SATHEE: Electrochemistry - Result Question 8

Electrochemistry - Result Question 8

####9. The equivalent conductance of $\frac{M}{32}$ solution of a weak monobasic acid is $8.0 m h o c m^{2}$ and at infinite dilution is $400 m h o c m^{2}$ . The dissociation constant of this acid is:

(a) $1.25 \times 10^{- 6}$

(b) $6.25 \times 10^{- 4}$

(d) $1.25 \times 10^{- 5}$

[2009]

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Solution:

(d) Degree of dissociation,

$\begin{aligned} α = \frac{Λ}{Λ_{\infty}} = \frac{8.0}{400} = 2 \times 10^{- 2} \\ K_{a} = \frac{C α^{2}}{(1 - α)} \approx C α^{2} = \frac{1}{32} \times (2 \times 10^{- 2})^{2} \end{aligned}$

$= 1.25 \times 10^{- 5}$

Electrochemistry - Result Question 8

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Electrochemistry - Result Question 8

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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