Question

The pH of 0.1 M C4H9NH2(aq) (butylamine) aqueous solution was measured to be 12.04. What is the value of pKb of butylamine

Answer 1

pkb =2.92

Step-by-step explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,

pH = - log [H⁺]

Given that:- pH = 12.04

Also, pH + pOH = 14

So, pOH = 14 - 12.04 = 1.96

The expression of the pOH of the calculation of weak base is:-

`pOH=-log(√(k_b* C))`

Where, C is the concentration = 0.1 M

`1.96=-log(√(k_b* 0.1))`

`\log _(10)\left(√(k_b0.1)\right)=-1.96`

`k_b=(1)/(10^(3.92)\cdot \:0.1)`

`k_b=0.00120`

pkb = -log kb =
`-\log 0.00120` = 2.92