(NH4)+ from NH4Cl will react with (OH)- to form NH3 and H2O
So, [tex]NH_4^+ + OH^- -\ \textgreater \ NH_3 + H_2O[/tex]
concentration of NH4+ = 0.12
and concentration of OH- = 0.03
Concentration of Remaining (NH4)+ = 0.12 - 0.03 = 0.09 mol /L
Concentration of Generated NH3 = 0.03 mol / L
The resulting solution is " NH4+ / NH3 Buffer System " { (NH4)+ is acid ; NH3 is Base ( " Salt " ) }
For a Buffer System, the formula for pH is :
[tex]pH = pKa + Log\frac{ [ Acid ] }{ [ Salt ]} [/tex]
[tex]pH = 9.25 + Log\frac{ [ NH_4^+ ] }{ [ NH_3 ]} [/tex]
[tex]pH = 9.25 + Log\frac{ [ 0.09 ] }{ [ 0.03 ]} [/tex]
= 9.25 + Log 3
= 9.25 + 0.4771
= 9.73 (after rounding off)
