Answer:
We need to add NaOH to increase the pH decreasing the H⁺ concentration.
0.4g of NaOH must be added
Explanation:
The pKa of HCHO2 is 3.74. Using H-H equation the initial pH of the buffer is:
pH = pKa + log [NaHCO2] / [HCHO2]
pH = 3.74 + log [0.025mol] / [0.025mol]
pH = 3.74
The initial pH is 3.74, as we want a pH of 4.10, we need to add NaOH to increase the pH decreasing the H⁺ concentration.
Total moles of HCHO2-NaHCO2 buffer are:
0.025mol + 0.025mol =
0.050mol = [HCHO2] + [NaHCO2] (1)
Using H-H equation:
4.10 = 3.74 + log [NaHCO2] / [HCHO2]
0.36 = log [NaHCO2] / [HCHO2]
2.29087 = [NaHCO2] / [HCHO2] (2)
NaOH reacts with HCHO2 as follows:
NaOH + HCHO2 → NaHCO2 + H2O
Replacing (1) in (2):
2.29087 = 0.050 - [HCHO2] / [HCHO2]
2.29087[HCHO2] = 0.050 - [HCHO2]
3.29087[HCHO2] = 0.050
[HCHO2] = 0.015 moles
As HCHO2 decreases from 0.025 moles to 0.015 moles, the moles of NaOH added are 0.010moles.
The mass is:
0.010moles NaOH * (40g/mol) =