# Chemistry Lab Report on standardization of acid and bases

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10 April 2016

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96 writers online Purpose: To prepare standardize solution of sodium hydroxide and to determine the concentration of unknown sulfuric acid solution. Data and Calculations: This experiment is divided into two parts (Part A and Part B). In the first part of experiment, the standardize solution of sodium hydroxide is prepared by titrating it with base Potassium hydrogen phthalate (KHP). The indicator Phenolphthalein is used to determine that whether titration is complete or not.

PART A: Standardization of a Sodium Hydroxide solution NaOH Sample Code = O Trial 1 Mass of KHP transferred = 0.42 g Volume of Distilled water = 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 water = 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 water = 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 water = 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 (mL) Final Volume of burette, Vf(mL) Volume of NaOH used (mL) 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

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 color crystals and has definite shape. NaOH is clear and transparent solution with no color. In the first trial, after adding 90 drops of NaOH solution there was repeatedly appearance and disappearance of light pink color. When the whole solution of KHP and water get titrated then, the color of solution becomes light pink and it stays permanently. The same color changes happen with the next three trials. Concentration of NaOH was almost similar for every trials. PART B: Concentration of Sulfuric Acid solution H2SO4 Sample Code = 34

Trial 1: Volume diluted acid = 25 mL Volume of NaOH used = 14.39 mL H2SO4 (aq) + 2NaOH (aq) 2H2O (l) + 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 acid = 25 mL Volume of NaOH used = 13.51 mL H2SO4 (aq) + 2NaOH (aq) 2H2O (l) + 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 acid = 25 mL Volume of NaOH used = 14.10 mL H2SO4 (aq) + 2NaOH (aq) 2H2O (l) + 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 acid = 25 mL Volume of NaOH used = 14.20 mL H2SO4 (aq) + 2NaOH (aq) 2H2O (l) + 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 (mL) Initial Volume of burette, Vi (mL) Final Volume of burette, Vf (mL) Volume NaOH used (mL) 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 M Observations: The H2SO4 is colorless and transparent liquid. The NaOH solution is colorless, odorless and transparent liquid. While doing the first trail, there were continuous appearance and disappearance of light pink color. After adding 10 mL of NaOH solution the pink color starts appearing. At certain volume the light pink color appeared, indicating that titration is done. The indictor phenolphthalein has no color and there was no specific odor of reagent. Discussion: Average concentration of NaOH solution was 4.35 M. There are many sources of error in this experiment as we got some percentage differences in the two different trials.

For the Trial 1 and Trial 2, the percentage difference is 3.7 % which is significant difference to be noted. This percentage difference could occur due to many reasons such as not measuring the KHP properly as we got 0.42 g for first trial and 0.4139 g for second trial of KHP for performing titration but it is more than required value as per literature value is concerned (0.40 g). The almost same percentagedifference occurs for next two trials (1.9 % and 2.5 %). The KHP is always 99.9 % pure, so the titration should give perfect results (Lab Manual). The other possible errors was due to the disturbance on the shelf by other students where analytical balance is placed in balance room, as it cause variability in the values in weight of KHP. In Part B of experiment, the average concentration of sulfuric was found to be 1.2 M and there was 100 % titration of both reagent (NaOH and H2SO4).

This 100 % results comes due to significant figures, if significant figures would not be concerned then there would be error of 1.0 % to 2.0 % in every two trials. There was identical difference of volume of NaOH used to titrate the acid for each trial due to some possible errors. The possible errors in this Part of experiment were same as for Part A, as the process is followed in the same way. The most significant error could occur by not shaking the flask properly while adding sodium hydroxide solution and not recognizing the pink color on the instant it appears and adding the NaOH solution vigorously into the sulfuric acid. Questions: The 10 mL volumetric pipette is rinse 2 or 5 times to make sure there is no bubble inside because air bubble can cause error in the measurement of concentration because the actual volume of unknown will be less.

The accuracy and precision for both sets of experiment was almost same as there were percentage difference of concentrations lies only in 2 % to 4 %. The endpoints of titration for each set of trails in both cases (Part A and Part B) were almost same but there is little difference in volume of NaOH used which cause errors in accuracy and precision of experiment. Using the analytical balance is really careful job as it is most accurate weighing machine with accuracy of 0.0004 g (Lab Manual) and we need to be précised using the balance but some few disturbance can cause big error such as disturbance other students on the shelf it is placed on and not reading the balance properly and taking measurements fastly.

The air bubble in burette can cause error in the true value of NaOH used. Few drops of liquid remain in burette and volumetric pipette which causes the error. Not shaking the flask properly while adding the NaOH solution. Adding the NaOH solution vigorously into the flask. Not recognizing the pink color instantly as it appears. Adding the more drops of indicator as needed (2 or 3 drops). The biggest error occur due to leaking of NaOH solution form burette, we lost 4 drops during every one trial and it cause the significant error in reading the volume of NaOH used. There is water left after washing the glass wares which can cause the error. This lab could be improved by improving the method of drying the graduated cylinder and beaker before filling it with the NaOH solution.

The glassware could be dried by small amount of acetone. Any acetone could be removed by evaporation. Finally, the experimenter should remove the clinging droplets to the neck of burette and volumetric pipette by using Kim Wipe. Conclusion: After careful consideration of all the results and all the possible concentration, it is concluded that the average concentration of NaOH (sample code O) was 4.35 M and average concentration of H2SO4 (sample code 34) was 1.2 M.

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