Specific conductance of 0-1 M HA is 3-75 - 10-4 ohm -1 cm -1- If - HA -250 ohm -1 cm 2 mol -1- the dissociation constant K a of HA is -A- 1-0 - 10-5B- 2-25 - 10-5C- 2-25 - 10-4D- 2-25 - 10-13

The correct option is B 2.25 × 10^ 5 λ^m = 1000k/0.1 = 1000 × 3.75 × 10^ 4/0.1 = 3.75; α = λ_m/λ^∞_m = 3.75/250 = 1.5 × 10^ 2; K_a = C α^2 = 0.1 × (1.5 × 10^ ...

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Standard XII

Mathematics

Particular Solution of a Differential Equation

Specific cond...

Question

Specific conductance of 0.1 M HA is $3.75 \times 10^{- 4} o h m^{- 1} c m^{- 1} .$ If $λ^{\infty} (H A) = 250 o h m^{- 1} c m^{2} m o l^{- 1},$ the dissociation constant $K_{a}$ of HA is :

$1.0 \times 10^{- 5}$

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$2.25 \times 10^{- 4}$

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$2.25 \times 10^{- 5}$

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$2.25 \times 10^{- 13}$

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Solution

The correct option is C
$2.25 \times 10^{- 5}$

$λ^{m} = \frac{1000 k}{0.1} = \frac{1000 \times 3.75 \times 10^{- 4}}{0.1} = 3.75; α = \frac{λ_{m}}{λ_{m}^{\infty}} = \frac{3.75}{250} = 1.5 \times 10^{- 2}; K_{a} = C α^{2} = 0.1 \times (1.5 \times 10^{- 2})^{2} = 2.25 \times 10^{- 5}$

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Similar questions

Q. Specific conductance of 0.1M HA is

3.75 \times 10^{- 4} o h m^{- 1} c m^{- 1}

. If

λ^{\infty}

of HA is

250 o h m^{- 1} c m^{2} m o l^{- 1}

, then dissociation constant Ka of HA is

Q. What is the specific conductance of 0.1 M HA solution, if

λ^{\infty} (H A) = 250 Ω^{- 1} c m^{2} m o l^{- 1}

and the dissociation constant

K_{a}

of HA is

2.25 \times 10^{- 5}

? (Assume

α << 1)

Specific conductance of 0.1 M HA is $3.75 \times 10^{- 4} o h m^{- 1} c m^{- 1} .$ If $λ^{\infty} (H A) = 250 o h m^{- 1} c m^{2} m o l^{- 1},$ the dissociation constant $K_{a}$ of HA is :

Q. The specific conductance of an acid HA of 0.1 M concentration is

3.75 \times 10^{- 4} Ω^{- 1} c m^{- 1}

. If

λ^{\infty} (H A) = 250 Ω^{- 1} c m^{2} m o l^{- 1}

. The dissociation constant

K_{a}

of HA is

y \times 10^{- 5}

. Find the value of y

Specific conductance of $0.1 M C H_{3} C O O H a t 25^{\circ}$ is $3.9 \times 10^{- 4} o h m^{- 1} c m^{- 1}$ . If $λ^{\infty} (H_{3} O^{+}) a n d λ^{\infty} (V H_{3} C O O^{-})$ at $25^{\circ} C$ are 349.0 and 41.0 $o h m^{- 1} c m^{2} m o l^{- 1}$ respectively, degree of ionization of $C H_{3} C O O H$ at the given concentration is :

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Specific conductance of 0.1 M HA is 3.75×10−4ohm−1cm−1. If λ∞(HA)=250 ohm−1cm2mol−1, the dissociation constant Ka of HA is :

The correct option is C 2.25×10−5 λm=1000k0.1=1000×3.75×10−40.1=3.75;α=λmλ∞m=3.75250=1.5×10−2;Ka=Cα2=0.1×(1.5×10−2)2=2.25×10−5

Specific conductance of 0.1 M HA is $3.75 \times 10^{- 4} o h m^{- 1} c m^{- 1} .$ If $λ^{\infty} (H A) = 250 o h m^{- 1} c m^{2} m o l^{- 1},$ the dissociation constant $K_{a}$ of HA is :

The correct option is C
$2.25 \times 10^{- 5}$

$λ^{m} = \frac{1000 k}{0.1} = \frac{1000 \times 3.75 \times 10^{- 4}}{0.1} = 3.75; α = \frac{λ_{m}}{λ_{m}^{\infty}} = \frac{3.75}{250} = 1.5 \times 10^{- 2}; K_{a} = C α^{2} = 0.1 \times (1.5 \times 10^{- 2})^{2} = 2.25 \times 10^{- 5}$