# Calculate the milliliters of 0.8916 M KOH needed to neutralize 25.00 mL of 0.6552 M H2SO4 (to produce K2SO4 and H2O)?

First let's write out a balanced chemical equation:

H2SO4 + 2KOH --> K2SO4 + 2H2O

So for each mole of sulfuric acid (H2SO4), you will need two moles of KOH to neutralize it.  So now let's calculate the moles of sulfuric acid that we have:

0.025 L * (0.6552 moles/L)...

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First let's write out a balanced chemical equation:

H2SO4 + 2KOH --> K2SO4 + 2H2O

So for each mole of sulfuric acid (H2SO4), you will need two moles of KOH to neutralize it.  So now let's calculate the moles of sulfuric acid that we have:

0.025 L * (0.6552 moles/L) = 0.01638 moles acid

So we need two moles of KOH per mole of acid, so that means we need 0.01638 * 2 = 0.03276 moles KOH.  So now we can calculate the volume of KOH solution that we need:

0.03276 moles * (1 L/0.8916 moles) = 0.03674 L KOH = 36.74 mL of KOH solution.

The answer is 36.74 mL of KOH solution.

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