Answer:
7.38g HCl
Explanation:
Using H-H equation for acetic buffer:
pH = pKa + log [NaC2H3O2] / [HC2H3O2]
<em>Where pKa is -log Ka = 4.74 and [] could be taken as moles of each compound.</em>
The initial moles of each specie is:
[NaC2H3O2]:
0.500L * (0.665mol/L) = 0.3325moles
[HC2H3O2]:
0.500L * (0.475mol/L) = 0.2375 moles
That means total moles are:
[NaC2H3O2] + [HC2H3O2] = 0.57 moles <em>(1)</em>
And solving H-H equation for a pH of 4.21:
4.21 = 4.74 + log [NaC2H3O2] / [HC2H3O2]
0.29512 = [NaC2H3O2] / [HC2H3O2] <em>(2)</em>
Replacing (1) in (2):
0.29512 = 0.57mol - [HC2H3O2] / [HC2H3O2]
0.29512 [HC2H3O2] = 0.57mol - [HC2H3O2]
1.29512 [HC2H3O2] = 0.57mol
[HC2H3O2] = 0.44 moles
The HCl reacts with NaC2H3O2 producing HC2H3O2, that means you need to add:
0.44 moles - 0.2375 moles =
0.2025 moles of HCl
Using molar mass of HCl (36.45g/mol), to convert these moles to grams:
0.2025 moles * (36.45g/mol) =
<h3>7.38g HCl</h3>
