Question

A certain weak acid has $Ka=1.0\times {10}^{-4}$. Calculate the equilibrium constant for its reaction with a strong Base.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$BOH+HA\rightleftharpoons BA+H_{2}O.....\left(i\right)B^{+}+O{H}^{-}+HA\rightleftharpoons B^{+}+A^{-}+H_{2}OO{H}^{-}+HA\rightleftharpoons A^{-}+H_{2}OK=\frac{\left[A^{-}\right]\left[H_{2}O\right]}{\left[O{H}^{-}\right]\left[HA\right]}=\frac{\left[A^{-}\right]}{\left[O{H}^{-}\right]\left[HA\right]}\text{Since it is an aqueous system,}\left[H_{2}O\right]\text{is negletected. For the weak acid}HA\rightleftharpoons H^{+}+A^{-}\frac{K}{K_a}=\frac{\left[A^{-}\right]}{\left[O{H}^{-}\right]\left[HA\right]}\frac{\left[HA\right]}{\left[H^{-}\right]\left[A^{-}\right]}=\frac{1}{\left[O{H}^{-}\right]\left[H^{+}\right]}\therefore K=\frac{K_a}{K_w}=\frac{1\times {10}^{-4}}{1\times {10}^{-14}}=1\times {10}^{10}\text{Equation (i) is a neutralization step and reverse of it is the hydrolysis. Hence}{K}_c=\frac{1}{K_h}=\frac{K_a}{K_w}eg:C{H}_{3}COOH+N{a}^{+}+OH.\rightleftharpoons C{H}_{3}CO{O}^{-}+N{a}^{+}+H_{2}O{K}_c=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[N{a}^{+}\right]\left[H^{+}\right]}{\left[H{K}_c\right]\left[O{H}^{-}\right]\left[N{a}^{+}\right]\left[H^{+}\right]}=\frac{K_a}{K_w}$