0.01 moles of a weak acid HA left( K _{ a }=2 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad\quad\quad HA \rightleftharpoons \quad H ^{+}+\quad A ^{-}$

Initial conc. $\quad0.01 M\quad \, 0.1 M \qua

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad\quad\quad HA ightleftharpoons \quad H ^{+}+\quad A ^{-}$

Initial conc. $\quad0.01 M\quad \, 0.1 M \quad\quad 0$

Equ. conc. $\,(0.01- x )\quad(0.1+ x )\quad xM$

Now, $K_{a}=\frac{\left[x^{+}ight]\left[A^{-}ight]}{[H A]} \Rightarrow 2 	imes 10^{-6}=\frac{0.1 	imes x}{0.01}$

$	herefore \quad x=2 	imes 10^{-7}$

Now, $\alpha=\frac{x}{0.01}=\frac{2 	imes 10^{-7}}{0.01}=2 	imes 10^{-5}$

Similar Questions

In $20\,\, ml \,\,0.4 \,M-HA$ solution, $80\,\, ml$ water is added. Assuming volume to be additive, the $pH$ of final solution is

$(K_a \,\,of\,\, HA = 4 \times 10^{-7} ,\, log\,2 = 0.3)$

The $pH$ value of decinormal solution of $N{H_4}OH$ which is $20\%$ ionised, is

Assuming that the degree of hydrolysis is small, the $pH$ of $0.1\, M$ solution of sodium acetate $(K_a\, = 1.0\times10^{- 5})$ will be

The dissociation constants of two acids $HA_1$ and $HA_2$ are $3.0 \times 10^{-4}$ and $1.8 \times 10^{-5}$ respectively. The relative strengths of the acids will be

Find $pH$ of $5 \times 10^{-3}\, M$ $H_2CO_3$ solution having $10\%$ dissociation