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March 30, 2017

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Calculate the theoretical pH after 2.50 mL and 9.50 mL of NaOH has been added in both the titration of HCl and of HC2H3O2. Indicate if the volume of NaOH is before or after the equivalence point. Compare these theoretical values with the actual values found on the titration curves created in lab.

(Hint: Use total volume of titrant (the initial and final volume on the buret) in conjunction with the time required to deliver that volume to convert the volumes given above to time.) Show a sample of all necessary calculations. Calculate the percent error from the theoretical values.

molarity is 0.1 for all substances, volume of HCl and HC2H3O2 is 8.00ml each, both diluted in 100ml of water

PLESASE HELP!!

  • General Chemistry - ,

    millimoles HCl = 8.00 mL x 0.1M=0.8mmols.
    mmoles NaOH = 2.50 mL x 0.1M = 0.25 mmols.
    mmoles NaOH = 9.50 x 0.1M = 0.95 mmoles.
    mmoles HAc = 8.00 mL x 0.1M = 0.8 mmoles.
    --------------------------------------

    ..........HCl + NaOH ==> H2O + NaCl
    initial...0.8....0........0......0
    add.............0.25.................
    change...-0.25..-0.25....+0.25..0.25
    equil...0.55......0.......0.25..0.25
    pH of the soln is determined by the excess HCl since there is no NaOH present and NaCl is not hydrolyzed.
    (HCl) = 0.55mmoles/(100+8.00+2.50)mL = ?? and convert to pH.

    initial......0.8..0......0.......0
    add...............0.95.............
    change.....-0.8...-0.8...+0.8..+0.8
    equil........0....0.15....0.8..0.8
    pH of the soln is determined by the excess NaOH since there is no HCl present and NaCl is not hydrolyzed.
    (NaOH) = 0.15mmoles/(100+8+9.5)mL = ?, convert to pOH and pH.

    ...........HAc + NaOH ==> NaAc + H2O
    initial...0.8.....0........0.......0
    add..............0.25..............
    change...-0.25..-0.25....+0.25...+0.25
    equil.....0.55.....0.......0.25...0.25

    Here you have a soln in HAc and NaAc so that is a buffer. Use the Henderson-Hasselbalch equation to solve for pH.
    pH = pKa + log (0.25/0.55) = ?

    ............HAc + NaOH ==> NaAc + H2O
    initial....0.8.....0........0......0
    add...............0.95.............
    change...-0.8...-0.8......+0.8..+0.8
    equil.......0.....0.15.....0.8....0.8

    Here you have a soln in NaOH and NaAc but that isn't a buffer and the pH will be determined by the excess NaOH present. It is true that the Ac^- will hydrolyze but it is small enough to ignore it.
    (NaOH) = mmoles NaOH/(100 + 9.5+8)mL = ?, convert to pOH and to pH.
    Post your work if you get stuck.

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