In the following reaction
`overset(ɵ)(O)H+S_(2)O_(3)^(2-)+Br_(2)toSO_(4)^(2-)+Br^(ɵ)`
Starting with 0.15 " mol of "`Br^(2)` 0.01 " mol of "`S_(2)O_(3)^(2-)` and 0.4 " mol of "`overset(ɵ)(O)H` ions. How many moles of `overset(ɵ)(O)H` ions are left in the solution after the reaction is complete.

Explanation

Balance the equation in basic medium
`10H+S_(2)O_(3)^(2-)to2SO_(4)^(2-)+5H_(2)O+8e^(-)`
`underline(2e^(-)+Br_(2)to2Br^(ɵ)]xx4)`
`underline(4Br_(2)+S_(2)O_(3)^(2-)+10overset(ɵ)(O)Hto2SO_(4)^(2-)+8Br^(ɵ)+5H_(2)O)`
`{:[(mol es),(befo re),(reaction)](0.15)(0.01)(0.4)(0)(0)(0)`
`{:[(mol es),(reacted),(and),(fo rmed)]:}(0.01xx4)(0.01)(0.01xx10)(0.01xx2)(0.01xx8)(0.01xx5)`
`{:[(mol es),(l eft)]:}[0.15-0.04]0(0.4-0.1)(0.02)(0.08)(0.05)(=0.11)(=0.3)`
So, molar ratio of combination is:
`Br_(2):S_(2)O_(3)^(2-):overset(ɵ)(O)H::4:1:10`
`[overset(ɵ)(O)H]` left after reaction `=0.3` mol