Question

Ammonia reacts with hypobromite, obr, by the reaction 2nh3 3obr n n2 3br 3h2o. what is the molarity of a hypobromite solution if 1.00 ml of the obr solution reacts with 1.69 mg of nh3

Answer 1

The balanced chemical equation representing the reaction of ammonia with hypobromite is:

$2NH_{3} (g)+3OBr^{-}(aq)-->N_{2}(g)+3Br^{-}(aq)+3H_{2}O(l)$

The mass of $NH_{3}$=1.69mg

Moles of $NH_{3}$=$1.69mg*\frac{1g}{1000mg}* \frac{1mol}{17.03g} =9.92*10^{-5}molNH_{3}$

Calculating the moles hypobromite that would react with $9.92*10^{-5}molNH_{3}$:

$9.92*10^{-5}molNH_{3}*\frac{3molOBr^{-} }{2molNH_{3}}= 1.488*10^{-4}mol OBr^{-}$

Volume of hypobromite required for titration = 1.00 mL

Molarity of hypobromite solution = $\frac{1.488*10^{-4}molOBr^{-}}{1.00mL} *\frac{1000mL}{1L} =0.1488mol/L$