Question

Conductivity of $ 0.001M $ acetic acid at a certain temperature is $ 5.07 \times {10^{^ - 5}}Sc{m^{ - 1}} $ . If limiting molar conductivity of acetic acid at the same temperature is $ 390{\text{ }}Sc{m^2}mo{l^{ - 1}} $ . The dissociation constant of acetic acid at that temperature is?

Answer 1

Hint: The question gives the limiting molar conductivity and molar conductivity, calculate the dissociation constant. It is not a direct question. You can use the formula of molar conductivity first:
$ {\Lambda _m} = \alpha \Lambda _m^0 $
After this find out the degree of dissociation using the formula: $ {K_a} = \dfrac{{C{\alpha ^2}}}{{1 - \alpha }} $ (where C is the concentration, and $ \alpha $ is the degree of dissociation).

Complete Step By Step Answer:
In the question we are given the limiting molar conductivity ( $ \Lambda _m^0 $ ) and the molar conductivity ( $ {\Lambda _m} $ ) and we have to find the value of dissociation constant. But before finding the dissociation constant we have to find the degree of dissociation.
At the given concentration the molar conductivity is:
 $ {\Lambda _m} = \alpha \Lambda _m^0 $
Also, note that the molar conductivity given is not in the proper units. Thus,
 $ 5.07 \times {10^{ - 5}}Sc{m^{ - 1}} \times \dfrac{{1{\text{ }}L}}{{0.001{\text{ }}mols}} \times \dfrac{{1000mL}}{{1{\text{ }}L}} \times \dfrac{{1{\text{ }}c{m^3}}}{{1{\text{ }}mL}} $
 $ \Rightarrow conductivity = 50.7{\text{ }}S \cdot c{m^2}mo{l^{ - 1}} $
So we get our conductivity in proper units. And now we will use the degree of dissociation in the $ \left( \alpha \right) $ formula and find out its value.
 $ \alpha = \dfrac{{{\Lambda _m}}}{{\Lambda _m^0}} $
 $ \Rightarrow \alpha = \dfrac{{50.7{\text{ }}S \cdot c{m^2}mo{l^{ - 1}}}}{{390{\text{ }}S \cdot c{m^2}mo{l^{ - 1}}}} $
 $ \Rightarrow \alpha = 0.13 $
Thus the degree of dissociation come out to be $ 0.13 $
Next we will find out the dissociation constant whose formula is: $ {K_a} = \dfrac{{C{\alpha ^2}}}{{1 - \alpha }} $
By definition, the dissociation of acetic acid is given by
 $ C{H_3}COOH(aq) \rightleftharpoons C{H_3}CO{O^ - }(aq) + {H^ + }(aq) $
The dissociation constant for this reaction will be written as:
 $ K_a = \dfrac{{[C{H_3}CO{O^ - }][{H^ + }]}}{{[C{H_3}COOH]}} $
Some fraction α of the starting concentration becomes the acetate and proton products, and that much is lost relative to 100% of the starting acid, so
 $ {K_a} = \dfrac{{(\alpha \cdot {{[C{H_3}CO{O^ - }]}_0})(\alpha \cdot {{[{H^ + }]}_0})}}{{(1 - \alpha ) \cdot {{[C{H_3}COOH]}_0}}} $
 $ \Rightarrow {K_a} = \dfrac{{{\alpha ^2} \cdot {{[C{H_3}CO{O^ - }]}_0}}}{{(1 - \alpha )}} $
 $ \Rightarrow {K_a} = \dfrac{{{{0.13}^2} \times 0.001{\text{ }}M}}{{1 - 0.13}} $
 $ \Rightarrow {K_a} = 1.94 \times {10^{ - 5}} $
Therefore the value of dissociation constant $ {K_a} $ of acetic acid at some temperature is $ 1.94 \times {10^{ - 5}} $ .

Note:
The actual value is around $ 1.76 \times {10^{ - 5}}\; $ at $ {25^ \circ }C $ , and dissociation is endothermic, so a higher $ {K_a} $ means this temperature is warmer than $ {25^ \circ }C $ . While solving this problem do not forget to convert the conductivity in proper units. The units should be the same for limiting molar and molar conductivity.