Question

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

Answer 1

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.