what is the concentration of nitrate ions in a 0.125 M Sr(NO3)2 solution?
The concentration of nitrate ions in a 0.125 M Sr(NO3)2 solution can be calculated by taking into account the stoichiometry of the compound.
Sr(NO3)2 is a salt that dissociates in water to give Sr2+ ions and two nitrate (NO3-) ions. Therefore, for every one mole of Sr(NO3)2 that dissociates, two moles of nitrate ions are produced.
Since the concentration of Sr(NO3)2 is 0.125 M, the concentration of nitrate ions will be twice this value:
Concentration of nitrate ions = 2 * 0.125 M = 0.25 M
Therefore, the concentration of nitrate ions in a 0.125 M Sr(NO3)2 solution is 0.25 M.