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Chemistry

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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

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