Question

How many grams of dry NH4Cl need to be added to 2.20 L of a
0.500 *M* solution of ammonia, NH3, to prepare a buffer
solution that has a pH of 8.62? *K*b for ammonia is
1.8×10−5.

Answer #1

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

POH = 14 - pH

= 14 - 8.62

= 5.38

use formula for buffer

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

5.380000000000001 = 4.7447 + log ([NH4Cl]/[NH3])

log ([NH4Cl]/[NH3]) = 0.6353

[NH4Cl]/[NH3] = 4.3179

[NH4Cl]/0.5 = 4.3179

[NH4Cl] = 2.1589

volume , V = 2.2 L

use:

number of mol,

n = Molarity * Volume

= 2.159*2.2

= 4.75 mol

Molar mass of NH4Cl,

MM = 1*MM(N) + 4*MM(H) + 1*MM(Cl)

= 1*14.01 + 4*1.008 + 1*35.45

= 53.492 g/mol

use:

mass of NH4Cl,

m = number of mol * molar mass

= 4.75 mol * 53.49 g/mol

= 254 g

Answer: 254 g

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0.500 M solution of ammonia, NH3, to prepare a buffer solution that
has a pH of 8.69? Kb for ammonia is 1.8×10−5?

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0.400 M solution of ammonia, NH3, to prepare a buffer solution that
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0.300 M solution of ammonia, NH3, to prepare a buffer solution that
has a pH of 8.64? Kb for ammonia is 1.8×10−5.

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0.700 M solution of ammonia, NH3, to prepare a buffer
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has a pH of 8.81? Kb for ammonia is 1.8×10−5.

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