The specific conductance of an acid HA of 0-1 M concentration is 3-75 - 10-4-1 cm -1- If - HA -250 -1 cm 2 mol -1- The dissociation constant K a of HA is y - 10-5- Find the value of y

Λ^0.1M(HA) = k × 1000/C = 3.75 × 10^ 4× 1000/0.1Ω^ 1 cm^2 mol^ 1 = 3.75 Ω^ 1 cm^2 mol^ 1 α = λ^0.1M/λ^∞ = 3.75/250 = 0.015 Ka = C α^2 = 0.1 (0.015)^2 = 2.25 × ...

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Standard XII

Chemistry

Conductivity of Ionic Solutions

The specific ...

Question

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

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Solution

$λ^{0.1 M} (H A) = \frac{k \times 1000}{C} = \frac{3.75 \times 10^{- 4} \times 1000}{0.1} Ω^{- 1} c m^{2} m o l^{- 1} = 3.75 Ω^{- 1} c m^{2} m o l^{- 1} α = \frac{λ^{0.1 M}}{λ^{\infty}} = \frac{3.75}{250} = 0.015 K a = C α^{2} = 0.1 (0.015)^{2} = 2.25 \times 10^{- 5}$

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The specific conductance of an acid HA of 0.1 M concentration is 3.75×10−4Ω−1cm−1. If λ∞(HA)=250Ω−1cm2mol−1. The dissociation constant Ka of HA is y×10−5. Find the value of y

λ0.1M(HA)=k×1000C=3.75×10−4×10000.1Ω−1cm2mol−1=3.75Ω−1cm2mol−1α=λ0.1Mλ∞=3.75250=0.015Ka=Cα2=0.1(0.015)2=2.25×10−5

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

$λ^{0.1 M} (H A) = \frac{k \times 1000}{C} = \frac{3.75 \times 10^{- 4} \times 1000}{0.1} Ω^{- 1} c m^{2} m o l^{- 1} = 3.75 Ω^{- 1} c m^{2} m o l^{- 1} α = \frac{λ^{0.1 M}}{λ^{\infty}} = \frac{3.75}{250} = 0.015 K a = C α^{2} = 0.1 (0.015)^{2} = 2.25 \times 10^{- 5}$