Question

The specific conductance of 2.5 × 10^-4 M formic acid is 5.25 × 10^-5 ohm^-1 cm^-1. Calculate its molar conductivity and degree of dissociation.

Given `λ°_("H"^+)` = 349.5 ohm^-1 cm² mol^-1 and

`λ°_("HCOO"^-)  = 50.5 " ohm"^-1 "cm"^2  "mol"^-1`

Answer 1

Given: K = 5.25 × 10^-5 ohm^-1 cm^-1

C = 2.5 × 10^-4 M

Formula used `lambda_m = (1000 xx "K")/"C", alpha = lambda_m/lambda_0`

Molar conductivity `(lambda_m) = (1000 xx "K")/"C"`

`(lambda_m) = (100 xx 5.25 xx 10^-5)/(2.5 xx 10^-4)`

= 210 s cm² mol^-1

\[\ce{\lambda^\circ_{{\phantom{..}}CH_3COOH} = (\lambda^\circ_{\phantom{..}H} + \lambda^\circ_{\phantom{..}HCOO^-})}\]

= (349.5 + 50.5)

Degree of dissociation:

`alpha = lambda_m/lambda_0 = 210/400` = 0.525

α = 52.5%