The side chain of tyrosine has a pKa of about 10. What percent of tyrosine side chains would be deprotonated at pH 8.5?

Explanation:

Let the tyrosine molecule be represented by TH . It will ionise in water as follows

TH ⇄ T⁻ + H⁺

Let C be the concentration of undissociated TH and α be the degree of dissociation

TH       ⇄       T⁻ + H⁺

c                    0       0   ( before )

c( 1-α )            αc      αc ( after ionisation)

Ka = α²c² / c( 1-α )

= α²c  ( neglect α in the denominator as it is very small )

pKa = 10

Ka  = 10⁻¹⁰

pH = 8.5

H⁺ = 10⁻⁸°⁵

αc = 10⁻⁸°⁵

α²c =Ka = 10⁻¹⁰

α x10⁻⁸°⁵ = 10⁻¹⁰

α = 10⁻¹⁰⁺⁸°⁵

= 10⁻¹°⁵ = 1 / 31.62

Percentage of dissociation = 100 / 31.62

= 3.16 %

percent of  tyrosine side chains   deprotonated

the answers are

c and c