How many moles of NH3CI must be added to 1.5 L of 0.2 M solution of NH3 to form a buffer whose PH is 9.00 (Kb = 1.8 x 10^-3) ?

V = 1,5 L

PH = 9.00

NH3 = 0,2 M

Kb = 1.8 x 10^-3

pOH = pKb + log (Csalt/Cbase)

14 - pH = -log Kb + log (Csalt/Cbase)

14 - 9 = -log (1.8x10^-3) + log (Csalt/0.2)

52.745 + log (Csalt/0.2)

log (Csalt/0.2) = 2.255

Csalt/0.2 = 5.555x10^-3

Csalt = 0.2 x 5.555x10^-3 

Csalt = 1.111 x 10^-3 M

NH3 = V . Csalt 

NH3 = 1,5 . 1.111x10^-3 M

So, How many moles of NH3CI must be added to 1,5 L of 0.2 M is 1.667×10^-3 M.