Question

Limiting molar conductance of H+ and CH3COO ions are 344 and 40 respectively. Molar conductance of 0.008 M CH3COOH is 48. What will be the value of Ka for CH3COOH:-
(1) 1.4 × 10–5 (2) 1.2 × 10–5 (3) 1.4 × 10–4 (4) 1 × 10–5

Answer 1

4 is a right answer because 4" (and any subsequent words) was ignored because we limit queries to 32 words

Answer 2

Answer:

(3)The value of Ka for acetic acid is

$1.4 \times 10 {}^{ - 4}$

Explanation:

Given that,

The limiting molar conductance of H+ ions=344

The limiting molar conductance of CH3COOH=40

Molar conductance of acetic acid is 48

Molarity of acetic acid is 0.008M

The limiting molar conductance of acetic acid is found by relation

$λ_{m(CH3COOH)∞}\\=λ_{m∞(H+)}+λ_{m(CH3COO−)∞}=344 + 40=384$

Hence,the dissociation constant is found as

$\alpha = \frac{48}{384} = 0.125$

Writing the chemical reactions and their dissociations as follows,we get our requited answer

$\\ CH3COOH⟷CH3COO {}^{ - } +H {}^{ + }   \\ c \: \: \: \: </p><p></p><p>                                \: 0        \: \: \: \: \: \: 0 \: \: \: \\</p><p></p><p>     C(1−α)                     C{ \alpha }        C { \alpha }</p><p></p><p></p><p>$

From the above reaction,we get

$k _{a} = \frac{c { \alpha } \times c { \alpha } }{c(1 - \alpha )} = \frac{c \alpha {}^{2} }{1 - \alpha }$

Substituting the known values,we get

$k _{ a } = \frac{0.008 \times 0.125 {}^{2} }{1 - 0.125} = 0.0001428 \\ = 1.4 \times 10 {}^{ - 4}$

Hence the value of Ka for acetic acid is

$1.4 \times 10 {}^{ - 4}$

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