Question

Given an equation:

$Cr{O}_4^{2-}+S_2{O}_3^{2-}+\stackrel{⊝}{O}H\to \left[Cr\right(OH{)}_4{]}^{-}+S{O}_4^{2-}$

What volume of $0.2M$ $N{a}_{2}Cr{O}_4$ solution is required to just react with $30$ mL of $0.2M$ $N{a}_2{S}_2{O}_3$ solution?

• A. $40$ mL
• B. $80$ mL
• C. $20$ mL
• D. $60$ mL

Answer 1

Answer: (B)

$80$ mL

$3e^{-}+\underset{\underset{x=6}{x-8=-2}}{Cr{O}_4^{2-}}\to \underset{\underset{x=3}{x-4=-1}}{\left[Cr\right(OH{)}_4{]}^{⊝}}$
Equivalent weight of $Cr{O}_4^{2-}=\frac{\text{Mw}}{3}$

$\underset{\underset{2x=4}{2x-6=-2}}{S_2{O}_3^{2-}}\to \underset{\underset{2x=12}{2x-6=-4}}{2S{O}_4^{2-}+8e^{-}}$
Equivalent weight of $S_2{O}_3^{2-}=\frac{\text{Mw}}{8}$

$N_1{V}_1\left(Cr{O}_4^{2-}\right)=N_2{V}_2\left(S_2{O}_3^{2-}\right)$

mEq. of $Cr{O}_4^{2-}=$ mEq. of $S_2{O}_3^{2-}$

$V_1\times 0.2\times 3=30\times 0.2\times 8$

$V_1=\frac{30\times 0.2\times 8}{0.2\times 3}=80$ mL