Question

How many grams of dry NH4Cl need to be added to 2.00 L of a 0.300...

How many grams of dry NH4Cl need to be added to 2.00 L of a 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.

Homework Answers

Answer #1

NH3(aq) + H2O ⇋ [NH4]+ + OH-
Kb = 1.8×10^-5
Kb = [NH4+][OH-]/[NH3] = 1.8×10^-5
-log both sides
pKb = pOH + log[NH3]/[NH4+]
pKb = -log(1.8×10−5) = 4.74 pOH = 14.0 - 8.64 = 5.36
4.74 = 5.36 + log[NH3]/[NH4+]
log[NH3]/[NH4+] = 4.74 - 5.36 = -0.62
[NH3]/[NH4+] = 0.24
[NH4+] = 0.300/0.24 = 1.25M
Now let me check doing it my way:
Kb = [NH4+][OH-]/[NH3] = 1.8×10-5
pH = 8.64 pOH = 14 - 8.64 = 5.36
[OH-] = 4.365×10-6
1.8×10-5 = [NH4+][4.365×10-6]/[0.300]
[NH4+] = [0.300]×(1.8×10-5)/[4.365×10-6] = 1.2371 M (so we are agreed!)
In 2.00 L we have 2.0×1.2371 = 2.5462 mol
X MWt NH4Cl = 2.5462 × 53.4917 = 136.2005 g NH4Cl

136.2005 grams of NH4Cl needed

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