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!
