Question

Consider a solution of monoprotic weak acid having dissociation constant $K_a$. What is the minimum concentration $C$ in terms of $K_a$ such that the concentration of the undissociated acid can be equated to $C$ within a $10$% limit of error?

• A. $90K_a$
• B. $50K_a$
• C. $70K_a$
• D. $100K_a$

Answer 1

Answer: (A)

$90K_a$

Let consider a weak acid $HA$

$\begin{array}{cccccc}& HA& \rightleftharpoons & H^{+}& +& A^{-}\\ initially& C& & 0& & 0\\ ateqn& C(1-\alpha )& & C\alpha & & C\alpha \end{array}$

$K_a=\frac{C^2{\alpha }^2}{C(1-\alpha )}=\frac{C{\alpha }^2}{1-\alpha }$

$\frac{K_a}{C}=\frac{\alpha }{1-\alpha }$

$\frac{C}{K_a}=\frac{1}{{\alpha }^2}-\frac{\alpha }{{\alpha }^2}=\frac{1}{{\alpha }^2}-\frac{1}{\alpha }$

$\alpha =0.1$

$\frac{C}{K_a}=\frac{1-0.1}{(0.1{)}^2}=\frac{0.9}{0.01}=90$

$C=90K_a$
Option A is correct.