Significant quantities of #H_2PO_4^(-)# and a little #HPO_4^(2-)# would be present at #"equilibrium"#......it is NOT a triacid.....For #"option B"#, there would be SOME #HSO_4^(-)# present at equilibrium, even tho' #"bisulfate"# is moderately strong. Whereas #"acetic acid"# is weak..........
On the other hand, #K_2SO_4# would give 3 equiv of ions......
#K_2SO_4(s) stackrel(H_2O)rarr 2K^(+) + SO_4^(2-)#
#A# is thus the best answer.
As for the second problem, your approach is correct......
#2AgNO_3(aq) +CaI_2(aq) rarr 2AgI(s)darr + Ca(NO_3)_2(aq)#
Silver iodide is as soluble as a brick, and precipitates from solution as a bright yellow powder. Calcium ion remains in solution as a spectator.........
#"Concentration of calcium ion"=("Moles of Ca"^(2+))/("Volume of solution")#
#=(1.25*Lxx1.00*mol*L^-1)/(2.00*L+1.25*L)~=0.4*mol*L^-1#
