College Chemistry

posted by .

Calculate the pH at the equivalence point if 25.00 mL of 0.010M barbituric acid (HC4H3N2O3) is titrated with 0.020M NaOH. (Ka barbituric acid = 1.0 x 10^-5)

  • College Chemistry -

    If we call barituric acid, HB, the equation is
    HB + NaOH ==> NaB + H2O, therefore, you can see that at the equivalence point we simply have the salt of barbituric acid so the pH will be determined by the hydrolysis of the salt.
    You need to know the concn of the salt. Since the NaOH is twice the concn of the acid you know the volume of the base will be 12.5 mL so the concn of the salt at the equivalence point will be M = mols/L = 0.01 x (25/37.5) = 0.00667
    ........B^- + HOH ==> HB + OH^-

    Kb for B^- = (Kw/Ka for HB) = (x)(x)/(0.00667-x)
    Solve for x = (OH^-) and convert to pH.
    Be careful with the algebra, you may need to use the quadratic equation since the concn of the salt is so small; i.e., 0.00667-x may not be equal to 0.00667

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. College Chemistry

    A 30.00 mL volume of a weak acid, HA, (Ka = 3.8 x 10^-6) is titrated with 39.00 mL of 0.0958 M NaOH to the equivalence point. A.) What is the pH of the acid solution, before any base is added to it?
  2. Chemistry

    In this case, the inflection point, and equivalence, occurs after 23.25mL of 0.40 M NaOH has been delivered. Moles of base at the equivalence point can be determined from the volume of base delivered to reach the equivalence point …
  3. College Chemistry

    1) A 25mL sample of the .265M HCI solution from the previous question is titrated with a solution of NaOH. 28.25mL of the NaOH solution is required to titrate the HCl. Calculate the molarity of the NaOH solution. 2) A 1.12g sample …
  4. chemistry

    if 0.4M NaOH is titrated with 0.4M HF, how do we calcualte the ph at equivalence. The book assumes each is 1L, but why do we use 1L * chemistry - Dr.Jim, Thursday, November 11, 2010 at 5:31am HF is a weak acid, so you need the dissociation …
  5. chemistry

    a. A 0.1 molal solution of a weak monoprotic acid was found to depress the freezing point of water 0.1930C. Determine the Ka of the acid. You can assume 0.1 molal and 0.1 molar are equivalent. b. 100 ml of a 0.1M solution of the above …
  6. Chemistry

    Amino acid solution was made by taking 3.58 g of the dry amino acid and making it up to 1.00L in water. A 25.0 mL sample was taken and titrated with NaOH of concentration of 0.0493M, with the first equivalence point at 18.15 mL and …
  7. Chemistry (College) URGENT

    1.6g of an unknown monoprotic acid (HA) required 5.79 mL of a 0.35 M NaOH solution to reach the equivalence point. Determine the molar mass of the acid and given that the pH at the half way point to the equivalent point is 3.86. Calculate …
  8. Chemistry

    When a diprotic acid, H2A, is titrated by NaOH, the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape. (a) Notice that the plot has essentially two titration …
  9. College Chemistry (DrBob222)

    A solution of an unknown weak acid, HA, is titrated with 0.100 M NaOH solution. The equivalence point is achieved when 36.12 mL of NaOH have been added. After the equivalence point is reached, 18.06 mL of 0.100 M HCl are added to the …
  10. Chemistry

    Citric acid is a tri-protic acid with a Ka1 =8.4x10-4 Ka2=1.8x10-5 and Ka3=4.0x10-6. calculate the pH at the 2nd equivalence point in the titration of 85.5 mL of a .21 M citric acid solution with .25 M NaOH. Calculate the pH, after …

More Similar Questions