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
