How many grams of dry nh4cl need to be added to 2.10l of a 0.700m solution of ammonia, nh3, to prepare a buffer solution that has a ph of 8.59? kb for ammonia is 1.8×10−5.

I got 9.25527 for the pKa 
then for the Henderson-Hasselbalch equation: 
9.00=9.25527+log(.600/acid) 
10^(-.25527)=(.600/acid)correct to this line. Then 
0.5556 = 0.6/acid and 
acid = 0.6/0.5556 = 1.08 which is the reciprocal of your number) 

.5556/.600M=acid 
acid=.925925...M 
(.925925)*2.10L= Molarity

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This is a solution, hope i could