2. Calculate the pH of a solution prepared by dissolving 0.850 mol of NH3 and 0.300 mol of NH4Cl in water sufficient to yield 2.00 L of solution. The Kb of ammonia is 1.77 x 10-5.

pH = 9.70

Explanation:

It is possible to answer this question using Henderson-Hasselbalch formula:

pH = pka + log [Base] /[Acid]

Base is NH₃ and acid NH₄⁺

Molarity of the compounds is:

NH₃: 0.850mol / 2.00L = 0.425M

NH₄⁺: 0.300mol / 2.00L = 0.150M

To find pka:

Ka×Kb = Kw

Ka = 1x10⁻¹⁴ / 1.77x10⁻⁵ = 5.65x10⁻¹⁰

pKa = -logKa = 9.25

Replacing:

pH = 9.25 + log [0.425] /[0.150]

pH = 9.70

I hope it helps!