[rtkt01]

 How many mL of 0.40 M HCl must be added to 900.0 ml of 0.25 M aqueous NH3 solution to produce a buffer with a pH = 9.00 ?

 NH3+ HCl -> NH4Cl

 This is a question, and What I've done so far is shown on the below.

 pH = PKa + log [ base]/[acid]

 Ka for NH4+= 5.56 x 10^(-10)
 pKa for NH4+= 9.26

 9 = 9.26 + log (nb/na)
 
 na= nHCl
 nb= nNH3start-na
 nNH3start = 0.25mol/L x 0.9L = 0.225 moles

 -0.26 = log ((0.225-na) / na )
 10^(-0.26) = (0.225/na ) - 1
   na = 0.1452
 
   0.1452 moles = 0.4 mol/L x X mL
   
    X = 363mL

 Is 363ml a correct answer?
 
   

[Borek]

Looks OK.