[H+] ion concentration of weak acid = (Ka*C)^1/2
                                                  Â
= (6.6*10^-4 *0.5)^1/2
                                                   Â
= 1.82*10^-2
Now we have to use M1V1 = M2V2
    M1 = 1.82*10^-2 , V1 = 50 ml , M2 = 1.0
M , V2 = ?
   1.82*10^-2*50.0 = 1.0*V2
     V2 = 0.91 ml
volume of base added is 0.91 ml
