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 $MnO_4^-$

$\text{Moles of }MnO_4^-=\text{Concentration of }MnO_4^-\times \text{Volume of solution}$

$\text{Moles of }MnO_4^-=0.1147M\times 0.01918L$

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 $\frac{5}{2}\times 0.01918=0.04795$ moles of $SO_3^{2-}$

Now we have to calculate the concentration of $SO_3^{2-}$

$\text{Concentration of }SO_3^{2-}=\frac{\text{Moles of }SO_3^{2-}}{\text{Volume of solution}}$

$\text{Concentration of }SO_3^{2-}=\frac{0.04795mol}{20.00mL}=0.002398M$

Therefore, the concentration of $SO_3^{2-}$ solution is 0.002398 M