We are given that there are 2 moles of NO3 from the 0.225 M Sr(NO3)2 solution. Therefore we simply use stoichiometry to solve this problem.
Concentration NO3 = 0.225 M Sr(NO3)2 * (2 moles NO3 / 1 mole Sr(NO3)2)
Concentration NO3 = 0.45 M
Answer : The concentration of nitrate ions is 0.450 m.
Solution : Given,
Molarity of [tex]Sr(NO_3)_2[/tex] = 0.225 m
The balanced ionic equation is,
[tex]Sr(NO_3)_2\rightarrow Sr^{+2}+2NO^-_3[/tex]
From the reaction, we conclude that the 1 mole of [tex]Sr(NO_3)_2[/tex] produces 2 moles of [tex]NO^-_3[/tex] ions. That means the concentration of nitrate ions is twice the value of [tex]Sr(NO_3)_2[/tex].
then, 0.225 m of [tex]Sr(NO_3)_2[/tex] gives 2 × 0.225 of [tex]NO^-_3[/tex] ions
Now the concentration of [tex]NO^-_3[/tex] ions is equal to 0.450 m.
