Question

If degree of ionization is $0.01$ of decimolar solution of weak acid $HA$, then $p{K}_a$ of acid is

• A. $2$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is C $5$

As we know,

$C{\alpha }^2=K_a=0.1\times (0.01{)}^2={10}^{-5}$

Now,

$pKa=-logKa=-log{10}^{-5}=5$

So, pKa is 5.