#HC(=O)OH + H_2O rightleftharpoons H_3O^+ + HC(=O)O^-#
#K_a=([H_3O^+][HCO_2^-])/[[HC(=O)OH]]#
#=# #x_1^2/(0.200-x_1)# #=# #1.77xx10^-4#.
Of course I had to look up #K_a" formic acid"# on the interwebz.
If, #x_1 <0.200#, then #(0.200-x_1)~=0.200#
Thus #x_1^2# #=# #1.77xx10^-4xx(0.200)#
#x_1# #=# #sqrt(1.77xx10^-4xx(0.200))##=# #5.95xx10^-3#
#x_2# #=# #sqrt(1.77xx10^-4xx(0.200-5.95xx10^-3))##=# #5.86xx10^-3#
#x_3# #=# #sqrt(1.77xx10^-4xx(0.200-5.86xx10^-3))##=# #5.86xx10^-3#
So successive approximations, leads us to #[H_3O^+]=5.86xx10^-3*mol*L^-1#.
#pH=-log_10[H_3O^+]=-log_(10){5.86xx10^-3}# #=# #2.23#
As carboxylic acids go, this is a fairly low #pH#. Why should formic acid be more acidic than say propionic acid?
