliliter Volume of NaOH used = 14. 39 milliliter H2SO4 ( aq ) + 2NaOH ( aq ) 2H2O ( cubic decimeter ) + 2Na2SO4 ( aq ) Average concentration of NaOH = 4. 35 M No. of moles of NaOH = ( Average concentration of NaOH ) * ( Volume of NaOH used ) = 4. 35 M * ( 14. 39 / 1000 ) L = 0. 0626 moles No. of moles of H2SO4 = 0. 0626 mol / 2 = 0. 0313 moles Concentration of H2SO4 = No. of moles / ( volume of diluted acid / 1000 ) = 0. 0313 mol / ( 25 / 1000 ) L = 1. 2 M Trial 2: Volume diluted acerb = 25 milliliter Volume of NaOH used = 13. 51 milliliter H2SO4 ( aq ) + 2NaOH ( aq ) 2H2O ( cubic decimeter ) + 2Na2SO4 ( aq ) Average concentration of NaOH = 4. 35 M No. of moles of NaOH = ( Average concentration of NaOH ) * ( Volume of NaOH used ) = 4. 35 M * ( 13. 51 / 1000 ) L = 0. 0588 moles No. of moles of H2SO4 = 0. 0588 mol / 2 = 0. 0294 moles Concentration of H2SO4 = No. of moles / ( volume of diluted acid / 1000 ) = 0. 0294 mol / ( 25 / 1000 ) L = 1. 2 M Trial 3: Volume diluted acerb = 25 milliliter Volume of NaOH used = 14. 10 milliliter H2SO4 ( aq ) + 2NaOH ( aq ) 2H2O ( cubic decimeter ) + 2Na2SO4 ( aq ) Average concentration of NaOH = 4. 35 M No. of moles of NaOH = ( Average concentration of NaOH ) * ( Volume of NaOH used ) = 4. 35 M * ( 14. 10 / 1000 ) L = 0. 0613 moles No. of moles of H2SO4 = 0. 0613 mol / 2 = 0. 0307 moles Concentration of H2SO4 = No. of moles / ( volume of diluted acid / 1000 ) = 0. 0307 mol / ( 25 / 1000 ) L = 1. 2 M Trial 4: Volume diluted acerb = 25 milliliter Volume of NaOH used = 14. 20 milliliter H2SO4 ( aq ) + 2NaOH ( aq ) 2H2O ( cubic decimeter ) + 2Na2SO4 ( aq ) Average concentration of NaOH = 4. 35 M No. of moles of NaOH = ( Average concentration of NaOH ) * ( Volume of NaOH used ) = 4. 35 M * ( 14. 20 / 1000 ) L = 0. 0618 moles No. of moles of H2SO4 = 0. 0618 mol / 2 = 0. 0309 moles Concentration of H2SO4 = No. of moles / ( volume of diluted acid / 1000 ) = 0. 0309 mol / ( 25 / 1000 ) L = 1. 2 M % Difference between Trail 1 and Trail 2 = [ 1. 2 M / 1. 2 M ] * 100 % = 100 % Difference = ( 100 – 100 ) % =0 % % Difference between Trail 1 and Trail 2 = [ 1. 2 M / 1. 2 M ] * 100 % = 100 % Difference = ( 100 – 100 ) % =0 % % Difference between Trail 1 and Trail 2 = [ 1. 2 M / 1. 2 M ] * 100 % = 100 % Difference = ( 100 – 100 ) % =0 % % Difference between Trail 1 and Trail 2 = [ 1. 2 M / 1. 2 M ] * 100 % = 100 % Difference = ( 100 – 100 ) % =0 %

Table 2: Trail 1 Volume diluted acid titrated ( milliliter ) Initial Volume of burette. Vi ( milliliter ) Final Volume of burette. Vf ( milliliter ) Volume NaOH used ( milliliter ) Concentration Of Sulfuric Acid Trail 2 Trial 3 Trial 4 25 25 25 25 2. 41 17. 20 8. 50 22. 60 16. 94 30. 71 22. 60 36. 80 14. 39 13. 51 14. 10 14. 20 1. 2 M 1. 2 M 1. 2 M 1. 2 Thousand Observations: The H2SO4 is colourless and crystalline liquid. The NaOH solution is colourless. odorless and crystalline liquid. While making the first trail. there were uninterrupted visual aspect and disappearing of light pink colour. After adding 10 milliliter of NaOH solution the pink colour starts looking. At certain volume the light pink colour appeared. bespeaking that titration is done. The indictor phenolphthalein has no colour and there was no specific olfactory property of reagent. Discussion: Average concentration of NaOH solution was 4. 35 M. There are many beginnings of mistake in this experiment as we got some per centum differences in the two different tests.

