Question

Ph of 1m ha (weak acid) is 2. Hence van't hoff factor is -

Answer 1

Answer: 1.01

Explanation:

Vant hoff factor is the ratio of observed colligative property to the calculated colligative property.

$i=\frac{\text {observed colligative property}}{\text {Calculated Colligative property}}$

$HA\rightarrow H^{+}++A^{-}$

initial     1M         0        0

eqm   $1-1\alpha$       $1\alpha$        $1\alpha$  

Total moles after dissociation =$1-1\alpha+1\alpha+1\alpha=1+\alpha$  

Thus $i=\frac{1+\alpha}{1}$

Given:  pH = 2

$pH=-log[H^+]$

$[H^+]=\alpha= 10^{-2}$

$i=\frac{1+10^{-2}}{1}$

$i=1.01$

Thus van't hoff factor is 1.01.