In the reaction given Mg(s)+NO_(3)^(ɵ)+H_(2)OtoMg(OH)_(2)(s)+overset(ɵ)(O)HO(aq.)+NH_(3)(g) 20 " mL of " sample of NO_(3)^(ɵ) solution is treated with Mg. The NH_(3)^(g) was passed into 50 " mL of " 0.1 M HCl. The excess HCl requried 30 " mL of " 0.1 M KOH for its neutralisation calculate the molarity of NO_(3)^(ɵ) ions in the original sample?

In the reaction given Mg(s)+NO_(3)^(ɵ)+H_(2)OtoMg(OH)_(2)(s)+overset

1.	In the reaction given Mg(s)+NO_(3)^(ɵ)+H_(2)OtoMg(OH)_(2)(s)+overset(ɵ)(O)HO(aq.)+NH_(3)(g) 20 " mL of " sample of NO_(3)^(ɵ) solution is treated with Mg. The NH_(3)^(g) was passed into 50 " mL of " 0.1 M HCl. The excess HCl requried 30 " mL of " 0.1 M KOH for its neutralisation calculate the molarity of NO_(3)^(ɵ) ions in the original sample?
Answer» Solution :Balance the EQUATION in basic medium `cancel(8e^(-))+6H_(2)O+NO_(3)^(ɵ)toNH_(3)+9overset(ɵ)(O)H` `underline(2(ɵ)(O)H+"Mg"to overset(+2)(M)g(OH)_(2)+cancel(2e^(-))]xx4)` `underline(6H_(2)O+NO_(3)^(ɵ)+4"Mg"toNH_(3)+4Mg+4Mg(OH)_(2)+overset(ɵ)(O)H)` `thereforem" Eq of "NH_(3) "formed"=m" Eq of "HCl" used for "NH_(3)` `=(50xx0.1xx1)-(30xx0.1xx1)` `=2 mEq` Thus, m" Eq of "`NH_(3)` for valence factor of `8=8xx2=16` Also, mEw of `NO_(3)^(ɵ)=m" Eq of "NH_(3)` `=8xx2=16mEq` `N_(NO_(3)^(ɵ))xxV_(mL)=16` `therefore N_(NO_(3)^(ɵ))xx20=16` `N_(NO_(3)^(ɵ))=(16)/(20)=0.8M` `M_(NO_(3)^(ɵ))=(0.8)M` `M_(NO_(3)^(ɵ))=(0.8)/(8)=0.1M`