Question

How many grams of dry NH4Cl need to be added to 1.90 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.74? Kb for ammonia is 1.8×10^-5.

Answer 1

as a poh p klet du - L mixture of NH3 and NHAd solution will acts basic buffer. According to Henderson's equation pot of basic buffer is - log [NHA d [NH3 Nowy given data Kle for NH3 = 1.8 X10-5 solution has pH = 8.79 We know that PH + poH = 14 pot = 14-879 poh = 5.26 190 L of 0.800 m NH3 1.90 KX 0.800 males Ammonia 1.52 males Now , putting the values in equation Рон pkle + log InHad] INH₂ 5:26= -log klet log NHAU 5.26 = :- log (1.8x10-5) + log whal 1019 4.745 + log NHAI 1.52 0.515 = long Nhael 3.273 NH 4 cl 1.52 : NHa el 9.975 moles Molar mass of whad = 4+4+35.5 = 53.5 gm NHA I needed = (4.975 X 53.5) gm dry NH4 l needed =1266•16 gm org 1.52 1.52 en 5:26 ཆུ\ ༡ 1.52 09