Question

A weak monobasic acid is 0.05% dissociated in 0.02 M solution.Calcute dissociation constant of the acid.

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Answer 1

⌬ According to the Question :

Given That-

$\\ \longrightarrow\sf \: Ka = {a}^{2}c$

★Need To Find-

• Calculate Dissociation Constant Of Acid

★Formula To Be Used-

$\small\longrightarrow \sf \: \boxed{\bf\pink{a = \frac{Percent \: Dissociation}{100} }}$

❒ Let's Find-

$= \sf \: \frac{0.5}{100}$

$\sf➪ \: 5 × 10⁻⁴$

$\sf➪ \: C = 0.02 M$

$\sf➪ \: 2 × 10⁻²M$

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✠Hence, Kₐ

$\longmapsto \sf \: (5 \times 10⁻⁴) ^{2} \times 2 \times 10 ^{ - 2} \\ \\ \longmapsto \sf \: 25 \times 10⁻⁸ \times 2 \times {10}^{ - 2} \: \: \: \\ \\ \longmapsto \sf \: 50 \times 10 ^{ - 10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longmapsto \sf \: \boxed{\bf\pink{ 5 \times {10}^{ - 9} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

❛❛ Therefore, The Dissociation Constant Of Acid Is = 5 × 10⁻⁹ ❜❜

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Answer 2

${ \underline {\large{ \sf{Solution}}}}$

The dissociation constant of acid is Given by $\rm{Ka=a^2c.}$

We Know,

• $\rm{a=\dfrac{Percent \: Dissosiation}{100}}$

Putting Values

$\dashrightarrow\rm{ = \dfrac{0.05}{100}}$

$\dashrightarrow \rm{= 5 \times 10^{-4}}$

$\dashrightarrow\rm{ \green{C = 0.02 M}}$

$\dashrightarrow \rm{=2 \times 10^{-2} {m}}$

Hence, K =

${ \dashrightarrow{\rm{(5 \times 10^{ -4})²×2×10^{-²} }}}$

${\dashrightarrow \rm{25 × 10^{ - 8} × 2 × 10^{-2}}}$

${ \dashrightarrow\rm{=50\times 10^{-10}}}$

${ \green{ \dashrightarrow\rm{-5 × 10^{⁹}}}}$

Therefore,

• Therefore, The Dissociation Constant Of Acid Is = 5 × 10⁻⁹

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