# What volume of 0.320 M KOH is needed to react completely with 23.4 mL of 0.220 M H_2SO_4?

Nov 13, 2016

32.2mL

#### Explanation:

First we need to write out the balanced equation:
2KOH + H_2SO_4 → 2H_2O + K_2SO_4

This shows us that we need 2 moles of KOH for each mole of sulfuric acid. Using milliequivalents we have:
(mL X M)2 = mL X M

0.00234L X 0.220mol/L = 0.000515 moles ${H}_{2} S {O}_{4}$

From our equation, this will require 0.00103 moles of KOH.

0.00103mole/0.320 mol/L = 0.00322L, or 32.2mL

Check:
(32.2 x 0.320)/2 = 23.4 x 0.220 ; 10.304/2 = 5.148 ; 5.15 = 5.15