`Lambda_(m)^∞` (weak mono basic HA acid)`=390.7Scm^(2)mol^(-1)``Lambda_(m)` of `HA` at `0.01M` is `3.907Scm^(2)mol^(-1)` Find `pH` of `0.01M HA ?

The conductivity of a 0.05 M solution of a weak monobasic acid is 10^(-3) S cm^(-1) , If lambda_(m)^(oo) for weak acid 500 Scm^(2) "mol"^(-1) , calculate Ka of weak monobasic acid :

Certain weak acid has P^(ka) = 4. Find pH of 0.01M NaA(ag)

Conductivity of an aqueous solution of 0.1M HX (a weak mono-protic acid) is 5 xx 10^(-4) Sm^(-1) Given: ^^_(m)^(oo) [H^(+)] = 0.04 Sm^(2) mol^(-1): ^^_(m)^(oo) [X^(-)] = 0.01 Sm^(2)mol^(-1) Find pK_(a)[HX]

Acetic acid is a weak acid. The molar conductances of 0.05 M acetic acid at 25^@C is 7.36 S " cm"^(2) "mol"^(-1) . If its ^^^_m^(oo) is 390.72 S cm^(2)"mol"^(-1) . The value of equilibrium constant is

wedge^(@)._(m) for CaCl_(2) and MgSO_(4) from the given data. lambda_(Ca^(2+))^(@)=119.0S cm^(2)mol^(-1) ltbr. lambda_(Cl^(c-))^(@)=76.3S cm^(2)mol^(-1) lambda_(Mg^(2+))^(@)=106.0S cm^(2)mol^(-1) lambda_(SO_(4)^(2-))^(@)=160.0 cm^(2)mol^(-1)

If the equivalent conductance of (M/20) solution of a weak monobasic acid is 25.0 S cm^2 eq^(-1) and at infinite dilution is 500 Scm^2eq^(-1) , then the dissociation constant of the acid is

Equivalent conductane of 0.1 M HA (weak acid) solution is 10 S cm^(2) "equivalent"^(-1) and that at infinite dilution is 200 S cm^(2) " equivalent"^(-1) Hence pH of HA solution is

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)

The molar conductivity of 0.05 M solution of weak acid is 16.6Omega^(-1)cm^(-2)mol^(-1). Its molar conductivity at infinite dilution is 390.7Omega^(-1)cm^(-2)mol^(-1) . The degree of dissociation of weak acid is

The P^(H) of a weak mono basic acid is 5. The degree of ionisation of acid in 0.1 M solution is