What volume of 0.500 m hno3(aq) must completely react to neutralize 100.0 milliliters of 0.100 m koh(aq)?
(1) 10.0 ml (3) 50.0 ml
(2) 20.0 ml (4) 500. ml

KOH+ HNO3--> KNO3+ H2O
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is (2) 20.0 mL.

Also, this problem can also be done by using dimensional analysis. 

Hope this would help~