Answer
See answer below.
Work Step by Step
a) $Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2\ H_2O(l)$
b)$Ba^{2+}(aq)+ 2\ OH^-(aq) +2\ H_3O^+(aq)+
SO_4^{2-}(aq)\rightarrow BaSO_4(s)+4\ H_2O(l)$
c) Since there are no ions as products, diagram A is the one representing the change in electrical conductivity.
