Answer:
3.97mL of 1.00M HCl must be added
Step-by-step explanation:
The buffer of ammonia and ammonium ion follows the equation:
pH = pKa + log [NH₃] / [NH₄⁺] (1)
Where pH is the pH we want to obtain: 10.00
pKa is the pKa of the buffer: 9.25
[NH₃] could be taken as moles of ammonia
And NH₄⁺ moles of ammonium ion
When NH₃ reacts with HCl:
NH₃ + HCl → NH₄⁺ + Cl⁻
NH₄⁺ is produced. That means moles of ammonia are Initial moles of ammonia - Moles of HCl added
Moles of HCl added = Moles of ammonium ion
The total moles of the buffer are:
0.100L * (0.263mol / L) = 0.0263moles = [NH₄⁺] + [NH₃] (2)
Rewriting (1):
10.00 = 9.25 + log [NH₃] / [NH₄⁺]
0.75 = log [NH₃] / [NH₄⁺]
Replacing (2) in (1):
0.75 = log 0.0263 - [NH₄⁺] / [NH₄⁺]
5.6234 = 0.0263 - [NH₄⁺] / [NH₄⁺]
5.6234[NH₄⁺] = 0.0263 - [NH₄⁺]
6.6234[NH₄⁺] = 0.0263
[NH₄⁺] = 0.00397 moles = Moles HCl
To obtain these moles we must add:
0.00397 moles HCl * (1L / mol) = 0.00397L =
3.97mL of 1.00M HCl must be added