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What volume of 0.40 M NaOH would be required to neutralize 100 cm3 of 0.25 M HCl?

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  1. What volume of 0.40 M NaOH would be required to neutralize 100 cm3 of 0.25 M HCl?

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First let's look at the equation for the reaction of HCl and NaOH.

HCl + NaOH -> H2O +NaCl

Since HCl is a strong acid and NaOH is a strong base, they react together to basically form saltwater.  More importantly, they react together in a 1:1 molar ratio.  So now let's find how many moles of HCl we have.  Please note that one cubic centimeter is equal to 1 mL.

0.25 moles HCl/liter * 0.1 liters = 0.025 moles HCl

Since we know that one mole of HCl requires one mole of NaOH for neutralization, we need 0.025 moles of NaOH.  Now divide by the molarity of the NaOH solution to find the volume required.

0.025 moles NaOH * (1 liter/0.4 moles NaOH) = 0.0625 liters

So you would need 62.5 liters of the 0.4 M NaOH solution.

 

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