Question

What volume of 0.1025 M HCl must be added to 15.64 ml of 0.0956 M NH3 to produce a solution of ph= 9.00 ?

Answer 1

we have formula pOH = pkb + log [NH4Cl]/[NH3]   whre pkb of NH3 is 4.75

given pH = 9 , hence pOH = 14-pH = 14-9 = 5

hence 5 = 4.75 + log [NH4Cl]/[NH3]

[NH4Cl] = 1.78 [NH3]

NH4Cl moles = 1.78 x NH3 moles    ( Moles = Molarity x vol , vol term from both sides cancells out)

NH3 + HCl ---> NH4Cl   is the equation

Initial NH3 moles = M x V ( in L) = 0.0956 x 15.64/1000 = 0.001495

Let HCl moles added = m

then after reactiong with HCl moles of NH3 = 0.001495-m , NH4Cl moles = m

we have relation   NH4Cl moles = 1.78 x NH3 moles

hence m = 1.78 ( 0.001495-m)

m = 0.00266 - 1.78m

m = 0.0009573 = HCl moles

now we have formula Molarity of HCl = moles /vol

0.1025 = 0.0009573 /vol

vol = 0.00933 L = 9.33 ml = 9.3 ml ( rounded value)