| (1) \[SeO_{3}^{2-}+BrO_{3}^{-}+{{H}^{+}}\to SeO_{4}^{2-}+B{{r}_{2}}+{{H}_{2}}O\] |
| (2) \[BrO_{3}^{-}+AsO_{2}^{-}+{{H}_{2}}O\to B{{r}^{-}}+AsO_{4}^{3-}+{{H}^{+}}\] |
A) \[1.6\times {{10}^{-3}}\]
B) \[1.25\]
C) \[2.5\]
D) None of these
Correct Answer: C
Solution :
[a] \[\overset{+4}{\mathop{SeO_{3}^{2-}}}\,+\overset{+5}{\mathop{BrO_{3}^{-}}}\,+{{H}^{+}}\to \overset{+6}{\mathop{SeO_{4}^{2-}}}\,+\overset{0}{\mathop{B{{r}_{2}}}}\,+{{H}_{2}}O\] [b] \[\overset{+5}{\mathop{BrO_{3}^{-}}}\,+\overset{+3}{\mathop{AsO_{2}^{-}}}\,+{{H}_{2}}O\to \overset{-1}{\mathop{Br}}\,+\overset{+5}{\mathop{AsO_{4}^{3-}}}\,+{{H}^{+}}\] In reaction (2) gm. eq. of \[Br{{O}_{3}}=\]gm. eq. of \[AsO_{2}^{-}\]. \[{{n}_{BrO_{3}^{-}}}\times 6={{n}_{AsO_{2}^{-}}}\times 2=\frac{12.5}{1000}\times \frac{1}{25}\times 2={{10}^{-3}}\]\[{{n}_{BrO_{3}^{-}}}=\frac{{{10}^{-3}}}{6}\] In reaction (1) moles of \[BrO_{3}^{-}\]-consumed \[=\frac{70}{1000}\times \frac{1}{60}-\frac{{{10}^{-3}}}{6}={{10}^{-3}}\] gm eq. of \[Se{{O}_{3}}^{2-}=\] gm. eq. of \[Br{{O}_{3}}^{-}\] \[{{n}_{SeO_{3}^{2-}}}\times 2={{10}^{-3}}\times 5;\,\,{{n}_{SeO_{3}^{2-}}}=2.5\times {{10}^{-3}}\]
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