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How many grams of dry NH4Cl need to be added to 2.30 L of a 0.100 M solution of ammonia,NH3, to prepare a buffer soluti...

How many grams of dry NH4Cl need to be added to 2.30 L of a 0.100 M solution of ammonia,NH3, to prepare a buffer solution that has a pH of 9.00? Kb for ammonia is 1.8x 10^-5.
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Answer #1

pH = 9
so, pOH = 14-pH = 5

now,
pOH = pKb + log([NH4Cl]/[NH3])

given, Kb = 1.8x10-5

so, pKb = -log(1.8x10-5) = 4.745

and

[NH3] = 0.1 M

so, put values

5 = 4.745 +  log([NH4Cl]/0.1)

solving, [NH4Cl] = 0.18 M

so,1 liter solution has 0.18 moles

so 2.3 liter solution must have 2.3*0.18 = 0.414 moles of NH4Cl

molar mass of NH4Cl = 53.49 gm/mol

so mass of NH4Cl to be added = 0.414*53.49 = 22.14486 gm

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