Home
Class 11
CHEMISTRY
Molar conductivity of aqueous solution o...

Molar conductivity of aqueous solution of `HA` is `200Scm^(2)mol^(-1),pH` of this solution is `4`
Calculate the value of `pK_(a)(HA)` at `25^(@)C`.
Given `^^_(M)^(oo)(NaA)=100scm^(2)mol^(-1),`
`^^_(M)^(oo)(HCl)=425Scm^(2)mol^(-1),`
`^^_(M)^(oo)(NaCl)=125 Scm^(2)mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 1 (Q.1 To Q.30)|1 Videos

  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 1 (Q.91 To Q.120)|1 Videos

  • DILUTE SOLUTION

    NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos

  • GASEOUS STATE

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos

Similar Questions

Explore conceptually related problems

Calculate the degree of dissociation (alpha) of CH_(3)COOH at 298K. ltbr Given that ^^_(CH_(3)COOH)^(oo)=11.75cm^(2)mol^(-1) ^^_(CM_(3)COO^(-))^(oo)=40.65cm^(2) mol ^(-1) ^^_(H^(+))^(0)=349.15 cm^(2)mol ^(-1)

The density of 3M solution of NaCL is 1.25 g mL^(-1) . Calculate molality of the solution.

Molarity of a 50 mL H_2SO_4 solution is 10.0 M. If the density of the solution is 1.4 g/cc, calculate its molality

Molar conductivity of a solution is 1.26 xx 10^(2)Omega^(-1) cm^(2) "mol""^(-1) . Its molarity is 0.01M. Its specific conductivity will be

The molar conductivity of 0.025mol L^(-1) methanoic acid is 46.1 S cm^(2)mol ^(-1). Calculate its degree of dissociation and dissociation constant. Given, lamda^(0)(H^(+))=349.6S cm^(2) mol ^(-1) and lamda^(0)(HCOO^(-))=54.6S cm^(2)mol^(-1)

The molar conductivity of 0.024 "mol" L^(-1) methanoic acid is 46.1 S cm^(-2)"mol"^(-1) . Calculate its degree of dissociation and dissociation constant. Given lambda^(0)(H^(+) = 349.6 S cm^(2) mol^(-1) and lambda^(0)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1) .

If the solubility product of MOH is 1xx10^(-10) mol^(2).dm^(-2) . Then the p^(H) of the its aqueous solution will be