Question

What is the concentration of nitrate ions in a 0.225 msr(no3)2 solution?

Answer 1

Answer : The concentration of nitrate ions is 0.450 m.

Solution : Given,

Molarity of $Sr(NO_3)_2$ = 0.225 m

The balanced ionic equation is,

$Sr(NO_3)_2\rightarrow Sr^{+2}+2NO^-_3$

From the reaction, we conclude that the 1 mole of $Sr(NO_3)_2$ produces 2 moles of $NO^-_3$ ions. That means the concentration of nitrate ions is twice the value of $Sr(NO_3)_2$.

then, 0.225 m of $Sr(NO_3)_2$ gives 2 × 0.225 of $NO^-_3$ ions

Now the concentration of $NO^-_3$ ions is equal to 0.450 m.

Answer 2

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