JEE Main 2021 — Chemistry Physical Chemistry

${H}_{2}{\mathrm{SO}}_{3}$H2​SO3​ $[$[Dibasic acid$]$]

$$c=0.588M$$c=0.588M

$\Rightarrow$⇒ $\mathrm{pH}$pH of solution is due to First dissociation only since ${{K}_{a}}_{1}>>{\mathrm{Ka}}_{2}$Ka​1​>>Ka2​

$\Rightarrow$⇒ First dissociation of ${H}_{2}{\mathrm{SO}}_{3}$H2​SO3​

$${H}_{2}{\mathrm{SO}}_{3}(\mathrm{aq})\rightleftharpoons {H}^{\oplus }(\mathrm{aq})+{\mathrm{HSO}}_{3}^{-}(\mathrm{aq}):{\mathrm{ka}}_{1}=1.7\times {10}^{-2}$$H2​SO3​(aq)⇌H⊕(aq)+HSO3−​(aq):ka1​=1.7×10−2

$$t=0$$t=0

$$tC-xxx$$tC−xxx

$$\Rightarrow {\mathrm{Ka}}_{1}=\frac{1.7}{100}=\frac{[{H}^{\oplus }][{\mathrm{HSO}}_{3}^{-}]}{[{H}_{2}{\mathrm{SO}}_{3}]}$$⇒Ka1​=1001.7​=[H2​SO3​][H⊕][HSO3−​]​

$$\Rightarrow \frac{1.7}{100}=\frac{{x}^{2}}{(0.588-x)}$$⇒1001.7​=(0.588−x)x2​

$$\Rightarrow 1.7\times 0.588-1.7x=100{x}^{2}$$⇒1.7×0.588−1.7x=100x2

$$\Rightarrow 100{x}^{2}+1.7x-1=0$$⇒100x2+1.7x−1=0

$$\Rightarrow [{H}^{\oplus }]=x=\frac{-1.7+\sqrt{(1.7{)}^{2}+4\times 100\times 1}}{2\times 100}=0.09186$$⇒[H⊕]=x=2×100−1.7+(1.7)2+4×100×1​​=0.09186

Therefore $\mathrm{pH}$pH of sol. is :

$$\mathrm{pH}=-\mathrm{log}[{H}^{\oplus }]$$pH=−log[H⊕]

$$\Rightarrow \mathrm{pH}=-\mathrm{log}(0.09186)=1.036\simeq 1$$⇒pH=−log(0.09186)=1.036≃1