The conductivity of 0.01 M solution of acetic acid at 25°C is 1.63 x 10–4 s cm–1. Given:
Λ°m (HCl) = 426 s cm2 mol–1, Δ°m (Na AC) = 91.5 cm2 mol–1
Λ°m (NaCl) = 126.5 cm2 mol–1 Calculate:
(a) the molar conductivity of acetic acid
(b) the degree of dissociation of acetic acid.
(c) the dissociation constant.
 (d) the pH of 0.01 M solution of acetic acid.

 

a) The molar conductivity of acetic acid given by;

Λm =103kM    =103 x1.63x10-40.01=16.3 S cm2mol-1b) The degree of dissociation (α) is α =ΛmΛm0Λm0 =Λm0(HCl) +Λm0(NaAc) -Λm0(NaCl)   =426 +91-126 =391S cm2 Mol-1Therefore α =ΛmΛm0 =16.3391 =0.042c) K=21-α = 0.01 x(0.042)21-0.042 =1.84 x 10-5d) [H+] =Cα =0.01 x0.042=4.2 x10-4MpH =-log [H+] =-log(4.2 x10-4) =3.38



207 Views

How much charge is required for the following reduction of
(i) 1 mol of Al3+ to Al
(ii) 1 mol of Cu2+ to Cu
(iii) 1 mol of MnO4– to Mn2+


(i) Al3+ + 3e → Al
1 mol Al3+ for reduction requires 3 mol e or 3F electricity
1F = 96500C
∴ 3F = 3 x 96500 = 289500 C
(ii) Reduction of 1 mol Cu2+ to Cu requires 2 mol electrons
2 mol electrons = 2F = 2 x 96500 = 193000 C
(iii) In the reduction of 1 mol MnO4 to Mn2+, there is net gain of 5e
MnO4 + 5e + 8H+ → Mn2+ + 4H2O
5F = 5 x 96500 = 482500 C

326 Views

Advertisement

Draw curves to show how the molar conductance of strong electrolytes varies with dilution.


When a plot is drawn between the molar conductivity versus square root of concentration for strong electrolytes, a curve (nearly straight line) shown in the given Fig is obtained. It is clear from the figure that molar conductance has higher value and increases linearly. Slight increase in the value of Λwith dilution is due to the decrease in interionic attraction with dilution i.e., decrease in concentration.

When a plot is drawn between the molar conductivity versus square roo

The molar conductivity at infinite dilution (Λm) can be obtained by extrapolating the above graph of strong electrolytes to zero concentration.
Kohlrausch Law of Independent migration of ions: The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Thus, if λ°Na+ and λCl are molar limiting conductivity of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation:
Λ°NaCl = λ°Na + λCl
In general, if an electrolyte on dissociation gives v+ cations and v anions then its limiting molar conductivity is given by
Λ° = v + λ°+ + V λ°
Here, λ°+ and λ° are the limiting molar conductivities of the cation and anion respectively.
Kohlrausch’s law helps us to calculate
(i) Determination of molar conductivities of weak electrolytes at infinite dilution.
(ii) Determination of the degree of dissociation of electrolytes.
Degree of dissociation: It is ratio of molar conductivity at a specific countraction ‘C’ to the molar conductivity at infinite dilution. It is denoted by α.
i.e.,                   straight alpha space equals space fraction numerator straight A subscript straight m superscript straight C over denominator straight A subscript straight m superscript infinity end fraction

458 Views

Advertisement
Calculate the standard electrode potential of the Mg2+ / Mg electrode for a cell in which the cell reaction is
Mg(s) + 2Ag+(aq) → Mg2+(aq) + 2Ag(s)
Given that, [Mg2+] = 0.1 M, [Ag+] = 0.01 M
Ag/Ag = + 0.80, Ecell = + 2.90 V


The cell is
Mg(s) | Mg2+ (aq) (0.1 M) || Ag+ (aq) (0.01 M) | Ag(s)
and the Ecell is given by

Ecell = E°cell - 0.0592log Mg2+Ag+22.90 = E°cell - 0.0592log(0.1)(0.01)22.90 = E°cell - 0.0592×3

                        2.90 = E°cell - 0.0885E°cell = 2.988 V

But               E°cell = E°Ag+/Ag - E°Mg2+/Mg
              E°Mg2+/Mg = + 0.80 - 2.988E°Mg2+/Mg = -2.188 V

991 Views

A Cell is prepared by dipping a copper rod in 0.01 M copper sulphate solution, and zinc rod in 0.02 M ZnSO4 solution. The standard reduction potentials of copper and zinc are + 0.34 V and – 0.76 V respectively.
(a) What will be the cell reaction?
(b) How will the cell be represented?
(c) What will be the emf. of the cell?


(a) As E°Zn2+/Zn < E°Cu2+/Cu . Zinc electrode acts as anode and copper electrode acts as a cathode. The cell reaction is
Zn(s) + Cu2+(aq) (0.01 M) → Zn2+(aq) (0.02 M) + Cu(s)
(b) The cell is represented as
Zn(s) | Zn2+(aq) (0.02 M) || Cu2+(aq) (0.01 M) + Cu(s)
(c) The emf of the cell can be calculated by using Nernst equation
E= E°-0.059nlogZn2+Cu2+E  = +1.10 - 0.0592log 0.020.01E =  +1.10 - 0.0592×0.301E = +1.10 - 0.0089  or E = + 1.091 V

384 Views

Advertisement