How many grams of dry NH4Cl need to be added to 1.70 L of a 0.100 mol L−1 solution of ammonia, NH3, to prepare a buffer solution?
How many grams of dry NH4Cl need to be added to 1.70 L of a 0.100 mol L−1 solution of ammonia, NH3, to prepare a buffer solution that has a pH of 9.00? Kb for ammonia is 1.8×10−5.
I get 17g but it says Im wrong.
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Kb = [NH4+][OH-] / [NH3]
pH = 9.0 ; pOH = 5.0 ; [OH-] = 10^-5
[NH4+](10^-5)/ 0.10 = 1.8 x 10^-5
[NH4+] = 1.8 x 10^-6/10^-5
[NH4+] = 0.18
0.18 moles NH4+/ liter x 1.7 liters = 0.306 moles NH4+
0.306 moles NH4+ x 1NH4Cl/1mole NH4+ x 53.5 grams NH4Cl/1mole NH4Cl = 16.4 grams NH4Cl (three significant figures)