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The molar conductivity of 2.5 xx 10^(-4)...

The molar conductivity of `2.5 xx 10^(-4)" mol L"^(-1)` methanoic acid is 46 S `cm^2" mol"^(-1).` Calculate its degree of dissociation and dissociation constant. Given `lambda_(H^+)^@"= 249.65 cm mol"^(-1)` and
`lambda_(HCOO^(-))^@"= 54.6S cm"^2" mol"^(-1)`

Text Solution

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Step I Calculation of degree of dissociation `(alpha)` of HCOOH.
`wedge_m="46 S cm"^2 "mol"^(-1)`
`wedge_(m)^@ (HCOOH)=lambda_(m(HCOO^(-)))^@ +lambda_(m)^@ (H^+)`
`=(54.6+249.6)"S cm"^2 "mol"^(-1)`
`=304.2"S cm"^(2)" mol"^(-1)`
`alpha=(wedge_(m))/(wedge_(m)^@)=((46)"S cm"^2 "mol"^(-1))/((304.2)"S cm"^2 "mol"^(-1))=0.1512`
Step II Calculation of dissociation constant
`underset(C)(HCOOH(aq)) overset("Water")Leftrightarrow underset(0)(HCOO^(-)(aq)) +underset(0)(H^+ (aq))`
Initial conc. `"C 0 0 "`
Equilibrium conc. `C(1-alpha)" "C alpha" "C alpha`
Dissociation constant `K_(alpha) =([HCOO^(-)] [H^+])/([HCOOH])`
`=(C alpha xx C alpha)/(C(1-alpha))=(C alpha^2)/((1-alpha))`
Putting values,
`K_(a)=((25 xx 10^(-4)"mol L"^(-1)) xx (0.1512)^2)/((1-0.1512))`
`=((5.715 xx 10^(-5)" mol L"^(-1)))/((0.848))`
`=6.73 xx 10^(-5)" mol L"^(-1)`
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