What volume of 0.410 M KOH is needed to react completely with 23.3 mL of 0.280 M H2SO4 ?

Question

The balanced equation for the neutralization reaction of aqueous H2SO4H2⁢SO4 with aqueous KOHKOH is shown.

H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2SO4(aq)

Answer 1

First determine how many moles of H2(SO4) are available for the reaction:

(23.3 mL / 1000 ml/L) x 0.280 mol/L = 0.006524 moles

1. Convert the volume from mL to L, then multiply volume times molarity to get moles of reactant

Next, H2(SO4) reacts with KOH in a 1:2 ratio. Therefore, we'll need twice as many moles of KOH as H2(SO4):

0.006524 moles x 2 = 0.013048 moles of KOH

Using the molarity of KOH provided in this problem and the moles necessary to fully react with the given H2(SO4), we can calculate the volume of KOH solution required:

Molarity = moles/Liter -> Liters = moles/Molarity

Volume KOH = 0.013048 mol / 0.410 mol/L = 0.031824 L

Finally, we should clean up our answer. To be consistent with the data from the problem, our volume should be converted to mL. And, it should be rounded to 3 SigFigs to match the given volume of H2(SO4):

0.031824 L x 1000 mL/L = 31.824 mL -> 31.8 mL KOH solution required

Answer 2

In order to answer this question you must first find the moles of H2SO4, then use mole to mole conversion from the balanced equation to find moles of KOH, and finally convert moles to volume using the molarity

moles of H2O4:

0.28M=xmol/).0233L

x=0.006524 mol H2SO4

Moles KOH:

0.006524 mol H2SO4 X (2mol KOH/1mol H2SO4)=0.013048 mol KOH

Volume KOH:

0.410M=0.013048/xL

x=0.0318L=31.8mL