A solution is 0.09 M in HCl and 0.09 M in CHCl2COOH. The pH of this solution is 1. Calculate Ka for CHCl2COOH.

Dear student ,

The dissociation of both acids are  ;   
                              CHCl₂COOH  ⇌ CHCl₂COO^-  + H⁺
Initially  ;                0.09 M                       0                0
At equlibrium ;     0.09 - 0.09x                      0.09  x        0.09 x 
             and  ;          HCl  $\stackrel{}{\to }$  H⁺  +     Cl^-
                                 -            0.09 M    0.09 M
    Now the Total [ H⁺]  =  0.09 ( 1 + x) 
And as per the given condition PH = 1 means [ H⁺] = 10^-1 M
   Then 0.09 ( 1+ x) = 0.1 then x = 0.1 or 10^-1 
    Now K_a = C x²  = 0.09 x 10^-2 m/l
   Regards