Plotting  $\frac{1}{{\Lambda }_m}$ against $c{\Lambda }_m$ for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio $\frac{P}{S}$ is

The correct option is K_aΛ⁰_m.

To find the ratio of the y-axis intercept (P) to the slope (S), we begin with Ostwald's dilution law for a weak monobasic acid (HX):

$K_a=\frac{c{\alpha }^2}{1-\alpha }$

Here, $\alpha$ is the degree of dissociation, given by the ratio of molar conductivity at concentration c (${\Lambda }_m$) to molar conductivity at infinite dilution (${\Lambda }_m^0$):

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}$

Substitute $\alpha$ into the expression for $K_a$:

$K_a=\frac{c{\left(\frac{{\Lambda }_m}{{\Lambda }_m^0}\right)}^2}{1-\frac{{\Lambda }_m}{{\Lambda }_m^0}}=\frac{c{\Lambda }_m^2}{{\left({\Lambda }_m^0\right)}^2\left(1-\frac{{\Lambda }_m}{{\Lambda }_m^0}\right)}=\frac{c{\Lambda }_m^2}{{\Lambda }_m^0\left({\Lambda }_m^0-{\Lambda }_m\right)}$

Rearranging this equation gives:

${\Lambda }_m^0-{\Lambda }_m=\frac{c{\Lambda }_m^2}{K_a{\Lambda }_m^0}$

Dividing both sides by ${\Lambda }_m{\Lambda }_m^0$:

$\frac{{\Lambda }_m^0-{\Lambda }_m}{{\Lambda }_m{\Lambda }_m^0}=\frac{c{\Lambda }_m}{K_a{\left({\Lambda }_m^0\right)}^2}$

$\frac{1}{{\Lambda }_m}-\frac{1}{{\Lambda }_m^0}=\frac{1}{K_a{\left({\Lambda }_m^0\right)}^2}\cdot \left(c{\Lambda }_m\right)$

Rearranging to match the standard equation of a straight line, $y=mx+c$:

$\frac{1}{{\Lambda }_m}=\frac{1}{K_a{\left({\Lambda }_m^0\right)}^2}\cdot \left(c{\Lambda }_m\right)+\frac{1}{{\Lambda }_m^0}$

Comparing this equation to $y=Sx+P$ where $y=\frac{1}{{\Lambda }_m}$ and $x=c{\Lambda }_m$:

$\text{y-axis Intercept (P)}=\frac{1}{{\Lambda }_m^0}$

$\text{Slope (S)}=\frac{1}{K_a{\left({\Lambda }_m^0\right)}^2}$

Now, calculate the ratio $\frac{P}{S}$:

$\frac{P}{S}=\frac{\left(\frac{1}{{\Lambda }_m^0}\right)}{\left(\frac{1}{K_a{\left({\Lambda }_m^0\right)}^2}\right)}=\frac{K_a{\left({\Lambda }_m^0\right)}^2}{{\Lambda }_m^0}=K_a{\Lambda }_m^0$

Thus, the ratio $\frac{P}{S}$ is equal to K_aΛ⁰_m.