Question

Hydrolysis constants of two salts $KA$ and $KB$ of weak acids $HA$ and $HB$ are ${10}^{-8}$ and ${10}^{-6}$. If the dissociation constant of third acid $HC$ is ${10}^{-2}$. The order of acidic strengths of three acids will be:

• A. $HA>HB>HC$
• B. $HB>HA>HC$
• C. $HC>HA>HB$
• D. $HA=HB=HC$

Answer 1

The correct option is C $HC>HA>HB$

$KA+H_{2}O\rightleftharpoons HA+KOH$

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

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

$K_h=\frac{\left[HA\right]\left[{OH}^{-}\right]}{\left[A^{-}\right]}=\frac{\left[HA\right]\left[O{H}^{-}\right]\left[H^{+}\right]}{\left[A^{-}\right]\left[H^{+}\right]}=\frac{K_w}{K_a}$

$K_a\left(HA\right)=\frac{K_w}{K_h}=\frac{{10}^{-14}}{{10}^{-8}}={10}^{-6}$

Similarly for $KB$ salt of weak acid $K_a$ is given by:
$K_a\left(HB\right)=\frac{K_w}{K_h}=\frac{{10}^{-14}}{{10}^{-6}}={10}^{-8}$
$K_a\left(HC\right)={10}^{-2}$
Greater the value dissociation constant $K_a$, more is the acidic strength.