Question

If 25 mL of a 1 M HI solution is added to 25 mL of a 1 M KOH solution, the resulting solution would be

Answer 1

The reaction equation is:

HI(aq) + KOH(aq) → KI(aq) + H2O

1 mol HI reacts with 1 mol KOH

In your problem you have:

Mol HI in 10 mL of 1M solution = 10mL/ 1000 mL/L * 1 mol /L = 0.01 mol HI

Mol KOH in 5 mL of 1 M solution = 5 mL / 1000 mL/L * 1 mol /L = 0.005 mol HI

The 0.005 mol of KOH will neutralise 0.005 mol of HI - leaving 0.005 mol HI unreacted in 15 mL solution

This solution will be acidic

I guess that that answers your question - But you may be interested in calculating the pH of the solution

Have a great day!