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Electrochemistry question

2019 · 12 Jan · Shift 2 · Q23

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Electrochemistry
2019 · 12 Jan · Shift 2 · Q23

Electrochemistry question

2019 · 12 Jan · Shift 2 · Q23

JEE MainChemistryElectrochemistryMCQ+4 / −1

\wedge _m^ \circ

for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1, respectively. If the conductivity of 0.001 M HA is- 5

\times

10–5 S cm–1, degree of dissociation of HA is -

A
0.50
B
0.125
C
0.25
D
0.75

View written solutionFree

Correct answer: B

Use Kohlrausch’s law to find $\Lambda_m^\circ$ of $\mathrm{HA}$ .

Given:

$\Lambda_m^\circ(\mathrm{NaCl}) = 126.4\ \mathrm{S\,cm^2\,mol^{-1}}$
$\Lambda_m^\circ(\mathrm{HCl}) = 425.9\ \mathrm{S\,cm^2\,mol^{-1}}$
$\Lambda_m^\circ(\mathrm{NaA}) = 100.5\ \mathrm{S\,cm^2\,mol^{-1}}$

By Kohlrausch’s law:

\Lambda_m^\circ(\mathrm{HA})=\Lambda_m^\circ(\mathrm{HCl})+\Lambda_m^\circ(\mathrm{NaA})-\Lambda_m^\circ(\mathrm{NaCl})

So,

\Lambda_m^\circ(\mathrm{HA})=425.9+100.5-126.4=400.0\ \mathrm{S\,cm^2\,mol^{-1}}

Find molar conductivity of $0.001\ \mathrm{M}$ $\mathrm{HA}$ .

Given conductivity:

\kappa = 5\times 10^{-5}\ \mathrm{S\,cm^{-1}}

Molar conductivity is:

\Lambda_m = \kappa\times \frac{1000}{C}

where $C=0.001\ \mathrm{mol\,L^{-1}}$ .

Thus,

\Lambda_m = 5\times 10^{-5}\times \frac{1000}{0.001}

\Lambda_m = 5\times 10^{-5}\times 10^6 = 50\ \mathrm{S\,cm^2\,mol^{-1}}

Degree of dissociation $\alpha$ for weak electrolyte:

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}

Hence,

\alpha = \frac{50}{400}=0.125

Check options

A: $0.50$ ❌
B: $0.125$ ✅
C: $0.25$ ❌
D: $0.75$ ❌

Therefore, the degree of dissociation is:

\boxed{0.125}

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