Chemistry

posted by .

While titrating 25.00mL of a weak acid, HA, with 0.1500M NaOH, you reach equivalence point after adding 27.00mL of the NaOH. The pH of the acid initially was 2.48. What is the dissociation constant of the acid?

  • Chemistry -

    HA ==> H^+ + A^-
    K = (H^+)(A^-)/(HA) = Ka
    If you know NaOH molarity and NaOH volume, can you calculate the molarity of HA initially? Then knowing (HA), use pH initially to determine (H^+). Of course, (A^-) will be the same (you know this from the ionization equation above. That leaves just Ka to calculate. Post your work if you get stuck.

  • Chemistry -

    The titration helps determine the molar concentration of HA:
    Moles of HA(aq) = moles of OH- used
    moles of OH- = (0.1500Mmol./L)(0.02500L) = 4.050x10^-3 = mol. HA
    4.050x10^-3 mol. HA / 0.02500L = 0.162 M HA
    Next we find the H+ ion concentration
    [H+] = 10^-pH = 10^-2.48 = 3.31x10^-3 M
    Finally we find the Ka:
    Ka = [H+][A-] / [HA]
    [H+] = [A-] = [HA]
    [HA] = 0.162 M - 3.31x10^-3 M
    Substitute into the Ka expression to get the value of Ka.

  • Chemistry -

    Correction of one more typo:

    [H+] = [A-] = [HA]
    should change to:
    [H+] = [A-] =3.31x10^-3 M

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chem

    What is the pH of the resulting solution if 30.00 mL of 0.100M acetic acid is added to 10.00mL of 0.100 M NaOH?
  2. Chemistry

    Part A: Unknown Acid use 1gram and mix with 120 mL of distilled water Determine the concentration of the acid by titrating with 0.0998 M of NaOH Concentration of NaOH= 0.0998M Volume of NaOH= 3.5 mL # moles of NaOH= Initial Concentration …
  3. 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 …
  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

    Hi, could someone help me with these calculations for my lab; Given: Volume of vinegar analyzed:5 mL Con'c of NaOH: 0.09890M Avg. V of NaOH from titration:43.75 mL Moles of NaOH required to reach the equivalence point: ?
  6. Chemistry

    Thanks for your help; I am still have a few problems, could you check my work: Given: Volume of vinegar analyzed-5mL Con'c of NaOH-0.09890M Avg. Volume of NaOH from titration-43.75mL Density CH3COOH: 1.049g/mL Calculations: 1.Moles …
  7. Chemistry

    Consider a weak acid-strong base titration in which 25.0 mL of 0.100 M acetic acid is titrated with 0.100 M NaOH. a) Calculate the pH after the addition of 3.00mL of NaOH. b) What is the pH of the solution before the addition of NaOH …
  8. chemistry

    You are titrating an unknown weak acid you hope to identify. Your titrant is a 0.0935 mol/L NaOH solution, and the titration requires 39.9 mL to reach the equivalence point. How many moles of acid were in your sample?
  9. AP Chemistry

    A solution is prepared by titrating a 100.0 mL sample of 0.10 M HF (Ka = 7.2 × 10-4) with 0.10 M NaOH a. Does the solution contain a strong acid with a strong base, a strong acid with a weak base, a weak acid with a strong base, or …
  10. AP Chemistry

    A solution is prepared by titrating a 100.0 mL sample of 0.10 M HF (Ka = 7.2 × 10-4) with 0.10 M NaOH a. Does the solution contain a strong acid with a strong base, a strong acid with a weak base, a weak acid with a strong base, or …

More Similar Questions