Question

50 mL of 1.50* 10^-2 M HI(aq) is mixed with 75.00 mL of 1.10* 10^-2 M KOH(aq).

What is the pH of the Final solution

Answer 1

First you must recognize that HI and KOH are strong acid + strong base respectively. The product is KI (a neutral salt) + H2O; therefore, the pH will depend entirely upon the reagent (HI or KOH) that is in excess. How do you determine the one in excess?

Calculate moles HI initially. That is M x L = ?
Calculate moles KOH initially. That is M x L = ?

Now subtract one from the other to see which one is in excess (obviously, the other one goes to zero since all of it will be used). Then moles/volume = M. Finally, if KOH is in excess, the pOH is the -log(OH^-) remaining or if HI is in excess, pH = -log(H^+) remaining.
Post your work if you get stuck.