JJ asked in Science & MathematicsChemistry · 2 months ago

Ka of acetic acid is 1.8×10^5 find the concentration of C2H3O2- if [H3O+] is 5.51x10^-5M  n the the concentration of acetic acid is 8.4M. ?

Update:

Based on the Ka value the products favored or the reactant is favored? 

Reason is 

A. Small Ka so reactant favored

B. Small Ka so product favored

C. Large Ka so reactant favored

D. Large Ka so product favored

Based on the Ka value describe acetic acids dissociation into ions 

A. Molecule does not dissociate into ions

B. Molecule partially dissociates into ions

C. Molecule completely dissociates into ions

3 Answers

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  • 2 months ago

    There seems to be some problem with the wording of this question:

    CH3COOH partially dissociates in water :

     CH3COOH + H2O ↔ CH3COO- + H3O+

     You will see that [CH3COO-] = [H3O+]

    You state that : [H3O+] = 5.51x10^-5M

     Therefore [CH3COO-] must also = 5.51*10^-5M

    BUT - let us see if what you have submitted is possibly true:

    Ka = 1.8*10^-5

     Ka = [H3O+]² / [CH3COOH]

     1.8*10^-5 = [H+]² /8.4

    [H3O+]² = ( 1.8*10^-5) * 8.4

    [H3O+]² = 1.512*10^-4

    [H3O+] = 0.0123 M or 1.23*10^-2 M

    Where does [H3O+] = 5.51*10^-5M originate?

    And as above , [CH3COO-] = 0.0123 M or 1.23*10^-2M

     You have also asked 2 questions:

     1) A. Small Ka so reactant favored

    2) Molecule partially dissociates into ions

     I suggest that you check the exact wording of the original question

  • 2 months ago

    First things first, you have a typo. The Ka for acetic acid is 1.8 * 10^-5, not 1.8 * 10^5.

    Anyway, the Ka of an acid is the equilibrium constant for its dissociation reaction, so for any acid HA, the reaction is as below:

    HA <--> H+ + A-

    Ka = [H+][A-]/[HA]

    So for acetic acid, CH3COOH, the Ka can be written as:

    Ka = [H+][CH3COO-]/[CH3COOH] = 1.8 * 10^-5

    Now, plug in the given values:

    (5.51*10^-5)[CH3COO-]/(8.4) = 1.8 * 10^-5

    [CH3COO-] = 8.4*1.8*10^-5/(5.51*10^-5) = 2.744 M (or 2.7 M to 2 sig figs)

    Ka is just an equilibrium constant, so if the equilibrium constant is less than 1, then the products are favored. So the answer is A.

    Finally, acetic acid is less acidic than the hydronium ion (Ka = 55.5 M, the molar concentration of liquid water) so it is a weak acid, and it only partially dissociates into ions. B.

  • 2 months ago

    In a solution of just acetic acid in water, [C2H3O2-] = [H3O+] = 5.51X10^-5. But that is not what [H3O+] would be in this solution. In just acetic acid in water, let x = [H+] = [C2H3O2-]. Then,

    Ka = 1.8X10^-5 = x^2 / 8.4

    x = [H3O+] = [C2H3O2-] = 0.012 M

    So, this cannot be a solution of just acetic acid in water.For this made up solution:Ka = [H+][C2H3O2-]/[C2H3O2H] = 1.8X10^-5(5.51X10^-5) [C2H3O2-]/8.4 = 1.8X10^-5[C2H3O2-] = 2.74 MReactants are favored because Ka is small (A.)B. Molecule partially dissociates into ions.

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