A 0.025 M solution of formic acid was taken in a conductance ceIl having a cell constant 0.5 cm^-1. The measured resistance was 605.0 ohms. The ionic mobilities of H⁺ and formate ions are 3.6 × 10^-3 and 5.7 × 10^-4 cm²sec^-1 volt^-l respectively at 298 K. Correct statements among the following is

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$Λ_{\infty} of formic acid (in units of {ohm}^{- 1} m^{- 2} {eq}^{- 1}) is 402.4 \times 10^{- 4}$

The equivalent conductivity of formic acid (in unit of ohm^-1 m²q^-1) at the given concentration is 3.304 × 10^-2

The dissociation constant of the acid is 1.84 × 10^-4

If 100 ml of 0.025 N formic acid is neutralized by 5 mI of 0.5N NaOH,-the specific conductivity of the resulting solution given, $λ_{{Na}^{+}}^{\circ} = 50 .0 {ohm}^{- 1} {cm}^{2} {eq}^{- 1}, λ_{H C O O^{-}}^{0} = 55 o h m^{- 1} c m^{2} e q^{- 1}$
is 2.62 × 10^-1 ohm^-1m^-1.

answer is A, B, C, D.

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Detailed Solution

(1) $Λ_{\infty} (HCOOH) = 96, 500 (36 \times 10^{- 4} + 5.7 \times 10^{- 4}) = 402.4 {ohm}^{- 1} {cm}^{2} {eq}^{- 1}$

$= 402.4 \times 10^{- 4} {ohm}^{- 1} m^{2} e q^{- 1}$

(2) $k = \frac{l}{aR} = \frac{0 .5}{605} = 8 .26 \times 10^{- 4} {Scm}^{- 1}$

$= 8.26 \times 10^{- 2} {sm}^{- 1}$

$Λ_{C} = \frac{k}{C} (\frac{{ohm}^{- 1} m^{- 1}}{{molm}^{- 3}}) = \frac{8.26 \times 10^{- 2}}{0.025 \times 10^{3}}$

$= 3.304 \times 10^{- 3} {Sm}^{2} {eq}^{- 1}$

(3) $K_{a} = \frac{α^{2} c}{1 - α}$

$α = \frac{Λ_{C}}{Λ_{\infty}} = \frac{3.304 \times 10^{- 3}}{402 \times 10^{- 4}} = 8.22 \times 10^{- 2}$

$∴ K_{a} = \frac{{(8.22 \times 10^{- 2})}^{2} \times 0.025}{0.9178} = 1.84 \times 10^{- 4}$

(4) Neglecting dilution, we have 100 mL of 0.025 N sodium formate

$\begin{aligned} k = \sum_{i} \frac{C_{i} λ_{i}}{1000} (or \sum C_{i} λ_{i}) = C_{{Na}^{+}} λ_{{Na}^{+}} + C_{{HCOO}^{-}} λ_{{HCOO}^{-}} \\ = 0.025 \times 10^{3} (50 \times 10^{- 4} + 55 \times 10^{- 4}) = 2.62 \times 10^{- 1} {Sm}^{- 1} \end{aligned}$