H2 ----> 2H++2e-
Ans. pH = 4.237
we have given that
electrolyitc conductivity =0.20mol/L
conductivity = 2.48 x 10-2 ohm-1cm-1
thus apply the formula
here M is molar conductivity.
Ans. 124 ohm–1 cm2 mol–1
Calculate the molar conductivity at infinite dilution of acetic acid from the following data:
Λ∞m (HCl) = 426 ohm–1 cm2 mol–1, Λ∞m CH3COONa = 91 ohm–1 cm2 mol–1 and Λ∞m(NaCl) = 126 ohm–1 cm2 mol–1.
Acetic acid is weak electrolyte than HCl and NaCl which is strong electrolyte.
Acoording to kohlransch's law
=91-126+426 =391
Ans. 391 ohm–1 cm2 mol-1
We have given
E°Ni2+/Ni = -0.25 V
E°Cu2+/Cu = + 0.34 V,
Ecell0 =
= 0.34-(-0.25)
= 0.34+0.25 =0.59
R = 8.314 J K–1 mol–1
F = 96500 C mol–1.
T=25celcius =273+25 =298 Kelvin
n= 2
= Antilog[19.956]
Ans. 9.07 x 1019