In a 10M solution of Lysine in water at pH 11, what it the concentration of the deprotonated side chain species?

3.16m

Explanation:

ph = 11

the question tells us that lysine concentration is equal to 10m

while in water lysine has a pka of 10.5

to get the concentration of deprotonated side species

we introduce the henderson hasselbalch equation

ph = pka + log[proton acceptor]/[proton donor]

inserting values into formula

11 = 10.5 +log[10m]/[x]

we take 10.5 to the left hand side

11-10.5 = log[10m]/[x]

0.5 = log[10m]/[x]

we take the log of both sides

10^0.5 = 10m/x

we cross multiply

x = 10m/10^0.5 = 10^0.5

x = 3.162

approximately 3.16

answer; economic order quantity;