The balanced chemical equation representing the reaction of ammonia with hypobromite is:
[tex]2NH_{3} (g)+3OBr^{-}(aq)-->N_{2}(g)+3Br^{-}(aq)+3H_{2}O(l)[/tex]
The mass of [tex]NH_{3}[/tex]=1.69mg
Moles of [tex]NH_{3}[/tex]=[tex]1.69mg*\frac{1g}{1000mg}* \frac{1mol}{17.03g} =9.92*10^{-5}molNH_{3}[/tex]
Calculating the moles hypobromite that would react with [tex]9.92*10^{-5}molNH_{3}[/tex]:
[tex]9.92*10^{-5}molNH_{3}*\frac{3molOBr^{-} }{2molNH_{3}}= 1.488*10^{-4}mol OBr^{-}[/tex]
Volume of hypobromite required for titration = 1.00 mL
Molarity of hypobromite solution = [tex]\frac{1.488*10^{-4}molOBr^{-}}{1.00mL} *\frac{1000mL}{1L} =0.1488mol/L[/tex]
