Question

$0.002\, M$ solution of a weak acid has an equivalent conductance $(\wedge) 60\, ohm^{-1}\, cm^{2}\, eq^{-1}$ . What will be the $pH$ ? (Given : $\wedge = 400 \, ohm^{-1} \, cm^{2} \, eq^{-1}$ ]

• A. $3.52$
• B. $2.52$
• C. $1.87$
• D. $2.7$

Answer 1

Answer: (A)

$3.52$

Answer 2

Degree of dissociation $\left(\alpha\right)=\frac{\wedge_{m}^{c}}{\wedge^{o}_{m}}$
Given $\wedge_{m}^{c}=60\,ohm^{-1}\,cm^{2}eq^{-1}$
$\wedge^{o}=400\,ohm^{-1}\,cm^{2}\,eq^{-1}$
On putting the values,
$\alpha=\frac{60}{400}=0.15$
Concentration of $H^{+}$ in the solution will be $c\alpha$, where $C= 0.002 \,M$
$\therefore \left[H^{+}\right]=\left(0.002\times0.15\right) M=0.0003$
Now, $pH=log \left[H^{+}\right]$
$=-log\left(0.0003\right)$
$3.52$