45.00 mL of 1.55×10^−2 M HI(aq) is mixed with 85.00 mL of 1.08×10^−2 M KOH(aq). What is the pH of the solution?

Answer 1

It depends upon which material (acid or base) is in excess.
So calculate moles HI. M x L = moles.
Calculate moles KOH. M x L = moles.
Since HI is a strong acid and KOH is a strong base, and the reaction is 1 mole to 1 mole, subtract the smaller number of moles from the larger number of moles. That will tell you which is in excess, then divide the answer by the number of L of solution (you will need to add volume of HI to volume of KOH to obtain final volume.)

Answer 2

i tried the -log of the solution
which is -log(0.001696153) = 2.77 pH but that is incorrect!

Answer 3

Thats cause you found POH, subtract that from 14

Answer 4

yes 11.23 is correct