A 497 mL sample of 0.9960 M HI is mixed with 443 mL sample of KOH (which has a pH of 13.51). What is the pH of the resulting solution?
What are the steps to solving this problem? I think you need to find the concentration of OH by taking 10^-pOH, and then do an ICE table from there but I keep getting an incorrect answer. Thank you.
Hi is an acid; KOH is a base. They partially neutralize each other.
HI + KOH ==> KI + H2O
You are right. Find the OH^-.
mols HI = M x L = about 0.495
mols KOH = M x L = ?
Subtract to find the one in excess and
M = mols excess/total volume in L.
Then pH = -log(H^+) if HI in excess or
14-pOH = pH if OH in excess.