Question

Which of the following expressions is correct for ‘$Ka$’ in terms of ${\Lambda }^{\circ }$ and $\Lambda$, where ‘$C$’ is the concentration.

• A. $K_a=\frac{C{\Lambda }_m^{\circ }}{{\Lambda }_m({\Lambda }_m^{\circ }-\Lambda )}$
• B. $K_a=\frac{C{\Lambda }_m^2}{{\Lambda }_m^{\circ }({\Lambda }_m^{\circ }-{\Lambda }_m)}$
• C. $K_a=\frac{C{\Lambda }_m^2}{{\Lambda }_m^{\circ }}$
• D. $K_a=\frac{C{\Lambda }_m^2}{({\Lambda }_m^{\circ }-{\Lambda }_m)}$

Answer 1

The correct option is B $K_a=\frac{C{\Lambda }_m^2}{{\Lambda }_m^{\circ }({\Lambda }_m^{\circ }-{\Lambda }_m)}$

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}{K}_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{C\times {\left(\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}\right)}^2}{1-\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}}=\frac{C\times {\Lambda }_m^2}{{\Lambda }_m^{\circ }({\Lambda }_m^{\circ }-{\Lambda }_m)}=\frac{C({\Lambda }_m^{\circ }{)}^2}{{\Lambda }_m^{\circ }({\Lambda }_m^{\circ }-{\Lambda }_m)}$