Respuesta :
Explanation:
Let us assume that volume of acetic acid added is V ml.
So, [tex][CH_{3}COOH] = \frac{0.05 \times 100}{100 + V}[/tex]
and, [tex][CH_{3}COONa] = \frac{0.05 \times V}{100 + V}[/tex]
Expression for the buffer solution is as follows.
pH = [tex]pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}[/tex]
5.1 = [tex]4.76 + log \frac{0.05 \times V}{0.05 \times 100}[/tex]
0.34 = log V - 2
log V = 2.34
or, V = 218.77 ml
Now, we will calculate the molarity of the buffer with respect to acetate as follows.
= [tex][CH_{3}COO^{-}] + [CH_{3}COOH][/tex]
= [tex]\frac{0.05 \times 218.77}{318.77} + \frac{0.05 \times 100}{318.77}[/tex]
= 0.0499 M
or, = 0.05 M (approx)
Thus, we can conclude that molarity of the resulting buffer with respect to acetate is 0.05 M.
