1.

Specific conductance of 0.1M HA is 3.75×10−4ohm−1cm−1. If λ∞ of HA is 250ohm−1cm2mol−1, then dissociation constant Ka of HA is

Answer»

Specific conductance of 0.1M HA is 3.75×104ohm1cm1. If λ of HA is 250ohm1cm2mol1, then dissociation constant Ka of HA is



Discussion

No Comment Found