Question

$0.1M$ aqueous solution of weak acid (HA) have $pH=2$. Van't Hoff factor for solution will be:

• A. 1.1
• B. 1.4
• C. 1.5
• D. 1.8

Answer 1

Answer: (A)

1.1

Solution:- (A) $1.1$

pH of aq. $HA=2$

$pH=-\log \left[{H_{3}O}^{+}\right]$

$\therefore \left[{H_{3}O}^{+}\right]={10}^{-2}$

$\underset{\text{At equilibrium}}{\underset{\text{Initially}}{}}\underset{1-\alpha }{\underset{1}{HA}}\rightleftharpoons \underset{\alpha }{\underset{0}{H^{+}}}+\underset{\alpha }{\underset{0}{A^{-}}}$

Total no. of moles $=1-\alpha +\alpha +\alpha =1+\alpha$

$\because \alpha =\frac{0.01}{0.1}=0.1$

$\therefore$ Total no. of moles $=1+0.1=1.1$

Therefore,

Van't Hoff factor $\left(i\right)=\frac{\text{Total no. of moles after dissociation}}{\text{No. of moles before dissociation}}$

$\therefore i=\frac{1+0.1}{1}=1.1$

Hence the value of Van't Hoff factor for solution is $1.1$.