Home
Class 12
CHEMISTRY
If the molar conductivity of a weak ...

If the molar conductivity of a weak monobasic acid at 0.1 M and zero concentration are 15.8 and 226. 4 S `cm^(2) mol^(-1)` respectively calculate the degree of dissociation and the dissociation constant for acid

Text Solution

Verified by Experts

`z= 0.06988 , K_(a) = 4.872 xx 10^(-4) `
Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

If the molar ionic conductivities of K^(+) and I^(-) at zero concentration are 73.4 and 76.3 Omega^(-1) cm^(2) mol^(-1) respectively calculate the molar conductivity of K1 at zero concentration

The molar conductance of ammonium hydroxide solution of concentration 0.1 M, 0.01M and 0.001 M are 3.6, 11.3 and 34.0 ohm^(-1)cm^(-2)mol^(-1) respectively. Calculate the degree of dissociation of NH_(4)OH at these concentrations. Molar conductance at infinite dilution for NH_(4)OH is 271.1 ohm^(-1)cm^(-1)mol^(-1) .

Knowledge Check

  • The pH of 0.1 M acetic acid is 3, the dissociation constant of acid will be

    A
    `1.0 xx 10^(-4)`
    B
    `1.0 xx 10^(-5)`
    C
    `1.0 xx 10^(-3)`
    D
    `1.0 xx 10^(-8)`

  • At T (K), the molar conductivity of 0.04 M acetic acid is 7.8 S cm^(2) mol^(-1) . If the limiting molar conductivities of H^(+) and CH_(3)COO^(-) at T (K) are 349 and 41 S cm^(2) mol^(-1) repsectively, the dissociation constant of acetic acid is :

    A
    `1.63xx10^(-5)`
    B
    `8.33xx10^(-5)`
    C
    `1.63xx10^(-4)`
    D
    `8.33xx10^(-4)`

    • Similar Questions

    Explore conceptually related problems

    (a) The molar conductivities at infinite dilution of potassium chloride, hydrochloric acid and potassium acetate are 130.1, 379.4 and 95.6 S cm^(2)mol^(-1) respectively. Calculate the value of molar conductivity at infinite dilution for acetic acid. (b) If the molar conductivity of given acetic acid solution is 48.5 S cm^(2)mol^(-1) at 25^(@)C , calculate the degree of dissociation of acetic for acetic acid.

    The molar conductivity of 0.025 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol L^(-1) . Calculate its degree of dissociation and dissociation constant. Given lambda^(@)(H^(+)) = 349.6 S cm^(2) mol^(-1) and lambda^(@) (HCOO^(-)) = 54.6 S cm^(2) mol^(-1) .

    The molar conductivity of "0.025 mol L"^(-1) methanoic acid is "46.1 S cm"^(2)" mol"^(-1) . Calculate its degree of dissociation and dissociation constant. Given Lambda^(@) (H^(+)) =" 349.6 S cm"^(2)" mol"^(-1), Lambda (HCOO^(-)) =" 54.6 S cm"^(2)" mol"^(-1) .

    The conducitivity of a 0.01 M solution of acetic acid at 298 K is 1.65 xx 10^(-4) S cm^(-3) . Calculate (i) Molar conductivity of the solution (ii) degree of dissociation of CH_3 COOH (iii) dissociation constant for acetic acid Given that lambda^(@) (H^(+)) = 349.1 and lambda^(@) (CH_(3) COO^(-)) = 40.9 S cm^(2) mol^(-1) .

    The molar conductivity of 0.025 M methanoic acid (HCOOH) is 46.15" S "cm^(2)mol^(-1) . Calculate its degree of dissociation and dissociation constant. Given lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1) and lambda_((HCOO^(-)))^(@)=54.6" S " cm^(2)mol^(-1) .

    Hydrofluoric acid is a weak acid. At 25^(@)C , the molar conductivity of 0.002 M HF is 176.2 ohm^(-1) cm^(2) mol^(-1) . If its Lamda_(m)^(oo) = 405.2 ohm^(-1) cm^(2) mol^(-1) , calculate its degree of dissociation and equilibrium constant at the given concentration.

    Home
    Profile