Question

Oxidation reduction

a) The concentration of SO32- in a solution is determined by titrating it with a 0.1147 M permanganatesolution. The balanced net ionic equation for the reaction is: 2MnO4-(aq) + 5SO32-(aq) + 6H3O+(aq) ebab9d75-6e9a-4c80-9bcd-b491b668f097.gif2Mn2+(aq) +5SO42-(aq) + 9H2O(l)

In one experiment, 19.18 mL of the0.1147 M permanganate solution is required to react completely with20.00 mL of the SO32- solution. Calculate the

concentration of theSO32- solution. _______ M

Answer 1

Answer : The concentration of
`SO_3^(2-)` solution is 0.002398 M

Explanation :

The given balanced net ionic equation for the reaction is:

`2MnO_4^-(aq)+5SO_3^(2-)(aq)+6H_3O^+(aq)\rightarrow 2Mn^(2+)(aq)+5SO_4^(2-)(aq)+9H_2P(l)`

First we have to calculate the moles of
`MnO_4^-`

`\text{Moles of }MnO_4^-=\text{Concentration of }MnO_4^-* \text{Volume of solution}`

`\text{Moles of }MnO_4^-=0.1147M* 0.01918L`

Now we have to calculate the moles of
`SO_3^(2-)`

From the balanced chemical reaction we conclude that,

As, 2 moles of
`MnO_4^-` react with 5 moles of
`SO_3^(2-)`

So, 0.01918 moles of
`MnO_4^-` react with
`(5)/(2)* 0.01918=0.04795` moles of
`SO_3^(2-)`

Now we have to calculate the concentration of
`SO_3^(2-)`

`\text{Concentration of }SO_3^(2-)=\frac{\text{Moles of }SO_3^(2-)}{\text{Volume of solution}}`

`\text{Concentration of }SO_3^(2-)=(0.04795mol)/(20.00mL)=0.002398M`

Therefore, the concentration of
`SO_3^(2-)` solution is 0.002398 M