How do I do this question step by step?Screenshot 2024-04-17 213027.png
First write the reaction of HCl (acid) and ammonia (base)
NH3 (aq) + HCl(aq) NH4+ (aq) + Cl-(aq)
Now you are mixing volume of two solutions so their concentrations are changing. Work with moles and millimoles to know how much of the reactants have reacted and how much of the product is formed.
Remember – molarity = mmol /V in mL = mol / Vol (L)
Means mol = M* V in L
Here volume is not given so let us assume the volume is 1L
Reaction
NH3(aq)+HCl(aq)-> NH4+(aq)+Cl-(aq)
Initial (Mol 0.12 0.10 0 0
Final (Mol)) 0.02 0.00 0.10 0.10
HCl is the limiting reactant and finally solution has NH3 (aq), NH4+ (aq) and Cl-(aq)
Cl-(aq) is a spectator here because it is a weak conjugate base of strong acid HCl.
pH is controlled by NH3 (aq) and NH4+ (aq) .However this is a base and its conjugate acid so it is a buffer and you can use Henderson - Hasselbalch equation to get the answer.
pH = pKa + log ([ NH3 ]/[NH4+ ] )
Since volume is doubled you can calculate new concentrations. New volume is 2L
[ NH3 ] = 0.02 /2 = 0.01 M
[NH4+ ] = 0.10 / 2 = 0.05 M
pH = 9.25 + log ( 0.01/ 0.05)
Or you can simply take the ratio of moles since volume finally is same in the buffer solution for both acid and its conjugate base.
pH = pKa + log (moles NH3 /moles NH4+ )
= 9.25 + log ( 0.02/ 0.10)
log ( 0.01/ 0.05) = log ( 0.02/ 0.10)