The isoelectric point is an average of all the pKas. 9 is not the pI.
Not quite.. Your right pI is not 9.. but its not the average of all 3 pKa's.. is the average of the two pKa's involved in the deprotonation to get Lysine to a neutral state.. so pI is (9+11)/2 =10 . but still, im not understanding how the question asks how many mol of HCl needed.. because in its most protonated form Lysine is at a low pH (much lower than the pI pH10).. adding HCl will lower the pH more, right?