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

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 ____

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 33.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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

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

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th