What is the value of pKb CH3 COO- if -m- - 390 -m - 7-8 for 0-04 M CH3 COOH at 25 oC-

The correct option is B 9.2As, dissociation factor α = λ_m/λ_m^∞ = 7.8/390 = 0.02 ⇒ K_a (CH_3 COOH) = C nbsp; α^2 = 0.4 × (0.02)^2 = (2)^4 × 10^ 6 ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

What is the v...

Question

What is the value of $p K_{b} (C H_{3} C O O^{-})$ if $λ_{m}^{\infty} = 390$ & $λ_{m} = 7.8$ for 0.04 M $C H_{3} C O O H$ at 25 $^{o}$ C?

9.3

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Solution

The correct option is B 9.2
As, dissociation factor
$α = \frac{λ_{m}}{λ_{m}^{\infty}} = \frac{7.8}{390} = 0.02$
$\Rightarrow K_{a} (C H_{3} C O O H) = C α^{2} = 0.4 \times (0.02)^{2} = (2)^{4} \times 10^{- 6}$
$\Rightarrow p K_{a} = 6 - l o g 2^{4} = 6 - 4 \times 0.3 = 4.8$
$\Rightarrow p K_{b} (C H_{3} C O O^{-}) = 14 - p k_{a} = 9.2$

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Similar questions

Q. What is the value of

p K_{b} (C H_{3} C O O^{-}) i f λ_{m}^{\infty} = 390 a n d λ_{m} = 7.8

for

0.04 M

of a

C H_{3} C O O H

solution at

25^{\circ} C

Q. What is the value of

p K_{b} (C H_{3} C O O^{-}) i f λ_{m}^{\infty} = 390 a n d λ_{m} = 7.8

for 0.04 M of a

C H_{3} C O O H

solution at

25^{\circ} C

Q. What is the value of

p K_{b} (C H_{3} C O O^{-}) i f λ_{m}^{\infty} = 390 a n d λ_{m} = 7.8

for

0.04 M

of a

C H_{3} C O O H

solution at

25^{\circ} C

Q. What is the value of

p K_{b} (C H_{3} C O O^{⊖})

Λ_{m}^{\circ} = 390 S c m^{- 1} m o l^{- 1} a n d Λ_{m} = 7.8 S c m^{- 2} m o l^{- 1}

for

0.04 M o f C H_{3} C O O H

25^{\circ} C

Q. Calculate

α

C H_{3} C O O H

Λ_{m}^{\infty}

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C H_{3} C O O N a

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426, 126, 91

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0.015

M concentration.

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What is the value of pKb(CH3COO−) if λ∞m=390 & λm=7.8 for 0.04 M CH3COOH at 25oC?

The correct option is B 9.2As, dissociation factorα=λmλ∞m=7.8390=0.02⇒Ka(CH3COOH)=Cα2=0.4×(0.02)2=(2)4×10−6⇒pKa=6−log24=6−4×0.3=4.8⇒pKb(CH3COO−)=14−pka=9.2

What is the value of $p K_{b} (C H_{3} C O O^{-})$ if $λ_{m}^{\infty} = 390$ & $λ_{m} = 7.8$ for 0.04 M $C H_{3} C O O H$ at 25 $^{o}$ C?