Question

The barbituate pentobarbitone, which is used as an anesthetic in vertebrates, has a single ionizable group with a pka of 8.0. what percentage of protonated pentobarbitone will be present in a solution that has a ph of 8.7?

Answer 1

Answer : The percentage of protonated pentobarbitone is 20%

Explanation :

Let's say B represents pentobarbitone and BH⁺ represents protonated pentobarbitone.

The protonation reaction can be written as

`B + H_(2)O  \leftrightarrow BH^(+) + OH^(-)`

When the above reaction reaches equilibrium, we get a mixture of B and BH⁺.

This mixture contains a weak base pentobarbitone and its protonated form which is a conjugate acid. Therefore the mixture acts as a buffer.

The pH of a buffer is calculated using Henderson formula which is written below.

`pH = pka + log [(Base)/(acid)]`

We have acid = BH⁺

base = B

pH = 8.7

pka = 8.0

Let us plug in the above values in Henderson equation.

`8.7 = 8.0 + log [(B)/(BH^(+))]`

`8.7 - 8.0 = log [(B)/(BH^(+)) ]`

`0.7 = log [(B)/(BH^(+))]`

`10^(0.7) =(B)/(BH^(+))`

`5.01 =(B)/(BH^(+))`

We want to find fraction of BH⁺ , which is written as BH⁺/B.

Therefore we will inverse the above equation.

`(1)/(5.01) =(BH^(+))/(B)`

`0.200 =(BH^(+))/(B)`

The percentage of protonated pentobarbitone is 0.200 x 100 = 20%