One litre of 0.15 M Na_2SO_3 aqueous soluti | vedclass

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(D) The balanced redox reaction in acidic medium is: $3SO_3^{2-} + Cr_2O_7^{2-} + 8H^+ \rightarrow 3SO_4^{2-} + 2Cr^{3+} + 4H_2O$. Moles of $Na_2SO_3$

Vedclass · Accepted Answer

$1/30$

Vedclass · Answer

(D) The balanced redox reaction in acidic medium is: $3SO_3^{2-} + Cr_2O_7^{2-} + 8H^+ ightarrow 3SO_4^{2-} + 2Cr^{3+} + 4H_2O$. Moles of $Na_2SO_3$ = $M 	imes V(L) = 0.15 	imes 1 = 0.15 \ mol$. Moles of $K_2Cr_2O_7$ = $M 	imes V(L) = 0.2 	imes 0.5 = 0.1 \ mol$. According to the stoichiometry,$3 \ mol$ of $SO_3^{2-}$ reacts with $1 \ mol$ of $Cr_2O_7^{2-}$. Therefore,$0.15 \ mol$ of $SO_3^{2-}$ will react with $0.15 / 3 = 0.05 \ mol$ of $Cr_2O_7^{2-}$. Remaining moles of $K_2Cr_2O_7$ = $0.1 - 0.05 = 0.05 \ mol$. Total volume of the solution = $1 \ L + 0.5 \ L = 1.5 \ L$. Concentration of unreacted $K_2Cr_2O_7$ = $0.05 \ mol / 1.5 \ L = 0.05 / 1.5 = 1/30 \ M$.

One litre of $0.15 \ M \ Na_2SO_3$ aqueous solution is mixed with $500 \ mL$ of $0.2 \ M \ K_2Cr_2O_7$ aqueous solution in acid medium. What is the concentration (in $mol \ L^{-1}$) of unreacted $K_2Cr_2O_7$ in the resultant solution?

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