Question

The pH of 0.5 m aqueous solution of weak base is observed to have a pH of 12.3.What is the value of Kb for this base?​

Answer 1

Answer:

• We know that, pH+pOH=14
• or pOH=14−pH=14−12−2
• So, [OH^−]=10^−2
• (C_2H_5)_2NH+H_2O⇌(C_2H_5)_2NH_2^++OH^−
• At equilibrium,[(C_2H_5)_2NH]=0.05−x,[(C_2H_5)_2NH_2^+]=x,[OH^−]=x
• where x=0.01mol
• Kb=[(C_2H_5)_2NH][(C_2H_5)_2NH