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

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Answer 1 · 2016-11-13T06:53:06

32.2mL

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_2SO_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