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March 26, 2017

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Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq)

(a) What is the initial pH of the 0.20 M C6H5COOH(aq)?

(b) What is the pH after the addition of 15.0 mL of 0.30 M KOH(aq)?


(c) What volume of 0.30 M KOH(aq) is required to reach halfway to the stoichiometric point?


(d) Calculate the pH at the halfway point.

(e) What volume of 0.30 M KOH(aq) is required to reach the stoichiometric point?


(f) Calculate the pH at the stoichiometric point.

---------attempted solutions:

C6H5COOH + H2O ==> C6H5COO^- + H3O^+
.2*.3
-x +x +x

[x^2]/[.6-x] = 6.5E-5
.00000039-6.5E-5x-x^2=0
x=.000593 moles
30 mL = .003L
[H+] = x/.003
pH=-log[H+]=0.704

Is this right? It seems really low...

b)
.20M * .03L = .006 mol C6H5COOH
.30 * .015L = .0045 mol KOH
.0045-.006 = .0015 excess

pKa C6H5COOH = 4.19

.03 + .015 = .045 L total volume

pH = pka + log([base]/[acid]) =
4.19 + log( (.0045/.045)/(.006/.045)) =
4.06

Is this right?

I haven't gotten to the other parts of the problem yet.

  • Chemistry - ,

    Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq)

    (a) What is the initial pH of the 0.20 M C6H5COOH(aq)?

    (b) What is the pH after the addition of 15.0 mL of 0.30 M KOH(aq)?


    (c) What volume of 0.30 M KOH(aq) is required to reach halfway to the stoichiometric point?


    (d) Calculate the pH at the halfway point.

    (e) What volume of 0.30 M KOH(aq) is required to reach the stoichiometric point?


    (f) Calculate the pH at the stoichiometric point.

    ---------attempted solutions:

    C6H5COOH + H2O ==> C6H5COO^- + H3O^+
    .2*.3
    -x +x +x

    [x^2]/[.6-x] = 6.5E-5
    .00000039-6.5E-5x-x^2=0
    x=.000593 moles
    30 mL = .003L
    [H+] = x/.003
    pH=-log[H+]=0.704

    Is this right? It seems really low...
    ....C6H5COOH ==> C6H5COO^- + H3O^+
    I.......0.20..........0..........0
    C........-x............x.........x
    E.......0.2-x.........x..........x
    Ka = (H3O^+)(C6H5COO^-)/(C6H5COOH)
    6.5E-5 = (x)(x)/(0.2-x)
    x = 0.00361 and pH = 2.44.


    b)
    .20M * .03L = .006 mol C6H5COOH
    .30 * .015L = .0045 mol KOH
    .0045-.006 = .0015 excess

    pKa C6H5COOH = 4.19

    .03 + .015 = .045 L total volume

    pH = pka + log([base]/[acid]) =
    OK to here
    4.19 + log( (.0045/.045)/(.006/.045)) =
    4.06

    Is this right?
    What you have labeled as excess is the benzoic acid. That's (acid) = 0.0015/0.045. The salt formed (base) = 0.0045/0.045. The volumes cancel so can plug into the HH equation as
    4.19 + log (0.0045/0.0015) = about 4.66

  • Chemistry - ,

    Thanks! I see what I did wrong. I used moles instead of molarity for part a... although I am still uncertain why I should use benzoic acid as the acid in part b? I thought the reaction was C6H5COOH + KOH ==> C6H5COO- + K + H2O ... and that I should use the acid and base (C6H5COOH + KOH )of the reactants for the HH equation? Am I supposed to use the base and its conjugate acid (KOH + C6H5COO-)? Does this hold for all cases of the HH equation? maybe I've been doing it wrong...

    Thanks again!

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