Question

A certain weak acid has a dissociation constant $\text{1 }\times 10^{- 4}$ The equilibrium constant for its reaction with a strong base is:

• A. $1 \text{.} 0 \times 1 0^{- 4 \, }$
• B. $1 \text{.} 0 \times 1 0^{- 10}$
• C. $1 \text{.} 0 \times 1 0^{- 14}$
• D. $1 \times 1 0^{10}$

Answer 1

Answer: (D)

$\text{HA} + \text{aq} \rightleftharpoons \underset{\text{aq}}{\text{H}}^{+} + \underset{\text{aq}}{\text{A}^{-}}$
$\text{K}_{\text{a}} = \frac{\left[\text{H}^{+}\right] \left[\text{A}^{-}\right]}{\left[\text{HA}\right]}$
Also $\text{HA} + \text{B}^{+} + \text{OH}^{-} \rightleftarrows \text{B}^{+} + \text{H}_{2} \text{O} + \text{A}^{-}$
$\text{HA} + \text{OH}^{-} \rightleftharpoons \text{H}_{2} \text{O} + \text{A}^{-}$
$\text{K}_{\text{eq}} = \frac{\left[\text{A}^{-}\right]}{\left[\text{HA}\right] \left[\text{OH}^{-}\right]} = \frac{\left[\text{A}^{-}\right] \left[\text{H}^{+}\right]}{\left[\text{HA}\right] \left[\text{OH}^{-}\right] \left[\text{H}^{+}\right]} = \frac{\text{K} a}{\text{K} w }$
$\therefore $ $\text{K}_{\text{eq}} = \frac{\text{K}_{\text{a}}}{\text{K}_{\text{w}}} = \frac{10^{- 4}}{10^{- 14}} = 10^{10}$