`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 `0.01MHA`?

Calcualte acid dissociatioin constant for 0.1 M HCOOH if its solution shows a resistance of 50 Omega filled in a cell having separation beween parallel electrodes 4 cm and cross section area of electrode 10 cm^(2) . [Given: Lambda_(m)^(infty)[Ca(HCCO)_(2)] = 230 Scm^(2)mol^(-) lambda_(m)^(infty)[CaCl_(2)] = 280 Scm^(2) mol^(-1) lambda_(m)^(infty) [HCl] = 425 Scm^(2) mol^(-1) ] In scientific notation, x xx 10^(-y) , find the value of y.

The conductivity of a 0.05 M solution of a weak monobasic acid is 10^(-3) 5 cm^(-1) , If lambda_(m)^(oo) for weak acid 500 5cm^(2) "mol"^(-1) , calculate Ka of weak monobasic acid :

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]

^^_(m) of 1MCH_(3)COOH is 20Scm^(2)mol^(-1) and ^^_(m)^( oo) is 400S cm^(2) mol- 1 The pH of solution will be : a) 1.30 b) 2.30 c) 3.30 (は) 1.70

Calculate Lambda_(m)^(oo) for acetic acid, given, Lambda_(m)^(oo)(HCl)=426 Omega^(-1)cm^(2)mol^(-1),Lambda_(m)^(oo)(NaCl)=126Omega^(-1)cm^(2)mol^(-1) , Lambda_(m)^(oo)(CHCOONa)=91Omega^(-1)cm^(2)mol^(-1)

The conductivity of a weak acid HA of concentration 0.001 mol L^(-1) is 2.0 xx 10^(-5)S cm^(-1) . If Lambda_(m)^(0)(H A) = 190 S cm^(2) mol^(-1) , the ionization constant (K_(a)) of HA is equal to ________ xx 10^(-6) . (Round off to the Nearest Integer)

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

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_((CHM_(3)COO^(c-)))^(@)=54.6Scm^(2)mol^(-1)