Question

Determine the pKb for the base B given that the equilibrium concentrations are [B]=2.64 M, [HB+]=1.5 M, and [OH−]=1.2 M. A) 3.12 B) 3.96 C) 1.58 D) 2.71

Answer 1

The answer is 0.36

Step-by-step explanation:

The equation describing the ionization of the base is as follows:

B+H2O↽−−⇀HB++OH−

The pKb can be determined from Kb according to the equation:

pKb=−log(Kb)

The Kb for the base can be written as:

Kb=[OH−][HB+][B]

The given equilibrium concentrations can be substituted into the expression above to solve for the value of Kb.

Kb=[OH−][HB+][B]=(0.75)(0.87)(1.5)=0.44

Now the pKb can be calculated.

pKb=−log(Kb)=−log(0.44)=0.36