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

Answer 1

Answer: (D)

$\text{HA}+\text{aq}\rightleftharpoons {\underset{\text{aq}}{H}}^{+}+\underset{\text{aq}}{{\text{A}}^{-}}$
${\text{K}}_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$
Also $\text{HA}+{\text{B}}^{+}+{\text{OH}}^{-}\rightleftarrows {\text{B}}^{+}+{\text{H}}_2\text{O}+{\text{A}}^{-}$
$\text{HA}+{\text{OH}}^{-}\rightleftharpoons {\text{H}}_2\text{O}+{\text{A}}^{-}$
${\text{K}}_{\text{eq}}=\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]\left[{\text{OH}}^{-}\right]}=\frac{\left[{\text{A}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[\text{HA}\right]\left[{\text{OH}}^{-}\right]\left[{\text{H}}^{+}\right]}=\frac{\text{K}a}{\text{K}w}$
$\therefore$ ${\text{K}}_{\text{eq}}=\frac{{\text{K}}_{\text{a}}}{{\text{K}}_{\text{w}}}=\frac{10^{-4}}{10^{-14}}=10^{10}$