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?