Question

What is the value of $p{K}_b\left(C{H}_{3}CO{O}^{\ominus }\right)$ if ${\Lambda }_m^{\circ }=390Sc{m}^{-1}mo{l}^{-1}and{\Lambda }_m=7.8Sc{m}^{-2}mo{l}^{-1}$ for $0.04MofC{H}_{3}COOH$ at ${25}^{\circ }C$ ?

• A. $9.3$
• B. $9.2$
• C. $4.7$
• D. $4.8$

Answer 1

The correct option is B $9.2$

As given,
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m{}^{\circ }}=\frac{7.8}{390}$

$=0.02$
$K_a\left(C{H}_{3}COOH\right)$

$=\frac{c\alpha \cdot c\alpha }{c-c\alpha }$
$=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.04\times (0.02{)}^2}{1-0.02}$

$\approx 1.6\times {10}^{-5}$
Use: $pK+p{K}_b=p{K}_w$

$=14$
$\Rightarrow p{K}_b=14-p{K}_a=14-4.8$

$=9.2$