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Stoichiometry

100 ml of N...

Question

100 ml of $N a_{2} S_{2} O_{3}$ solution is divided into two equal parts A and B. A part requires 12.5 ml of $0.2 M l_{2}$ solution (acidic medium) and part B is diluted x times and 50 ml of diluted solution requires 5 ml of $0.8 M l_{2}$ solution in basic medium. What is value of x?

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Solution

$+ 4 S_{2} O_{3}^{2 -} + I_{2} ⟶ + 12 2 S O_{4}^{2 -} + 2 I^{-}$
$50 m l 12.5 m l$
$0.2 M$
$n_{f} = 12 - 4$ $n_{f} = 2$
Equivalent of $S_{2} O_{3}^{2 -} =$ Equivalent of $I_{2}$
Equivalent $=$ moles $\times n_{f}$
Moles of $S_{2} O_{3}^{2 -} \times 8 = 2 \times \frac{12.5}{1000} \times 0.2$
mili moes of $S_{2} O_{3}^{2 -} = 0.825$
2nd part:
$S_{2} O_{3}^{2 -} + I_{2} ⟶ 2 S O_{4}^{2 -} + 2 I$
$50 m l 5 m l$
$0.8 M$
Equivalent of $S_{2} O_{3}^{2 -} =$ Equivalent of $I_{2}$
Moles $\times 8 = \frac{5}{1000} \times 0.8 \times 2$
mili moles = 1
Initial concentration $\times$ x=Final concentration.
Concentration $= \frac{m o l e s}{V o l u m e}$
$\frac{0.825}{50} \times x = \frac{1}{50} \Rightarrow x = 1.21$

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