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How many grams of dry NH4Cl need to be added to 2.40 L of a 0.300...

How many grams of dry NH4Cl need to be added to 2.40 L of a 0.300 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.65? Kb for ammonia is 1.8×10−5.

Expert Solution

no of moles of NH3 = molarity * volume in L

= 0.3*2.4 = 0.72 moles

PH = 8.65

POH = 14-PH

= 14-8.65 = 5.35

Kb = 1.8*10^-5

PKb = -logKb

= -log1.8*10^-5

= 4.75

POH = Pkb + log[NH4Cl]/[NH3]

5.35 = 4.75 + log[NH4Cl]/0.72

5.35-4.75 = log[NH4Cl]/0.72

log[NH4Cl]/0.72 = 0.6

[NH4Cl]/0.72 = 10^0.6

[NH4Cl]/0.72 = 3.981

[NH4Cl] = 3.981*0.72 = 2.866

no of moles of NH4Cl = 2.866

mass of NH4Cl = no of moles * gram molar mass

= 2.866*53.5 = 153.33g of NH4Cl >>>>>answer

queen_honey_blossom answered 2 years ago

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