Final answer:
Out of the given options, option (d) Pb(C2H3O2)2 + NH4Cl will form a precipitate. When lead (II) reacts with a chloride, as in the reaction of Pb(NO3)2 with KI forming PbI2, it results in the formation of an insoluble salt, which in this case will be lead (II) chloride (PbCl2).
Step-by-step explanation:
When combining 0.10 M aqueous solutions to determine which pair forms a precipitate, one should refer to the solubility rules. In this case, option (a) NH₄NO₃ + Pb(NO₃)₂ does not form a precipitate because all nitrates are soluble. Option (b) NaNO₃ + KCl also doesn't lead to a precipitate because nitrates and chlorides (except those of Ag+, Pb2+, and Hg₂2+) are soluble.
Option (c) HNO₃ + Ca(OH)₂ does not form a precipitate as nitrates are soluble and Ca(OH)₂ is soluble in the presence of strong acids like HNO₃. Lastly, option (d) Pb(C₂H3O₂)₂ + NH₄Cl will result in a precipitate. According to the solutions provided, when Pb(NO₃)₂ reacts with KI, a precipitate of PbI₂ forms. By analogy, replacing KI with NH₄Cl in option (d), where the lead (II) compound reacts with a chloride, a similar reaction will occur, forming PbCl₂ as a precipitate.