Question

How many grams ammonium chloride and how many milliliters 3.0 M sodium hydroxide should be added to 200 ml of water and diluted to 500 ml to prepare a buffer of 9.50 with a salt concentration of 0.1 M?
Ka=9.24

Answer 1

I THINK this problem is asking for the amount NH4Cl to add to 3M NaOH so that YOU WILL END UP WITH 0.1M NH4^+ in 500 mL. That will be 0.05 mol NH4Cl in 500 mL.
......NH4^+ + OH^- ==> NH3 + H2O
I.....x........0........0
E...0.05.......0........y

9.50 = 9.26 log (base)/(acid)
1.82 = x/0.05
x = base = OH^- = 0.091
Therefore, if you want the final NH4^+ to base 0.05 mols (0.1M) you must start with 0.05+0.091 = 0.141 mols NH4Cl (7.54g) and add 0.091 mol NaOH.
Since M = mols/L then L = mols/M = 0.091/3M = 0.03033 or 33.33 mL 3M NaOH.

I would suggest you convert g NH4Cl to mols, convert 30.33 mL 3M NaOH to mols, set up an ICE chart and check to see if that will produce 500 mL of solution with pH = 9.50 and a final NH4Cl of 0.1M.