Question

A weak acid, $HA$, has $K_a$ of $1.00\times {10}^{-5}$. If $0.100$ mole of this acid is dissolved in $1L$ of water, the percentage of acid dissociated at equilibrium is closest to:

• A. $99.0$%
• B. $1.00$%
• C. $99.9$%
• D. $0.100$%

Answer 1

The correct option is D $0.100$%

For the equilibrium reacton of acid dissociaton:

$HA\rightleftharpoons H^{+}+A^{-}$

Initial concentration of $HA$ is 0.1 mol/L

$0.100$

At equilibrium:

$0.1-\alpha \alpha \alpha$

$\alpha$ = degree of dissociation= $\frac{{\left[HA\right]}_{\text{dissociated}}}{{\left[HA\right]}_{\text{undissociated}}}$

Given: $K_a={10}^{-5}$

Using the relation: $K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

${10}^{-5}=\frac{\alpha .\alpha }{(0.1-\alpha )}$

Since ita a weak acid, $\alpha <<0.1,\therefore 0.1-\alpha \approx 0.1$

${10}^{-5}=\frac{\alpha .\alpha }{0.1}$

$\alpha ={10}^{-3}$

Percent dissociated = $100\times \alpha$= 0.1 %

Hence, the correct option is $D$