The concentration of iodide ions in a solution made by mixing 0.193 m barium iodide and 0.250 m sodium iodide 0.443 m 0.636 m 0.0643 m 0.579 m 0.193 m

The iodide  ion  concentration in the  mixture is=iodide concentration in dissociation  of  barium  iodide  plus  iodide concentration in  dissociation  of sodium iodide

The  ionic  equation  for  dissociation  of  barium  iodide  is
BaI2 ---> Ba2+   +  2 I-
Since  the  ratio  of  Bal2  to l- is 1:2  hence concentration  of iodide  ions  is  2  x0.193 =0.386M

the  ionic equation  for dissociation  of sodium  iodide  
Nal--->Na+  +  l-
since the ratio of  Nal  to  l-  is 1:1 the concentration  of iodide  ion  is o.250M

The  concentration of  iodide  in  the mixture  is  therefore  0.386M +0.250M=0.636M

Show and image of the mixtures or write them

The answer is not c