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Question: Conductivity of \( 0.001M \) acetic acid at a certain temperature is \( 5.07 \times {10^{^ - 5}}Sc{m...

Conductivity of 0.001M0.001M acetic acid at a certain temperature is 5.07×10−5Scm−15.07 \times {10^{^ - 5}}Sc{m^{ - 1}} . If limiting molar conductivity of acetic acid at the same temperature is 390 Scm2mol−1390{\text{ }}Sc{m^2}mo{l^{ - 1}} . The dissociation constant of acetic acid at that temperature is?

Explanation

Solution

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:
Λm=αΛm0{\Lambda _m} = \alpha \Lambda _m^0
After this find out the degree of dissociation using the formula: Ka=Cα21−α{K_a} = \dfrac{{C{\alpha ^2}}}{{1 - \alpha }} (where C is the concentration, and α\alpha is the degree of dissociation).

Complete Step By Step Answer:
In the question we are given the limiting molar conductivity ( Λm0\Lambda _m^0 ) and the molar conductivity ( Λm{\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:
Λm=αΛm0{\Lambda _m} = \alpha \Lambda _m^0
Also, note that the molar conductivity given is not in the proper units. Thus,
5.07×10−5Scm−1×1 L0.001 mols×1000mL1 L×1 cm31 mL5.07 \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}}
⇒conductivity=50.7 S⋅cm2mol−1\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 (α)\left( \alpha \right) formula and find out its value.
α=ΛmΛm0\alpha = \dfrac{{{\Lambda _m}}}{{\Lambda _m^0}}
⇒α=50.7 S⋅cm2mol−1390 S⋅cm2mol−1\Rightarrow \alpha = \dfrac{{50.7{\text{ }}S \cdot c{m^2}mo{l^{ - 1}}}}{{390{\text{ }}S \cdot c{m^2}mo{l^{ - 1}}}}
⇒α=0.13\Rightarrow \alpha = 0.13
Thus the degree of dissociation come out to be 0.130.13
Next we will find out the dissociation constant whose formula is: Ka=Cα21−α{K_a} = \dfrac{{C{\alpha ^2}}}{{1 - \alpha }}
By definition, the dissociation of acetic acid is given by
CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)C{H_3}COOH(aq) \rightleftharpoons C{H_3}CO{O^ - }(aq) + {H^ + }(aq)
The dissociation constant for this reaction will be written as:
Ka=[CH3COO−][H+][CH3COOH]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
Ka=(α⋅[CH3COO−]0)(α⋅[H+]0)(1−α)⋅[CH3COOH]0{K_a} = \dfrac{{(\alpha \cdot {{[C{H_3}CO{O^ - }]}_0})(\alpha \cdot {{[{H^ + }]}_0})}}{{(1 - \alpha ) \cdot {{[C{H_3}COOH]}_0}}}
⇒Ka=α2⋅[CH3COO−]0(1−α)\Rightarrow {K_a} = \dfrac{{{\alpha ^2} \cdot {{[C{H_3}CO{O^ - }]}_0}}}{{(1 - \alpha )}}
⇒Ka=0.132×0.001 M1−0.13\Rightarrow {K_a} = \dfrac{{{{0.13}^2} \times 0.001{\text{ }}M}}{{1 - 0.13}}
⇒Ka=1.94×10−5\Rightarrow {K_a} = 1.94 \times {10^{ - 5}}
Therefore the value of dissociation constant Ka{K_a} of acetic acid at some temperature is 1.94×10−51.94 \times {10^{ - 5}} .

Note:
The actual value is around 1.76×10−5  1.76 \times {10^{ - 5}}\; at 25∘C{25^ \circ }C , and dissociation is endothermic, so a higher Ka{K_a} means this temperature is warmer than 25∘C{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.