Question

The conductivities at infinite dilution of nh4cl naoh and nacl are 130 ,128 ,120 of equivalent conductance of n/100 solution of nh4oh is 10 then degree of dissociation of nh4oh at this

Answer 1

At infinite dilution of conductivitiy of NH₄OH,

$\Lambda ^{0}_{NH_{4}OH}=\Lambda ^{0}_{NH^{+}_{4}} +\Lambda ^{0}_{OH^{-}}$

$\Lambda ^{0}_{NH_{4}OH}=\Lambda ^{0}_{NH_{4}Cl} +\Lambda ^{0}_{NaOH} -\Lambda ^{0}_{NaCl}$

$\Lambda ^{0}_{NH_{4}OH}=130 +128 -120$

$\Lambda ^{0}_{NH_{4}OH}=138$

Equivalent conductance of solution of

$\frac{N}{100}$ NH₄OH is 10

$\Lambda_{eq} =10$

$\alpha =\frac{\Lambda _{eq}}{\Lambda ^{0}_{NH_{4}OH}}$

$\alpha$,

is the degree of dissociation of NH₄OH

SO,

$\alpha =\frac{10}{\Lambda ^{0}_{138}$

$\alpha =\frac{10}{\Lambda ^{0}_{138}$

$\alpha =0.072$

That is degree of dissociation of NH₄OH is 0.072.

Answer 2

∧°(NH₄OH) = ∧°(NH₄Cl) + ∧°(NaOH) - ∧°(NaCl)

or, ∧°(NH₄OH) = 130 + 128 - 120

or,∧°(NH₄OH) = 138

Now given, equivalent conductance of N/100 NH₄OH solution = 10

Required formula,

degree of dissociation(α) = Equivalent conductance(∧eq) / Conductance at infinite dilution(∧°)  

Thus for NH₄OH(N/100),

α = 10/138 = 0.072