Question

How many grams of dry NH4Cl need to be added to 1.80 L of a 0.400 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.60? Kb for ammonia is 1.8×10−5. Express your answer with the appropriate units.

Answer #1

pH of buffer = 8.60

pOH = 14-pH = 14-8.60 = 5.4

K_{b} for ammonia = 1.8*10^{-5}

pK_{b} = -log(K_{b}) = -log(1.8*10^{-5})
= 4.74

Using Henderson-Hasslebach equation,

pOH = pK_{b} + log[salt]/[base]

5.4 = 4.74 + log[NH4Cl]/[NH3]

log[NH4Cl]/[NH3] = 5.4-4.74 = 0.66

[NH4Cl]/[NH3] = 10^{0.66} = 4.57

[NH4Cl] = 4.57*[NH3 = 4.57*0.400M

[NH4Cl] = 1.828M

volume = 1.80L

moles of NH4Cl = Molarity*Volume in L = 1.828mol/L * 1.80L = 3.291moles

molar mass of NH4Cl = 53.491g

Mass of NH4Cl = moles*molar mass = 3.291moles*53.491g/mol = 176g

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