Question

The ionization constant of base ${\left(C_2{H}_5\right)}_3$ is $6.4\times {10}^{-5}$. Calculate its degree of dissociation in its $0.1M$ solution when it is mixed with $0.01M$ $NaOH$ solution.

Answer 1

Consider $\alpha$ , be very small ,by applying formula directly.

$\alpha =\sqrt{\frac{K_a}{C}=\sqrt{\frac{(6.4X10^{-5}{)}^{}}{0.1}}}=2.53X10^{-2}$

In the presence of$NaOH$ , equilibrium will be shift in the backward i.e, the concentration of base will increase.

If C is the initial concentration,

$X$ is the amount dissociated, then equilibrium will be:

[${(C}_2{H}_5){}_{3}N$] = $c-x,$

${\left[\right(C}_2{H}_5){}_{3}N{H}^{+}$ =$X,$

[O$H^{-}$] =$0.01+X$

$\therefore K_a=\frac{X(0.01+X)}{c-x}=\frac{X\left(0.01\right)}{X}$

$\frac{K_a}{0.01}=\frac{x}{c}$

but $\frac{X}{C}$ means $\alpha$

$\alpha =\frac{K_a}{0.01}=\frac{6.4X10^{-5}}{0.01}=6.4X10^{-3}$

Hence, $6.4X10^{-3}$ is the answer.