What is the value of pK b CH 3 COO -if - m -390 and - m -7-8 for 0-04 M of a CH 3 COOH solution at 25- C -A- 9-3B- 9-2C- 4-8D- 4-7

The correct option is B 9.2α =λ _mλ _m=7.8390=0.02 K_a(CH_3COOH)=Cα ^2=0.04× 4 nbsp;× 10^ 4 = 16 ×10^ 6 pK_a=4.8 pK_a+pK_b =14 so pK_b=9.2 Hence, optio ...

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Standard XII

Mathematics

Integration by Parts

What is the v...

Question

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 $25^{\circ} C$ ?

9.3

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Solution

The correct option is B 9.2
$α = \frac{λ_{m}}{λ_{m}} = \frac{7.8}{390} = 0.02$

$K_{a} (C H_{3} C O O H) = C α^{2} = 0.04 \times 4 \times 10^{- 4} = 16 \times 10^{- 6}$

$p K_{a} = 4.8$

$p K_{a} + p K_{b} = 14$

so $p K_{b} = 9.2$

Hence, option B is correct.

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Similar questions

Q. What is the value of

p K_{b} (C H_{3} C O O^{-})

λ_{m}^{\infty} = 390

λ_{m} = 7.8

for 0.04 M

C H_{3} C O O H

at 25

^{o}

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. Which of the following expressions is correct for ‘

K a

’ in terms of

Λ^{\circ}

and

Λ

, where ‘

C

’ is the concentration.

Q. Given the limiting molar conductivity as

_{m}

(HCl)

= 425.9 Ω^{- 1} c m^{2} m o l^{- 1}

_{m}

(NaCl)

= 126.4 Ω^{- 1} c m^{2} m o l^{- 1}

_{m}

(CH3 COONa)

= 91 Ω^{- 1} c m^{2} m o l^{- 1}

The molar conductivity, at infinite dilution, of acetic acid (in

Ω^{- 1} c m^{2} m o l^{- 1}

) will be

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What is the value of pKb(CH3COO−)ifλ∞m=390andλm=7.8 for 0.04 M of a CH3COOH solution at 25∘C?

The correct option is B 9.2α=λmλm=7.8390=0.02 Ka(CH3COOH)=Cα2=0.04×4×10−4=16×10−6 pKa=4.8 pKa+pKb=14 so pKb=9.2 Hence, option B is correct.

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 $25^{\circ} C$ ?

The correct option is B 9.2
$α = \frac{λ_{m}}{λ_{m}} = \frac{7.8}{390} = 0.02$

$K_{a} (C H_{3} C O O H) = C α^{2} = 0.04 \times 4 \times 10^{- 4} = 16 \times 10^{- 6}$

$p K_{a} = 4.8$

$p K_{a} + p K_{b} = 14$

so $p K_{b} = 9.2$

Hence, option B is correct.