Volumetric Analysis: Determination of Mineral Acid Concentration and Element Identity through Acid-Base Titration

The objective is to determine the precise concentration of the mineral acid, HX, and calculate the relative atomic mass of the associated element. Acid-base titration is employed as a method for determining solution molarity, standardizing solutions, and identifying the molecular mass of unknown substances. The experiment focuses on achieving equivalence points where the amounts of acid and base neutralize each other, leading to the formation of a solution consisting solely of salt and water.

Theory: Acid-base titration, a form of neutralization reaction, is pivotal for quantifying unknown concentrations of acids or bases.

The equivalence point is reached when the moles of acid equal the moles of base, resulting in no excess of either. The selection of an appropriate indicator is crucial as it signals the endpoint, aligning with the equivalence point. For instance, titrating hydrochloric acid with sodium hydroxide results in the formation of NaCl and H2O.

Apparatus:

1. 25 cm3 pipette
2. Pipette filler
3. Three titration flasks
4. 50 cm3 burette
5. White tile
6. Wash bottle filled with distilled water

Chemicals:

1. KA1 - Mineral acid, HX
2. KA2 - Solution containing 1.
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   70g of OH- ions per dm3

3. Phenolphthalein as an indicator

Procedure:

1. Pipette 25.0 cm3 of KA2 into a titration flask, add a few drops of phenolphthalein, and titrate with KA1.
2. Record the readings.
3. Repeat the titration until accurate results are obtained.

Formulae for Calculations: M1​V1​=M2​V2​

This formula facilitates the calculation of molarity, where M1 and V1 are the molarity and volume of the titrant (KA1), and M2 and V2 are the molarity and volume of the analyte (KA2).

This experiment enables precise determination of the concentration of the mineral acid and provides insights into the relative atomic mass of the associated element.

Results :

Questions:

1. Calculate the average titration value. Average Titre=2.5+2.6+2.43=7.5 cm3Average Titre=32.5+2.6+2.4​=7.5cm3
2. Calculate the concentration of solution KA2 in mol dm^-3. No. of moles=MassMolar Mass=1.7 g17=0.1 molNo. of moles=Molar MassMass​=171.7g​=0.1mol Molarity of KA2=MolesVolume=0.1 mol25 cm31000=4.0 mol dm−3Molarity of KA2=VolumeMoles​=100025cm3​0.1mol​=4.0mol dm−3
3. Write a balanced ionic equation for the reaction between solution KA1 and solution KA2. H++OH−→H2OH++OH−→H2​O

1. Calculate the concentration (mol dm^-3) of mineral acid HX in solution KA1. 2M1​V1​=M2​V2​ M1​(0.025dm−3)=(4.0mol dm−3)(0.025dm−3) 1=(4.0 mol dm−3)(0.025 dm−3)0.025 dm−3=4.0 mol dm−3M1​=0.025dm−3(4.0mol dm−3)(0.025dm−3)​=4.0mol dm−3

1. Molarity, M=RMMConcentration​ 4.0=20.1RMM4.0=RMM20.1​ RMM of HX=5.025RMM of HX=5.025
2. Using the answer to (5), determine the relative atomic mass of element X. RAM of (H+X)=5.025RAM of (H+X)=5.025

1. 1+X=5.025 X=5.025−1=4.025
2. Suggest an identity for element X. Element X: FluorineElement X: Fluorine

Discussion:

Titration is a valuable laboratory technique involving the precise analysis of one solution by another. A standard solution, delivered from a burette, reacts with a known volume of another solution in a conical flask with a known stoichiometry. The completion of the reaction is determined by the addition of the standard solution, and the endpoint is detected using an indicator like phenolphthalein. The average titration value is calculated using recorded values.

Conclusion: Volumetric analysis of acid and base through titration allows for the determination of the endpoint, marked by the indicator's color change.