Solution :
The buffer is the one that contains weak acid (the 
is nearly equal to the required pH) and the salt of its conjugate base.
The values of the given weak acids are as follows :
HCOOH : = 3.744
HCN : = 9.21
Since the required pH is 5.03, the suitable buffer is the mixture of the acetic acid and the salt of its conjugate base.
Let suppose the volumes of 
and 
are x and y mL respectively.
So the total volume of the buffer is 1000 mL.
∴ x+ y = 1000 ................(1)
Writing the Henderson-Hasselbalch equation for the given buffer solution :
![$pH = pK_a + \log \ \frac{[CH_3COONa]}{[CH_3COOH]}$](https://tex.z-dn.net/?f=%24pH%20%3D%20pK_a%20%2B%20%5Clog%20%5C%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%24)
.............(2)
y = 1.9 x
Substituting the values of y in equation (1), we get
x + 1.9 x = 1000
x = 345
Putting the value of x in (1), we get
345 + y = 1000
y = 655
Therefore the volume of is 345 mL and the volume of is 655 mL.
Answer:
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Explanation:
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