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Local Maxima

Given an equa...

Question

Given an equation:
$C r O_{4}^{2 -} + S_{2} O_{3}^{2 -} + ⊝ O H \to [C r (O H)_{4}]^{-} + S O_{4}^{2 -}$
What volume of $0.2 M$ $N a_{2} C r O_{4}$ solution is required to just react with $30$ mL of $0.2 M$ $N a_{2} S_{2} O_{3}$ solution?

40

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80

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20

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60

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Solution

The correct option is C $80$ mL
$3 e^{-} + C r O_{4}^{2 -} x - 8 = - 2 x = 6 \to [C r (O H)_{4}]^{⊝} x - 4 = - 1 x = 3$
Equivalent weight of $C r O_{4}^{2 -} = \frac{Mw}{3}$

$S_{2} O_{3}^{2 -} 2 x - 6 = - 2 2 x = 4 \to 2 S O_{4}^{2 -} + 8 e^{-} 2 x - 6 = - 4 2 x = 12$
Equivalent weight of $S_{2} O_{3}^{2 -} = \frac{Mw}{8}$

$N_{1} V_{1} (C r O_{4}^{2 -}) = N_{2} V_{2} (S_{2} O_{3}^{2 -})$

mEq. of $C r O_{4}^{2 -} =$ 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

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