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Chemistry problem solving (neutralisation)

  1. Jan 2, 2005 #1
    On se propose de préparer une solution de pH= 4.4 en utilisant 100 ml d'une solution d'acide benzoïque de concentration 2,5.10^-2 mol.L^-1
    pKa C6H5COOH/C6H5COO- = 4.20

    a)quel volume V d'une solution d'hydroxyde de sodium NaOH de concentration 0.1 mol.L^-1 faut-il lui ajouter?

    b)si à la place de la solution d'hydroxyde de sodium on utilise une solution de benzoate de sodium de même concentration, quel volume V' de cette solution faut-il ajouter?

    I hope someone will answer me :)
    thanks :)
  2. jcsd
  3. Jan 4, 2005 #2


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    This might be of help - http://chimge.unil.ch/Fr/ph/1ph34.htm [Broken]

    from - http://chimge.unil.ch/Fr/ph/1ph1.htm [Broken]

    It appears that on is to prepare a solution with pH=4.4 using 0.1 L (liter) of 0.025 M benzoic acid (pKa = 4.2),

    In part a, one is asked 'what volume, V, of NaOH of 0.1 M.

    So moles of NaOH = 0.1 V, and this added to a solution 0.1 L containing 0.0025 moles of C6H5COOH.

    the resulting volume will be (0.1+V).

    Remember, the strong base will react completely with molar equivalent of a weak acid, so not much NaOH is needed.

    Perhaps then you can solve b.
    Last edited by a moderator: May 1, 2017
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