Finding Ka of a monoprotic weak acid from pH and concentration of added sodium salt
I don't know why shouldn't it work, this is my approach, leading to final result, what's the fallacy?

Consider the reaction
$$\ce{A-} + \ce{H2O} \ce{&lt;=&gt;} HA + \ce{OH-}$$
It can be broken into two steps
$$\ce{A-} + \ce{H+} \ce{&lt;=&gt;} HA$$

$K = \dfrac{1}{K_a}$
$$\ce{H2O} \ce{&lt;=&gt;} \ce{H+} + \ce{OH-}$$

$K = K_w = 10^{-14}$ at 25°C

If h is hydrolysis constant, approximating $h\ll 1$, then [$\ce{A-}$] = c(1 - h) $\approx$ c and [$\ce{OH-}$] = [$\ce{HA}$] = ch.

$\therefore$, $K_h = K_b = \dfrac{K_w}{K_a} = ch^2$. 

$\implies K_a = \dfrac{K_w}{ch^2} = K_\mathrm{a} = K_\mathrm{w} \frac{[\ce{A-}]}{[\ce{OH-}]^2}$

$h = \sqrt{\dfrac{K_w}{K_{a}.c}}$

$[\ce{OH-}] = ch = \sqrt{\dfrac{K_{w}.c}{K_a}}$

pH = 7 + $\dfrac{1}{2} log {c} + \dfrac{1}{2} log {pK_a}$

Which is a well known result...