Mg can reduce `NO_(3)^(-)" to " NH_(3)` in basic medium.
`NO_(3)^(-) + Mg (s) + H_(2)O to Mg (OH)_(2)(s) + OH^(-) (aq) + NH_(3)(g)`
A `25.0` ml sample of `NO_(3)^(-)` solution was treated with Mg. The `NH_(3)(g)` was passed into 50 ml of `0.15` N HCl . The excess of HCl required `32.10` ml of `0.10` g NaOH for neutralization .
What was the molarity of `NO_(3)^(-)` ions in the original sample ?

Explanation

Meq . of `NH_(3)` formed = Meq. of HCl used
`= 50 xx 0.15 - 32.10 xx 0.10`
` = 4.29`
Here, n-factor of `NH_(3)` is 1 (acid - base reaction)
For redox change,

(n- factor = 8)
` :. ` Meq. of `NH_(3)` for n- factor `8 = 8 xx 4.29`
` :. ` Normality of `NO_(3)^(-) = (34.32)/25 = 1.37`
Molarity of `NO_(3)^(-) = (1.37)/8 = 0.176`