How many grams of dry NH4Cl need to be added to 2.40L of a 0.300M solution of ammonia,NH3..?

How many grams of dry NH4Cl need to be added to 2.40L of a 0.300M solution of ammonia,NH3 , to prepare a buffer solution that has pH of 8.53? Kb for ammonia is 1.8x10^-5 .

Answer:
pH = 8.53
pOH = 14-8.53 = 5.47
[OH-] = 10^-5.47 = 3.39 x 10^-6 M
NH3 + H2O <--> NH4+ + OH-
1.8 x 10^-5 = (x) 3.39 x 10^-6 / 0.300 -3.39 x 10^-6
x = 1.59 M
1.59 moles/ L x 2.4 L = 3.82 moles
Molecular weight NH4Cl = 53.4 g/mol
3.82 mol x 53.4 g/mol =204.2 g

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