How many grams of dry NH4Cl need to be added to 2.10 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.76? Kb for ammonia is 1.8×10−5.
Express your answer with the appropriate units.
Use the Henderson-Hasselbalch equation to find out what the NH3
/ NH4+ ratio should be.
pH = pKa + log ([NH3]/[NH4+]) which is equivalent to pH = pKa + log
(moles NH3 / moles NH4+)
Kb NH3 * Ka NH4+ = 1 * 10^-14
Ka NH4+ = 1 * 10^-14 / 1.8 * 10^-5 = 5.6 * 10^-10
pKa = -log (5.6 * 10^-10) = 9.25
pH = pKa + log (moles NH3 / moles NH4+)
8.76 = 9.25 + log (moles NH3 / moles NH4+)
-0.49 = log (moles NH3 / moles NH4+)
10^-0.49 = moles NH3 / moles NH4+ = 0.323
moles NH3 = M NH3 * L NH3 = (0.500)(2.10) = 1.05 moles NH3
0.323 = 1.05 / moles NH4+
moles NH4+ = 1.05 / 0.323 = 3.25
3.25 moles NH4Cl * (53.5 g NH4Cl / 1 mole NH4Cl) = 173.8 g
NH4Cl
How many grams of dry NH4Cl need to be added to 2.10 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solu...
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