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A buffer solution with a pH of 3.80 is prepared with the same volumes of 0.72 M formic acid and ________ M sodium formate. The Ka of formic acid is

(A) 1.8x10-4.
(B) 0.82
(C) 4.0×10−8
(D) 3.4×10−5
(E) 0.41 1.6

User Frank N
by
7.9k points

1 Answer

7 votes

Answer : The correct option is, (B) 0.82 M

Explanation : Given,

The dissociation constant for formic acid =
K_a=1.8* 10^(-4)

Concentration of formic acid (weak acid)= 0.72 M

pH = 3.80

First we have to calculate the value of
pK_a.

The expression used for the calculation of
pK_a is,


pK_a=-\log (K_a)

Now put the value of
K_a in this expression, we get:


pK_a=-\log (1.8* 10^(-4))


pK_a=4-\log (1.8)


pK_a=3.745

Now we have to calculate the concentration of sodium formate (conjugate base or salt).

Using Henderson Hesselbach equation :


pH=pK_a+\log ([Salt])/([Acid])

Now put all the given values in this expression, we get:


3.80=3.745+\log (([Salt])/(0.72))


[Salt]=0.82M

Therefore, the concentration of sodium formate is 0.82 M.

User Enea Dume
by
8.8k points
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