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Posted by on Wednesday, April 7, 2010 at 5:55pm.

10.0 mL aliquot of 0.100M Na3AsO4 is titrated with 0.100M HCl.
Sketch the titration graph by calculating the pH at 0mL, 5mL, 10mL, 15mL, 20mL, 25mL, 30mL, 40mL of HCl.

  • chemistry lab titration - , Wednesday, April 7, 2010 at 8:14pm

    Which of these do you need help with. Explain what you don't understand about each that you need help with. Two hints to get you started.
    Step 1. You should determine where the three equivalence points are (with regard to mL HCl).
    Step 2. The pH at the beginning (0 mL HCl) must be the pH of the hydrolysis of the 0.1 M salt. You can show the hydrolysis as
    AsO4^-3 + HOH ==> HAsO4^-2 + OH^-

    Set up an ICE chart, and substitute into the Kb expression. Kb=(Kw/k3)

  • chemistry lab titration - , Wednesday, April 7, 2010 at 8:20pm

    I don't understand how I would have to find the pH at 0mL or at 5 mL added or at 10 mL added.
    are the equivalence points at 10mL, 20mL, 30 mL?

    and the k3 that you are referring to did you get that from the pK3?

  • chemistry lab titration - , Wednesday, April 7, 2010 at 8:26pm

    so for 0mL added
    would it be
    k3 = [(HAsO4^-2)(OH^-)/(AsO4^-)]
    K3 = (x^2)/(0.1M of AsO4)
    pk3 = -log k3.
    then solve for x which would give you [OH^-] which you could use to find
    pOH and find pH by using pH + pOH = pKw??

  • chemistry lab titration - , Wednesday, April 7, 2010 at 10:47pm

    Yes, equivalence points are at 10, 20, and 30 mL.

    For the 0 mL, you almost have it but not quite.
    In my post I said
    Kb = (Kw/k3) = (HAsO4^-)(OH^-)/(AsO4^-3)
    and solve for (OH^-) as you indicated, convert to pOH, then to pH.
    At 5 mL you have a mixture of AsO4^-3 and HAsO4^-2 so you use the Henderson-Hasselbalch equation,
    pH = pKa + log[(base)/(acid)]. For pKa that is pK3, (base) is concn AsO4^-3 and concn acid is concn (HAsO4^-2).
    At 10 mL, the first equivalence point, hydrolyze the HAsO4^-2 in a similar manner to the way you did at 0 mL. The only real difference is that you use a different k (k2 this time in place of k3).
    At 15 mL you have a buffer again and use the H-H equation.
    20 mL is 2nd equivalence point and hydrolysis.
    25 mL is H-H equation.
    30 mL equilvence point

  • chemistry lab titration - , Thursday, April 8, 2010 at 2:42am

    thank you!!

    for 25 mL would i still be able to use the h-h equation since k1>10^-5?

  • chemistry lab titration - , Monday, October 4, 2010 at 11:49pm

    Are the pH's at 5mL, 15mL, and 25mL equal the respective pKa's (11.60, 6.77, and 2.25) since they are all half-equivalence points?

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