Final answer:
The concentration of nitrate ions ([NO3-]) in a 0.25 M solution of Sn(NO3)4 is 1.00 M, as each unit of Sn(NO3)4 provides four NO3- ions upon complete dissociation in water.
Step-by-step explanation:
To determine the concentration of nitrate ions ([NO3-]) in a solution of Sn(NO3)4, you must consider that each formula unit of Sn(NO3)4 will dissociate completely in water to form one tin ion (Sn4+) and four nitrate ions (NO3-). Therefore, if the initial concentration of Sn(NO3)4 is 0.25 M, the concentration of NO3- ions will be four times that amount because each unit of Sn(NO3)4 provides four NO3- ions. Using this stoichiometry, we can calculate the concentration of NO3- ions as follows:
[NO3-] = 4 × (Initial concentration of Sn(NO3)4) = 4 × 0.25 M = 1.00 M.
Thus, the concentration of nitrate ions in a 0.25 M solution of Sn(NO3)4 is 1.00 M.