Question

Percentage ionization of weak acid can be calculated using the formula:

• A. $\frac{100}{1+{10}^{(p{K}_a-pH)}}$
• B. $100\sqrt{\frac{K_a}{C}}$
• C. Both (a) and (b)
• D. none

Answer 1

The correct option is C Both (a) and (b)

For weak acid dissociation equilibrium, degree of dissociation $\alpha$ is given as
$\alpha =\sqrt{\frac{K_a}{C}}\therefore \%\alpha =100\sqrt{\frac{K_a}{C}}\text{Also,}{K}_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}=\frac{\left[H^{+}\right]C\alpha }{C(1-\alpha )}=\frac{\left[H^{+}\right]\alpha }{(1-\alpha )}\log K_c=\log H^{+}+\log \frac{\alpha }{(1-\alpha )}orp{K}_a=pH+\log \frac{(1-\alpha )}{\alpha }p{K}_a-pH=\log \frac{(1-\alpha )}{\alpha }\frac{(1-\alpha )}{\alpha }={10}^{K_a-pH},or\frac{1}{\alpha }={10}^{K_a-pH}+1or\alpha =\frac{1}{(1+{10}^{p{K}_a-pH})}$