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

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