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Solubility Product

0.96 gm of HI...

Question

0.96 gm of HI were heated to attain equation $2 H I ⇌ H_{2} + I_{2}$ . The reaction mixture on titration requires 15.7 mL, $\frac{N}{200}$ hypo solution. Calculate % dissociation of HI?

22.5%

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23.67%

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4.1 %

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None of thse

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Solution

The correct option is C 4.1 %
$H I ⇌ \frac{1}{2} H_{2} P \frac{1}{2} I_{2} \frac{0.36}{128} m o l e s$
$0.0075 (1 - α) \frac{α}{2} \frac{\frac{α}{2}}{0.0075} \Rightarrow 0.0075$
$H y p o \to N a_{2} S_{2} O_{3}$
$2 N a_{2} S_{2} O_{3} + I_{3} \to N a_{2} S_{4} O_{6} + 2 N a I$
$(\frac{15.7}{200}) \times 10^{- 3} = \frac{α}{2} = \frac{0.0075}{2}$
$\Rightarrow α = \frac{4 \times 15.7}{200 \times 0.0075} = \frac{4}{3} \times \frac{4}{2} \times 15.7 \times 10^{- 3}$
$= \frac{8}{3} \times 15.7 \times 10^{- 3}$
$\Rightarrow 0.041$
$% d i s s = 4.1 %$

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0.96 gm of HI were heated to attain equation 2HI⇌H2+I2. The reaction mixture on titration requires 15.7 mL, N200 hypo solution. Calculate % dissociation of HI?

The correct option is C 4.1 %HI⇌12H2P12I20.36128moles0.0075(1−α)α2α20.0075⇒0.0075Hypo→Na2S2O32Na2S2O3+I3→Na2S4O6+2NaI(15.7200)×10−3=α2=0.00752⇒α=4×15.7200×0.0075=43×42×15.7×10−3=83×15.7×10−3⇒0.041%diss=4.1%

0.96 gm of HI were heated to attain equation $2 H I ⇌ H_{2} + I_{2}$ . The reaction mixture on titration requires 15.7 mL, $\frac{N}{200}$ hypo solution. Calculate % dissociation of HI?