0 . 01 moles of a weak acid HA   K a = 2 . 0 × 10 - 6 is dissolved in 1 . 0   L of 0 . 1 M…

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

Solution

HA                    H+               +   A-Initial conc.          0.1M0M           Equ. conc. (0.01-x)             (0.1+x)xM0.01M               0.1M

Now, Ka=H+A-[HA]2×10-6=0.1×x0.01

x=2×10-7

Now, α=x0.01=2×10-70.01=2×10-5

Asked in: JEE Main 2021 (17 Mar Shift 1)

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