Answer:
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
