Question

How many moles of NH4Cl must be added to 3.0 L of aqueous 0.10 M NH3 to form a buffer whose pH is 9.00? Kb of NH3 is 1.8x10 (Assume addition of NH4Cl to the solution does not alter its volume). 0.36 O 0.54 0.18 O 0.42 0.11

Answer 1

Henderson. Hamelbalch eduation can be willen an follown! pot = pky + log 1.I BHOJ ..... 0 [B] " the given solution contain a base (NH) and it's conjugate acid (MHP) So abore equation can be applied for this buffer solution. here given [B] = [NH3] = 0.10 m 1 kb (NH3) = 1.8x105 wipkb = -log kb = -log("8X105) = 4.74 here, pH = 9.00 ke Клон pH +рон - 14 W, POH = 14-pH = 14-9.00 = 5 Now putting the above values in equation we get 5 = 4.74 +log <8H67 A log Botol = 5-4-14 = 0.26 or [BH] 0.26 = 1.819 0.10 0.10 0.10 = 1026 = leo a, [BH] = 0.10 x 1.819 .. [BH0] = [ WW7 = 0.1819 M me know molarity (M) = No of moles of solute volume of the solution in like " No of moles of NH₃ C = molarity x volume of the solution in like = 0.1819 M X 32 = 0.1819 moleele X:34 ('M: moles/L) = 0.5457 males No of moles of NH d = o.sast moles, ". The correct option is = option & de o.sa Canawer)