Question

Sebanyak 100 mL larutan NH3 0,8 M dicampurkan dengan 50 mL larutan HCl 0,2 M . Apabila Kb NH3 = pH campuran adalah ?

Answer 1

Mol NH3 = 0,8M x 100ml = 80mmol
mol HCl = 0,2M x 50ml = 10mmol

   NH3 + HCl ==> NH4Cl
m  80......10.............-
r   -10.....-10..........+10
s   70......-...............10

basa lemah + asam kuat ==> sisa basa lemah ==> larutan buffer

[OH-] = kb x sisa basa/garam
[OH-] =  10^-4 x 70/10 = 7x 10^-4
pOH = -log [OH-] = -log 7x10^-4 = 4-log 7
pH = 14-(4-log 7) = 10+log 7

Answer 2

NH3 = basa
HCL = asam (maka bereaksi)

     NH3 + HCL --> NH4CL
m   80       10     
r     10       10          10
_____________ -________+
s     70       -             10        (buffer)

maka :[OH-] = kb nNH3
                       nNH4+

                  = 10^-4 x 70
                           10
pOH = 4-log7
pH    = 10+log 7