Home
Class 12
CHEMISTRY
Calculate degree of dissociation of 0.02...

Calculate degree of dissociation of 0.02 M acetic acid at 298 K given that `mho_(m)(CH_(3)COOH)=17.37 cm^(2) mol^(-1),lambda_(m)^(@)(H+)=345.8 S cm^(2) mol^(-1), lambda_(m)^(@)(CH_(3)COO^(-))=40.2 Scm^(2) mol^(-1)`

Text Solution

AI Generated Solution

To calculate the degree of dissociation (α) of 0.02 M acetic acid (CH₃COOH) at 298 K, we can follow these steps: ### Step 1: Identify the given values - **Concentration of acetic acid (C)** = 0.02 M - **Molar conductivity of acetic acid (λ_m)** = 17.37 cm² mol⁻¹ - **Molar conductivity of H⁺ (λ_m^∞(H⁺))** = 345.8 S cm² mol⁻¹ - **Molar conductivity of acetate ion (λ_m^∞(CH₃COO⁻))** = 40.2 S cm² mol⁻¹ ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION A) Objective Type Questions (One option is correct)|43 Videos

  • ELECTROCHEMISTRY

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION B) Objective Type Questions (One option is correct)|25 Videos

  • ELECTROCHEMISTRY

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION - J) (Aakash Challengers Questions)|10 Videos

  • COORDINATION COMPOUNDS

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (Section-J Aakash Challenger Questions )|10 Videos

  • ENVIRONMENTAL CHEMISTRY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (SECTION-D) (Assertion - Reason Type Questions)|4 Videos

Similar Questions

Explore conceptually related problems

Calculate the degree of dissociation (alpha) of acetic acid if its molar conductivity is 39.05 Scm^(2)mol^(-1) . Given lamda^(@)(H^(+))=349.6S" "cm^(2)mol^(-1) and lamda^(@)(CH_(3)CO O^(-))=40.9" S "cm^(2)mol^(-1) .

Calculate the degree of dissociation (alpha) of acetic acid if its molar conductivity (^^_(m)) is 39.05 S cm^(2) mol^(-1) Given lamda^(@) (H^(+)) = 349.6 cm^(2) mol^(-1) and lamda^(2) (CH_(3)COO^(-)) = 40.9 S cm^(2) mol^(-1)

(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 xx 10^(-5)" S "cm^(-1) . Given . lambda_(H^(+))^(@) = 349.6" S " cm^(2)" mol"^(-1), lamda_(CH_(3)COO^(-))^(@)= 40.9" S " cm^(2) " mol "^(-1) (b) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases to a smaller extent while that of ‘A’ increases to a much larger extent comparatively. Which of the two is a strong electrolyte? Justify your answer.

Calculate pH of 0.05 M Acetic acid solution , if K_(a)(CH_(3)COOH) = 1.74 xx 10^(-5) .

The equivalent conductivites of acetic acid at 298K at the concentration of 0.1M and 0.001M are 5.20 and 49.2S cm^(2) eq^(-1) respectively. Calculate the degree of dissociation of acetic acid at the these concentrations. Given that, lambda^(prop) (H^(+)) and lambda^(prop) (CH_(3)COO^(-)) are 349.8 and 40.9 ohm^(-1) cm^(2) mol^(-1) respectively.

The resistance of 0.01 M CH_(2)COOH solution was found to be 2220 ohm in a conductivity cell having cell constant 0.366 cm^(-1) . Calculate: (i) molar conductivity (wedge_(m))" of "0.01 M CH_(3)COOH (ii) wedge_(m)^(oo) (iii) degree of dissociation, alpha and (iv) dissociation constant of the acid. [lambda^(0) (H^(+))=349.1"ohm"^(-1)cm^(2)"mol"^(-1), lambda^(0)(CH_(3)COO^(-)) =40.9 "ohm"^(-1)cm^(2) "mol"^(-1)]

The conductivity of 0.001 "mol" L^(-1) solution of CH_(3)COOH " is " 3.905 xx 10^(-5) "S" cm^(-1) . Calculate its molar conductivity and degree of dissociation (alpha) . "Given" lambda^(@) (H^(+)) = 349.6 "S" cm^(2) "mol"^(-1) " and " lambda^(0) (CH_(3)COO^(-)) = 40.9 "S" cm^(2) per mol)

Molar conductance of 1M solution of weak acid HA is 20 ohm^(-1) cm^(2) mol^(-1) . Find % dissocation of HA: (^^_(m)^(o)(H^(+)) =350 S cm^(2) mol^(-1)),(^^_(m)^(o)(A^(-)) =50 S cm^(2) mol^(-1))

From the following molar conductivities at infinite dilution, Lamda_(m)^(@) " for " Al_(2) (SO_(4))_(3) = 858 S cm^(2) mol^(-1) Lamda_(m)^(@) " for " NH_(4)OH = 238.3 S cm^(2) mol^(-1) Lamda_(m)^(@) " for " (NH_(4))_(2)SO_(4) = 238.4 S cm^(2) mol^(-1) Calculate Lamda_(m)^(@) " for " Al(OH)_(3)

The molar conductivity of 0.25 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol^(-1) . Calculate the degree of dissociation constant. Given : lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) and lambda_(HCOO^(-))^(@)=54.6Scm^(2)mol^(-1)

Home
Profile