Explanation:
The given data is as follows.
[HCOOH] = 0.2 M, [NaOH] = 2.0 M,
V = 500 ml, [Benzoic acid] = 0.2 M
First, we will calculate the number of moles of benzoic acid as follows.
No. of moles of benzoic acid = Molarity × Volume
= ![2 \times 0.475](https://tex.z-dn.net/?f=2%20%5Ctimes%200.475)
= 0.095 mol
And, moles of NaOH present in the solution will be as follows.
No. of moles of NaOH = Molarity × Volume
= ![2 \times 0.025](https://tex.z-dn.net/?f=2%20%5Ctimes%200.025)
= 0.05 mol
Hence, the ICE table for the chemical equation will be as follows.
![C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O](https://tex.z-dn.net/?f=C_%7B6%7DH_%7B5%7DCOOH%20%2B%20NaOH%20%5Crightarrow%20C_%7B6%7DH_%7B5%7DCOONa%20%2B%20H_%7B2%7DO)
Initial: 0.095 0.05 0 0
Equlbm: (0.095 - 0.05) 0 0.05
pH =
= ![4.2 + log \frac{0.05}{0.045}](https://tex.z-dn.net/?f=4.2%20%2B%20log%20%5Cfrac%7B0.05%7D%7B0.045%7D)
= 4.245
For,
![HCOOH + NaOH \rightarrow HCOONa + H_{2}O](https://tex.z-dn.net/?f=HCOOH%20%2B%20NaOH%20%5Crightarrow%20HCOONa%20%2B%20H_%7B2%7DO)
Initial: 0.2x 2(0.5 - x) 0
Equlbm: 0.2x - 2(0.5 - x) 0 2(0.5 - x)
As,
pH =
4.245 = 3.75 +
= 0.5
= 3.162
Now,
= 3.162
x = 0.464 L
Volume of NaOH = (0.5 - 0.464) L
= 0.036 L
= 36 ml (as 1 L = 1000 mL)
And, volume of formic acid is 464 mL.