25.0 mL of 0.20 M NaOH is neutralized by 12.5 mL of an HCl solution. The molarity of the HCl solution is

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A titration problem is just a stoichiometry problem in which involves the reaction of an acid and a base.  The first thing to do in any stoichiometry problem is to write a balanced chemical equation so that we know the mole ratios among all the substances involved.

NaOH + HCl --> NaCl + H2O

When we write a titration reaction, the products are salt and water.  We can also look at it like a double displacement reaction since Cl- and OH- can't form a compound and H+ and Na+ can't either.

If we look at the reaction above, we see that all the elements are there in the same quanitity on both sides so we have a balanced reaction.  That means that for every 1 mole of NaOH, we need exactly 1 mol of HCl to react with it.  We'll take advantage of this relationship to find the answer to our question.

25.0 mL (1 L/1000 mL)(0.20 mol NaOH/L)(1 mol HCl/1 mol NaOH)

= 0.00500 mol HCl

Which results in the moles of HCl in the 12.5 mL of solution so we can now find molarity (moles solute/L solution)

0.00500 mol / 0.0125 L = 0.400 M HCl solution

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