Question

If in the given reaction, $3{I_2} + O{H^ - } \to IO_3^ - + 5{I^ - }$2 moles of iodine are taken, then the ratio of iodate and iodide formed in the alkaline medium is
(1).1: 5
(2).3: 5
(3).5: 1
(4).5: 3

Answer 1

In order to approach the solution of the given question one should have prior knowledge about balancing the equation and also remember first to balance the given chemical reaction then find the moles of iodate and iodide.

Complete answer:
According to the given information we have a chemical reaction $3{I_2} + O{H^ - } \to IO_3^ - + 5{I^ - }$ where 2 moles of iodine are taken
Before finding the ratio of iodide and iodate we have to balance the reaction
So, the balanced equation in an alkaline medium will be; $3{I_2} + 6O{H^ - } \to IO_3^ - + 5{I^ - } + 3{H_2}O$
Now let’s calculate the number of moles of iodide and Iodate in the given reaction in an alkaline medium
After considering the balanced chemical reaction it shows that 3 moles of ${I_2}$ react and give 1 mole of iodate ($IO_3^ -$)
So, 2 moles of ${I_2}$ will react to give $\dfrac{2}{3} = 0.66{\text{ }}moles$of iodate ($IO_3^ -$)
Thus, the moles of iodate ($IO_3^ -$) = 0.66 moles
Now, again considering the balanced equation we can say that
As 3 moles of ${I_2}$ reacts and give 5 moles of iodide (${I^ - }$)
So, 2 moles of ${I_2}$ will react to give $\dfrac{2}{3} \times 5 = 0.33{\text{ }}moles$ of iodide (${I^ - }$)
Thus, the moles of iodide (${I^ - }$) = 0.33 moles
Now the ratio of iodate and iodide will be $0.66:0.33 = 1:5$
Therefore, ratio of iodate and iodide formed in the alkaline medium is 1: 5

Hence, option (1) is the correct option.

Note:
In the above question we came across the term alkaline medium which can be defined as the medium which has pH greater than 7.0 which tells that the solution is basic in nature. The alkaline solution can be formed by dissolving the base into water.