The equivalent conductances of CH3COONa, HCl and NaCl at infinite dilution are 91,426 and 126 S cm2 eq-1 respectively at 25°C. The equivalent conductance of 1 MCH3COOH solution is 19.55Scm2 eq-1 . The pH of solution is

ΛCH3COONa∞=λCH3COO−0+λNa+0ΛHCl00=λH+0+λCl−0ΛNaCl2∞=λNa+∞+λC−0ΛCH3COOH∞=ΛCH3COONa∞+ΛHCl−∞−ΛNaCl2∞⇒91+426+(−126)=391=ΛCH3COOH∞Λc=19.55α=ΛCΛ∞=19.55391=0.05H+=αC=1×0.05MpH=−log⁡H+=−log⁡5×10−2⇒2−log⁡5=2−0.7=1.3

The equivalent conductances of CH3COONa, HCl and NaCl at infinite dilution are 91,426 and 126 S cm2 eq-1 respectively at 25°C. The equivalent conductance of 1 MCH3COOH solution is 19.55Scm2 eq-1 . The pH of solution is

ΛCH3COONa∞=λCH3COO−0+λNa+0ΛHCl00=λH+0+λCl−0ΛNaCl2∞=λNa+∞+λC−0ΛCH3COOH∞=ΛCH3COONa∞+ΛHCl−∞−ΛNaCl2∞⇒91+426+(−126)=391=ΛCH3COOH∞Λc=19.55α=ΛCΛ∞=19.55391=0.05H+=αC=1×0.05MpH=−log⁡H+=−log⁡5×10−2⇒2−log⁡5=2−0.7=1.3

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The equivalent conductances of CH₃COONa, HCl and NaCl at infinite dilution are 91,426 and 126 S cm² eq^-1 respectively at 25°C. The equivalent conductance of 1 MCH₃COOH solution is 19.55Scm² eq^-1 . The pH of solution is

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5.3

1.3

4.3

2.3

answer is D.

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Detailed Solution

$\begin{array}{l} Λ_{{CH}_{3} COONa}^{\infty} = λ_{{CH}_{3} {COO}^{-}}^{0} + λ_{{Na}^{+}}^{0} Λ_{{HCl}^{0}}^{0} = λ_{H^{+}}^{0} + λ_{{Cl}^{-}}^{0} Λ_{{NaCl}^{2}}^{\infty} = λ_{{Na}^{+}}^{\infty} + λ_{C^{-}}^{0} Λ_{{CH}_{3} COOH}^{\infty} = Λ_{{CH}_{3} COONa}^{\infty} + Λ_{{HCl}^{-}}^{\infty} - Λ_{{NaCl}^{2}}^{\infty} \\ \Rightarrow 91 + 426 + (- 126) = 391 = Λ_{{CH}_{3} COOH}^{\infty} Λ_{c} = 19 .55 α = \frac{Λ_{C}}{Λ^{\infty}} = \frac{19 .55}{391} = 0 .05 [H^{+}] = αC = 1 \times 0 .05 M \\ pH = - \log [H^{+}] = - \log (5 \times 10^{- 2}) \Rightarrow 2 - \log 5 = 2 - 0 .7 = 1 .3 \end{array}$