Explanation:
As it is given that
= 4.20 and pH is 4.0.
Now, we use Henderson-Hasslbach equation as follows.
pH = ![pK_{a} + log \frac{[salt]}{acid}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%20%5Cfrac%7B%5Bsalt%5D%7D%7Bacid%7D)
So, it is given that salt is sodium benzoate and acid is benzoic acid. Hence, the given equation will become as follows.
pH =
........ (1)
Now, substitute the given values into equation (1) as follows.
pH =
4.0 = ![4.20 + log \frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}](https://tex.z-dn.net/?f=4.20%20%2B%20log%20%5Cfrac%7B%5B%7BC_%7B6%7DH_%7B5%7DCOONa%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DCOOH%5D%7D)
= 4.0 - 4.20
= - 0.20
= 
= 0.613
or,
=
............. (2)
Since, the total volume of (acid + base) is given as 100.0 mL. So, let us assume volume of acid is x and volume of base is y.
Hence, x + y = 100 mL ....... (3)
And, equation (2) will become as follows.
= 
y = 0.255x
Substituting value of y into equation (3) as follows.
x + 0.255x = 100 mL
1.255 x = 100 mL
x = 79.68 mL
So, value of y will be as follows.
y = 0.255 x
= 0.255 × 79.68 mL
= 20.31 mL
Thus, we can conclude that volume of benzoic acid is 79.68 mL and volume of sodium benzoate is 20.31 mL.