Answer:
75.15g of NH₄Cl
Explanation:
It is possible to answer this question using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻] / [HA] <em>(1)</em>
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 → <em>9.26</em>
Moles of NH₃ are:
1.53L × (0.08mol / L) = <em>0.1224 moles of NH₃</em>.
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) = <em>75.15g of NH₄Cl</em>
I hope it helps!
