How many mL of a 1.00 M HCl solution would need to be added to 100 mL of 0.263 M ammonia to produce a buffer with a pH of 10.00. The pKa of ammonia is 9.25.

3.97mL of 1.00M HCl must be added

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