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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

5.3

1.3

4.3

2.3

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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