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Posted by on Friday, June 1, 2012 at 5:13pm.

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.
(a) 100.0 mL of 0.14 M HC7H5O2 (Ka= 6.4 multiplied by 10-5) titrated by 0.14 M NaOH
halfway point

equivalence point

(b) 100.0 mL of 0.29 M C2H5NH2 (Kb = 5.6 multiplied by 10-4) titrated by 0.58 M HNO3
halfway point

equivalence point

(c) 100.0 mL of 0.28 M HCl titrated by 0.14 M NaOH
halfway point

equivalence point

I'm pretty unclear as far as how to go about it and would greatly appreciate help. Thanks in advance!

  • Chemistry - , Friday, June 1, 2012 at 5:39pm

    The pH at the half-way point of a monoprotic acid is just pKa. For a monoprotic base (C2H5NH2) it is pKa but remember they give you pKb in the problem so pKa = 14-pKb.

    The pH at the equivalence point of a monoprotic acid or monoprotic base is calculated from the hydrolysis of the salt. Use (salt) = C = mols salt/L soln.
    For the acid the anion is hydrolyzed:
    ........A^- + HOH ==>HA + OH^-

    Kb for the A^- = Kw/Ka for the acid = (x)(x)/(C-x) and solve for x = OH^-, then convert to pH.

    For the C2H5NH2 it is the salt C2H4NH3 that is hydrolyzed. I'll call that BNH3^+. Find (BNH3^+) at the equivalence point following the same procedure for the weak acid above.
    ........BNH3^+ + H2O ==> H3O^+ + BNH2

    Ka for BNH3^+ = (Kw/Kb for BNH2) = (x)(x)/(C-x) and solve for x = (H3O^+) and convert to pH.

    For the NaOH/HCl (strong base/strong acid),
    mols acic to start - mols base added = mols HCl remaining. That gives H^+ when corrected for dilution and pH from that.
    The equivalence pint is the hydrolysis of the salt but for SA/SB titrations neither the anion nor the cation are hydrolyzed so the pH = 7.0.
    Post your work if you get stuck.

  • Chemistry - , Friday, June 1, 2012 at 7:09pm

    So for the first portion I came to Kb=1.5x10^-10
    x^2/.14-x= 1.56x10^-10
    x^2-(2.2x10^-11)x - 1.56x10^-10=0
    solving with the quadratic equation I came to x= 1.25x10^-5 (I'm disregarding i)
    So -log(1.25x10^-5)= 4.9

    But somewhere I've done something wrong as the homework program I'm using is saying neither 4.9 or 9.1 are correct.

  • Chemistry - , Friday, June 1, 2012 at 8:13pm

    The concn of the salt is not 0.14. Note my instructions were the C = mols/L. At the equivalence point you have (let's call the acid HA)
    HA + NaOH ==> NaA + H2O
    You have 100 mL x 0.14M acid = 0.014 mols.
    You have added 100 mL of 0.14M NaOH so
    (salt) = 0.14mol/0.200 L = 0.07M.
    You should have ended up with
    Kb = 1.56E-10 = (x)(x)/(0.07-x)
    If we assume 0.07-x = 0.07, then
    x^2 = 1.09E-11 and
    x = 3.31E-6 = (OH^-). First we check to see that the assumption is ok. Since 0.07-3.31E-6 = essentially 0.07 the assumption is ok and we need not solve a quadratic.
    If OH^- = 3.31E-6 then pOH = 5.48 so pH = 8.52

  • Chemistry - , Friday, June 1, 2012 at 8:44pm

    Alright, I was able to figure them all out. Thank you!

  • Chemistry - , Friday, June 1, 2012 at 11:38pm

    Hi, I need help with a chemistry question.

    Question: Iron ore is treated with carbon monoxide to extract and purify the iron.

    a) Calculate the minimum mass of carbon monoxide that must be ordered by a refining company for every metric tonne of iron ore that is processed.

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