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The molar conductivity of acetic at infi...

The molar conductivity of acetic at infinite dilution is `390.7` and for `0.01 M` acetic acid is `3.9.7 S cm^(2)mol^(-1).` Calculate `(a) alpha ` and `(b) pH` of solution.

Text Solution

Verified by Experts

`alpha=(wedge_(m)^(c))/(wedge_(m)^(@))=(3.907).(390.7)=0.01`

`:.[H^(o+)]=calpha=0.01xx0.01=10^(-4)M`
`pH=-log (10^(-4))=4`
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Knowledge Check

  • The molar conductivity of acetic acid at infinite dilution is 390.7 Scm^(2) "mol"^(-1) . Conductivity of 0.1 M acetic acid solution is 5.2 S cm^2 "mol"^(-1) , find out degree of dissociation of acetic acid :-

    A
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    B
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    C
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    D
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  • The molecular conductivity of acetic acid at infinite dilution is 390 S cm^(2) "mole"^(-1) and for 0.1 acetic and solution is 5.8 cm^(3) "mole"^(-1) .The hydrogen ion concentration of the solution is

    A
    `15 xx 10^(-3)`
    B
    `15 xx 10^(-4)`
    C
    `66 xx 10^(-3)`
    D
    `66 xx 10^(-4)`

  • The molar conductance of acetic acid at infinite dilution is lambda_(oo) . If the conductivity of 0.1M acetic acid is S, the apparent degree of ionisation is

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    D
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