A weak acid, HA, has a K_a of 1.00 times 10^{ | vedclass

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Question

$H A \rightleftharpoons H^{+}+A$

At equilibrium $\left[H^{+}=A^{-}\right]$

$K_{a} =\frac{\left[H^{+}\right]\left[A^{-}\right]}{[H A]}=\frac{\lef

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$H A ightleftharpoons H^{+}+A$

At equilibrium $\left[H^{+}=A^{-}ight]$

$K_{a} =\frac{\left[H^{+}ight]\left[A^{-}ight]}{[H A]}=\frac{\left[H^{+}ight]^{2}}{[H A]}$

$\left[H^{+}ight]=\sqrt{K_{a}[H A]} =\sqrt{1 	imes 10^{-5} 	imes 0.1}$

$= \sqrt{1 	imes 10^{-6}}=1 	imes 10^{-3}$

$\alpha =\frac{A 	ext {ctual ionisation}}{	ext {Molar concentration}}$

$\%$ of acid dissociated $=10^{-2} 	imes 1.00$

$=1 \%$

A weak acid, $HA,$ has a $K_a$ of $1.00 \times 10^{-5}.$ If $0.100 \,mol$ of this acid is dissolved in one litreof water, the percentage of acid dissociated at equilibrium is closest to.....$\%$

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