Selenious acid H2SeO3 a diprotic acid has ka1=3.0×10−3 and ka2=5.0×10−8 the OH− of a 0.30 M solution of a selenious acid is x×10−14, then x is ____

ka1>>ka2 ∴ neglect ka2 ka1=H+HSO3−H2SO3⇒x⋅x0.3−x=3×10−30.3−x≃0.3∴x20.3=3×10−3(or)9×10−4 (or) x=3×10−2 H+=3×10−2kw=H+OH−∴OH−=13×10−1410−2=0.33×10−12

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Selenious acid $(H_{2} {SeO}_{3})$ a diprotic acid has ${ka}_{1} = 3 .0 \times 10^{- 3}$ and ${ka}_{2} = 5 .0 \times 10^{- 8}$ the $[{OH}^{-}]$ of a 0.30 M solution of a selenious acid is $x \times 10^{- 14},$ then x is ____

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answer is 33.

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Detailed Solution

${ka}_{1} >> {ka}_{2} ∴ neglect {ka}_{2}$

$\begin{array}{l} {ka}_{1} = \frac{[H^{+}] [{HSO}_{3}^{-}]}{H_{2} {SO}_{3}} \Rightarrow \frac{x \cdot x}{0 .3 - x} = 3 \times 10^{- 3} 0 .3 - x ≃ 0 .3 \\ ∴ \frac{x^{2}}{0 .3} = 3 \times 10^{- 3} (or) 9 \times 10^{- 4} (or) x = 3 \times 10^{- 2} [H^{+}] = 3 \times 10^{- 2} \\ k_{w} = [H^{+}] [{OH}^{-}] ∴ [{OH}^{-}] = \frac{1}{3} \times \frac{10^{- 14}}{10^{- 2}} = 0 .33 \times 10^{- 12} \end{array}$