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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