How many grams of dry NH4Cl need to be added to 1.53 L of a 0.08 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.2? Kb for ammonia is 1.8×10−5. Express your answer in grams.

75.15g of NH₄Cl

Explanation:

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

pH = pka + log₁₀ [A⁻] / [HA] (1)

Where A⁻ is weak base, NH₃, NH₄Cl is HA, conjugate acid.

Using Kb, it is possible to find pka, thus:

Kw / Kb = Ka → 5.56x10⁻¹⁰.

pKa is -log Ka → 9.26

Moles of NH₃ are:

1.53L × (0.08mol / L) = 0.1224 moles of NH₃.

Replacing these values in (1):

8.2 = 9.26 + log₁₀ [0.1224] / [HA]

[HA] = 1.405moles of NH₄Cl

As molar mass of NH₄Cl is 53.491g/mol, you need to add:

1.405moles of NH₄Cl ₓ (53.491g / 1mol) = 75.15g of NH₄Cl

I hope it helps!

lol idk man i guess u gotta ask bill nye the science guy