chemistry lab titration

posted by .

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 -

    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 -

    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 -

    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 -

    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 -

    thank you!!

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

  • chemistry lab titration -

    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?

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chem Titration and pH

    Calculate the pH at 0mL, 5mL,...40mL for a 10.0mL aliquote of 0.100M Na3AsO4 (weak base) titrated with 0.100M HCl pKa1 = 2.25 pKa2 = 6.77 pKa3 = 11.60 Thank you First you should write equations to know where we are in the titration. …
  2. chemistry

    what is ph during titration of 10.00ml of 0.1000m triethylamine, ch3ch2)3n (kb=2.3x10^-, with 0.1000m HCL solution after addition of titrant (HCL) 1, 0ml, 5ml,10ml, 20ml,30ml
  3. Chemistry - titration lab

    Hello. For my chemistry of solutions class, we had a titration lab where we titrated NaOH into an HCl + H2O mix. This was a preparatory titration for the next one consisting of the hydrolysis of ethyl acetate into acetic acid with …
  4. CHemistry- For Dr.Bob

    Hey Dr.Bob, I asked two days ago a question about a titration lab I had that you answered, but I still have one more question about that lab. Here's the lab summary: For my chemistry of solutions class, we had a titration lab where …
  5. chem II

    starting with 15.0 ml of 0.370M of organic base pyridine C5H5N a titration is carried out using 0.100M HCL. how do i find the pH before titration, after adding 22.0mL of the acid , at equivalent point and after 41.0mL is of acid is …
  6. Chemistry

    A 50mL solution of Histidine-HCl (Histidine: pKa1=1.8, pKa2=9.2, pKa3=6.0)is titrated with 0.500M NaOH. What is the expected pH of the histidine solution at the points in the titration when 14.0mL, 26mL, and 38mL of the NaOH titrant …
  7. Chemistry

    Calculate the volume of .100M NaOH that must be added to reach ph of 3 in the titration of 25.00 mL of .100M HCl
  8. Chemistry - Solubility

    Given: Concentration of HCl is 0.1388M 0.5 g of Ca(OH)2 was placed in a flask. 100mL of 0.05M NaOH was poured into the flask. 25mL aliquot was filtrated and used for titration. Suppose to calculate the OH^- equilibrium concentration …
  9. Chemistry

    In a titration experiment involving an unknown concentration of HCl and a 0.100M solution of NaOH(aq), it was determined that it took 12.5mL of NaOH to neutralize 20.0mL of the HCl(aq). Write a balanced chemical equation for this reaction …
  10. Chemisty

    Based on the standard analytical practice during a titration of taking pH-volume measurements every 0.3 pH units, how many measurements should you take between the addition of 5mL and 7.5mL of titrant, a) in the titration of a 10mL …

More Similar Questions