0.01 moles of a weak acid HA (Ka=2.0×10−6) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is ______ ×10−5 (Round off to the Nearest Integer). [Neglect volume change on adding HA. Assume degree of dissociation <<1]

ka=[H+][A−][HA]2×10−6=0.1[A−]0.01[A−]=2×10−7=cα2×10−7=10−2αα=2×10−5=2

Book Online Demo

Check Your IQ

Try Test

Courses

0.01 moles of a weak acid HA $(K_{a} = 2.0 \times 10^{- 6})$ is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is ______ $\times 10^{- 5}$ (Round off to the Nearest Integer). [Neglect volume change on adding HA. Assume degree of dissociation <<1]

see full answer

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

An Intiative by Sri Chaitanya

answer is 2.

(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

$k_{a} = \frac{[H^{+}] [A^{-}]}{[H A]}$

$2 \times 10^{- 6} = \frac{0.1 [A^{-}]}{0.01}$

$[A^{-}] = 2 \times 10^{- 7} = c α$

$2 \times 10^{- 7} = 10^{- 2} α$

$α = 2 \times 10^{- 5} = 2$

Watch 3-min video & get full concept clarity