0.01, M ,HA(aq.) is 2% ionized, [OH^-] of sol | vedclass

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Question

$\left[\mathrm{H}^{+}ight]_{\mathrm{wea} \mathrm{k} 	ext { acid }}=\mathrm{C} \alpha=0.01 	imes\left(\frac{2}{100}ight)$

$=2 	imes 10^{-4}$

Vedclass · Accepted Answer

$5 	imes 10^{-11}$

Vedclass · Answer

$\left[\mathrm{H}^{+}ight]_{\mathrm{wea} \mathrm{k} 	ext { acid }}=\mathrm{C} \alpha=0.01 	imes\left(\frac{2}{100}ight)$

$=2 	imes 10^{-4}$

$\left[\mathrm{H}^{+}ight] 	imes\left[\mathrm{OH}^{-}ight]=\mathrm{Kw}=10^{-14}$

$	herefore \quad\left[\mathrm{OH}^{-}ight]=\frac{10^{-14}}{2 	imes 10^{-4}}=5 	imes 10^{-11}$

$0.01\, M \,HA(aq.)$ is $2\%$ ionized, $[OH^-]$ of solution is :-

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