Question

The slope of the graph $\frac{{\Lambda }_v^2}{{\Lambda }_{\infty }({\Lambda }_{\infty }-{\Lambda }_v)}vs.\frac{1}{c}$, if ${\Lambda }_v$ and ${\Lambda }_{\infty }$ are molar conductivity of weak electrolyte having dissociation constant $K_a$ at concentration c and zero respectively is:

• A. concave downwards
• B. concave upwards
• C. straight line with intercept $K_a$
• D. straight line with slope $K_a$

Answer 1

The correct option is D straight line with slope $K_a$

The slope of the graph $\frac{{\Lambda }_v^2}{{\Lambda }_{\infty }({\Lambda }_{\infty }-{\Lambda }_v)}vs.\frac{1}{c}$, If ${\Lambda }_v$ and ${\Lambda }_{\infty }$ are molar conductivity of weak electrolyte having dissociation constant $K_a$ at concentration c and zero respectively is straight line with slope $K_a$

This is explained below.
The relationship betwen the dissociation constant and the degree of dissociation is given below.

$K_a=\frac{c{\alpha }^2}{(1-\alpha )}=\frac{c{\Lambda }_v^2}{{\Lambda }_{\infty }^2(1-\frac{{\Lambda }_v}{{\Lambda }_{\infty }})}$

Rearrange this equation

$K_a=\frac{c{\Lambda }_v^2}{{\Lambda }_{\infty }({\Lambda }_{\infty }-{\Lambda }_v)}$

or $\frac{{\Lambda }_v^2}{{\Lambda }_{\infty }({\Lambda }_{\infty }-{\Lambda }_v)}=\frac{K_a}{c}$

This is an equation of straight lin with slope m .

$y=mx(x=\frac{1}{c})$