1. Calculate the pH of a solution of 0.2M acetic acid and 0.35M acetate ion. The pk of acetic acid is 4.8. pH = pk + log ([A] :

Question

1. Calculate the pH of a solution of 0.2M acetic acid and 0.35M acetate ion. The pk
of acetic acid is 4.8.
pH = pk + log ([A] : [HA])
A.
5.10
B.
5.04
c.
5.25
D.
6.10
E.
6.00
of which

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Orla Orla 5 years 2021-08-12T00:33:17+00:00 1 Answers 25 views 0

Answers ( )

  1. Xavia
    0
    2021-08-12T00:34:49+00:00
    Reply

    Answer:

    The correct answer is option C.

    Explanation:

    The pH of the solution with weak acid and its conjugate base is given by the Henderson-Hasselbalch equation:

    pH=pK_a+\log[\frac{[A^-]}{[HA]}]

    Where:

    pK_a= The negative logarithm of the dissociation constant of a weak acid

    [A^-]= Concentration of conjugate base of a weak acid

    [HA]= Concentration of weak acid

    We are given a solution with acetic acid and acetate ion.

    HAc(aq)\rightleftharpoons H^+(aq)+Ac^-(aq)

    The concentration of acetic acid in a solution= [HAc]=0.2M

    The concentration of acetate ion in a solution = [Ac^-]=0.35M

    The pK_A of the acetic acid = pK_a=4.8

    The pH of the solution:

    pH=4.8+\log[\frac{0.35 M}{0.2M}]=5.04

    5.04 the pH of a solution of 0.2M acetic acid and 0.35M acetate ion.

    Hence, the correct answer is option C.

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