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    lll
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    مُساهمة  lll في الخميس أكتوبر 30, 2014 5:54 am


    Experiment -3-
    Using dilution in an acid-base titration and analysis of vinegar

    Aim
    1- To prepare volumetric solutions by dilution of more concentrated solutions.
    2- To determine an unknown concentration of sodium hydroxide solution.
    3- To acquire experience in applying solution stoichiometry rules to an acid – base titration.
    4- To determine the concentration of acetic in commercials vinegar.

    Introduction
    This experiment begins with the dilution of more concentrated solution of 1.0 M HCl
    And unknown concentration of NaOH and then we will use phenolphthalein indicator to know the concentration of base NaOH .
    HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
    CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) +H2O(l)
    The requirements are :
    Pipettes/burette/volumetric flask /conical flask / beaker



    Data and results
    Part i
    dilution
    • 4ml of concentrated HCl add sufficed distilled water to make 100 ml of solution
    • 10ml of concentrated NaOH add sufficed distilled water to make 100 ml of solution


    Part ii
    Titration of HCl and NaOH
    Clean the burette , put diluted NaOH in it to full , measure exactly 10 ml of diluted HCl in 250 ml conical flask ,then add 3 or 4 drops of phenolphthalein indicator to HCL solution , put it under the burette and open it carefully and add NaOH solution slowly while swirl the HCl until the whole color be pink stop the titration . determine the volume of NaOH dropped from the start point to stop point , repeat the titration steps 3 times to find the average .
    Part iii
    Analysis of vinegar
    • 5ml of concentrated vinegar add sufficed distilled water to make 100 ml of solution
    Clean the burette , put diluted NaOH in it to full , measure exactly 10 ml of diluted vinegar in 250 ml conical flask ,then add 3 or 4 drops of phenolphthalein indicator to vinegar solution , put it under the burette and open it carefully and then add NaOH solution slowly while swirl the vinegar until the whole color be pink stop the titration . determine the volume of NaOH dropped from the start point to stop point , repeat the titration steps 3 times to find the average.


    Calculation
    Part a:
    nmole(concentrated HCl)=nmole(diluted HCl)
    M1xV1 = M2xV2
    1*4*10-3 = M2*100*10-3
    M2=4/100 = 0.04M
    Dilution factor = 100/10 = 10
    Part b:
    10.0 mL Volume of titrant HCl Titrant (known concentration)
    Phenolphthalein Indicator NaOH Analyze (unknown concentration)
    Volume used= F-I (mL) Reading of the burette Titrations
    Final (F) mL Initial (I) mL
    4.6-0=4.6 4.6 0 1.
    9.8-4.6=5.2 9.8 4.6 2.
    15-9.8=5.2 15 9.8 3.
    5 ml Average volume

    HCl + NaOH → NaCl + H2O
    n mole(NaOH) = n mole(HCl)
    n mole (NaOH diluted) = 0.04 x 10*10-3 = 4*10-3 mol
    M (NaOH diluted) = 8*10-3
    M(NaOH stock) = M1(NaOH diluted) x dilution factor
    M(NaOH concentrated) = 8*10-3 x 10 = 8*10-2 M
    MM NaOH = 40.00 g\mole
    M(NaOH concentrated in g/L) = M1(NaOH diluted) x dilution factor x MM
    M(NaOH concentrated in g/L) = 8*10-2 x 40 = 3.2 g/L



    Part c:
    10.0 mL Volume of analyte NaOH Titrant
    Phenolphthalein Indicator CH3COOH Analyte
    Volume used= F-I (mL) Reading of the burette Titrations
    Final (F) mL Initial (I) mL
    5.7-0=5.7 5.7 0 1.
    11-5.7=5.3 11 5.7 2.
    16.2-11=5.2 16.2 11 3.
    5.4 Average volume


    CH3COOH + NaOH → CH3COONa +H2O
    n moles CH3COOH = n moles NaOH = 4.32*10-5
    M1(diluted CH3COOH) = 4.32*10-3
    M(stock CH3COOH) = M1(CH3COOH diluted) x dilution factor
    M(stock CH3COOH) = 4.32*10-3*20 = 0.0864 M
    MM(CH3COOH) = 60.00 g\mole
    M(CH3COOH concentrated in g/L)= M1(CH3COOH diluted) x dilution factor x MM
    M(CH3COOH concentrated in g/L)= 0.0864 * 60 = 5.184 g/L

    Discussion:
    From this experiment, we added water to all the solutions we used (NaOH, CH3COOH, HCl) to dilution them, and even by doing that the number of moles still the same, so that helped us to determine the molarity of NaOH in part b, and the molarity of CH3COOH in part c.

    Conclusion:
    In part a of this experiment, we found that the molarity of diluted HCl is 0.04M and the dilution factor is 10.
    In part b, we found that the average volume of NaOH is 5 mL and we used it to determine the molarity of the original NaOH and it was 8*10-2 M.
    In part c, we used our result in part b to determine the molarity of the acetic acid in the vinegar and we it was 0.0864 M.



      الوقت/التاريخ الآن هو السبت سبتمبر 23, 2017 3:38 pm