One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buff

Question

One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of 0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that “one solution is a buffer” is not sufficient.

H3O+(aq) + C2H3O2-(aq) HC2H3O2(aq) +H2O

Ka of acetic acid = 1.8 x 10-5

Acetic acid/sodium acetate buffer: Acetic acid = 0.1M, Sodium acetate = 0.1M

Explanation / Answer

     CH3COONa    +      H+ ----------->          CH3COOH + NaCl

    1 millimole            0.025 millimole                     1 millimole

(1 -.025millmoles)/vtotal        0                             (1+0.025 millmoles)/vtotal

Ka = [H+][CH3COO-]/CH3COOH] =1.8x10-5

     [ H+] = 1.8x10-5 (1.025)/0.975 = 1.892 x 10-5

-------------

For water,

[ H+]    = 0.025/(10+.25) = 2.43x10-3

water has strong acid HCl which disscoiates 100% so gives all added to H+,

where as the solution above was buffer solution which does not allow major change in pH , Buffer has reserve acid and base within its composition so the added acid gets consumed,

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