Question

The conjugate base of a weak acid has $K_b={10}^{-9}$. The equilibrium constant for the reaction of an acid with a strong base is ${10}^x$ then x is:

Answer 1

The reaction for the base dissociation is

$A^{-}+H_{2}O\rightleftharpoons HA+O{H}^{-}$

Where, $k_b=\frac{\left[HA\right]\left[O{H}^{-}\right]}{\left[A^{-}\right]\left[H_{2}O\right]}={10}^{-9}$

The reaction of an acid with a strong base is

$HA+O{H}^{-}\rightleftharpoons A^{-}+H_{2}O$

$k_a=\frac{\left[A^{-}\right]\left[H_{2}O\right]}{\left[HA\right]\left[O{H}^{-}\right]}=\frac{1}{k_b}=\frac{1}{{10}^{-9}}={10}^9$

${10}^x$ $={10}^9$

So, the value of x is 9.