For the Trial 1 and Trial 2. the per centum difference is 3. 7 % which is important difference to be noted. This per centum difference could happen due to many grounds such as non mensurating the KHP decently as we got 0. 42 g for first test and 0. 4139 g for 2nd test of KHP for executing titration but it is more than required value as per literature value is concerned ( 0. 40 g ) . The about same percentagedifference occurs for following two tests ( 1. 9 % and 2. 5 % ) . The KHP is ever 99. 9 % pure. so the titration should give perfect consequences ( Lab Manual ) . The other possible mistakes was due to the perturbation on the shelf by other pupils where analytical balance is placed in balance room. as it cause variableness in the values in weight of KHP. In Part B of experiment. the mean concentration of sulfuric was found to be 1. 2 M and there was 100 % titration of both reagent ( NaOH and H2SO4 ) .

This 100 % consequences comes due to important figures. if important figures would non be concerned so there would be mistake of 1. 0 % to 2. 0 % in every two tests. There was indistinguishable difference of volume of NaOH used to titrate the acid for each test due to some possible mistakes. The possible mistakes in this Part of experiment were same as for Part A. as the procedure is followed in the same manner. The most important mistake could happen by non agitating the flask decently while adding sodium hydroxide solution and non acknowledging the pink colour on the blink of an eye it appears and adding the NaOH solution smartly into the sulphuric acid. Questions: The 10 milliliter volumetric pipette is rinse 2 or 5 times to do certain there is no bubble indoors because air bubble can do mistake in the measuring of concentration because the existent volume of unknown will be less.

The truth and preciseness for both sets of experiment was about same as there were per centum difference of concentrations lies merely in 2 % to 4 % . The end points of titration for each set of trails in both instances ( Separate A and Part B ) were about same but there is small difference in volume of NaOH used which cause mistakes in truth and preciseness of experiment. Using the analytical balance is truly careful occupation as it is most accurate deliberation machine with truth of 0. 0004 g ( Lab Manual ) and we need to be precised utilizing the balance but some few perturbation can do large mistake such as perturbation other pupils on the shelf it is placed on and non reading the balance decently and taking measurings fastly.

Using Volumetric glasswork is other method to be more accurate in experiment but there are some possible random mistake while reading the values such as non reading the lower semilunar cartilage of liquid cause mistake and non taking the air bubble from the burette and utilizing the beaker in topographic point graduated cylinder in instance of volume as calibrated cylinder is more accurate as compared to beaker ( 0. 02 g ) ( Lab Manual ) . Beginnings of Experimental Error include: The Analytical balance could give incorrect reading because of the perturbation due to other pupils on the shelf it is placed on. Besides. taking the reading quickly and non sing the reading when balance show gets steady. The possible mistake can happen utilizing incorrect glasswork like utilizing beaker alternatively of calibrated cylinder. The mistake could happen while taking reading through graduated cylinder and non sing the lower semilunar cartilage of liquid.

The air bubble in burette can do mistake in the true value of NaOH used. Few beads of liquid remain in burette and volumetric pipette which causes the mistake. Not agitating the flask decently while adding the NaOH solution. Adding the NaOH solution smartly into the flask. Not acknowledging the pink colour immediately as it appears. Adding the more beads of index as needed ( 2 or 3 beads ) . The biggest mistake occur due to leaking of NaOH solution signifier burette. we lost 4 beads during every one test and it cause the important mistake in reading the volume of NaOH used. There is H2O left after rinsing the glass wares which can do the mistake. This lab could be improved by bettering the method of drying the graduated cylinder and beaker before make fulling it with the NaOH solution.

The glasswork could be dried by little sum of propanone. Any propanone could be removed by vaporization. Finally. the experimenter should take the cleaving droplets to the cervix of burette and volumetric pipette by utilizing Kim Wipe. Decision: After careful consideration of all the consequences and all the possible concentration. it is concluded that the mean concentration of NaOH ( sample codification O ) was 4. 35 M and mean concentration of H2SO4 ( sample codification 34 ) was 1. 2 M.

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