Question

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

Answer 1

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 2

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.