Sulphurous acid left( H _{2} SO _{3}right) ha | vedclass

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Question

$H _{2} SO _{3}$ [Dibasic acid]

$c=0.588 \,M$

$\Rightarrow \quad pH$ of solution $P$ due to First dissociation only

since $K _{ a },>>

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$H _{2} SO _{3}$ [Dibasic acid]

$c=0.588 \,M$

$\Rightarrow \quad pH$ of solution $P$ due to First dissociation only

since $K _{ a },>> Ka _{2}$

$\Rightarrow$ First dissociation of $H _{2} SO _{3}$

$H _{2} SO _{3}( aq ) ightleftharpoons H ^{\oplus}( aq )+ HSO _{3}^{-}( aq ): ka _{1}=1.7 	imes 10^{-2}$

$t=0$ $\;\;C$

$t$ $\quad\;\;\;\;C - x\;\; \, x \;\;\, x$

$\Rightarrow \quad Ka _{1}=\frac{1.7}{100}=\frac{\left[ H ^{\oplus}ight]\left[ HSO _{3}^{-}ight]}{\left[ H _{2} SO _{3}ight]}$

$\Rightarrow \frac{1.7}{100}=\frac{ x ^{2}}{(0.58- x )}$

$\Rightarrow 1.7 	imes 0.588-1.7 x =100 x ^{2}$

$\Rightarrow \quad 100 x^{2}+1.7 x-1=0$

$\Rightarrow \quad\left[ H ^{\oplus}ight]= x =\frac{-1.7+\sqrt{(1.7)^{2}+4 	imes 100 	imes 1}}{2 	imes 100}=0.09186$

Therefore $pH$ of sol. is $: pH =-\log \left[ H ^{\oplus}ight]$

$\Rightarrow \quad pH =-\log (0.09186)=1.036 \simeq 1$

Sulphurous acid $\left( H _{2} SO _{3}\right)$ has $Ka _{1}=1.7 \times 10^{-2}$ and $Ka _{2}=6.4 \times 10^{-8} .$ The $pH$ of $0.588 \,M\, H _{2} SO _{3}$ is ..... . (Round off to the Nearest Integer)

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