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You want to determine the nitric acid content in a sample using an acid-base titration. You take 10.00 mL of the sample and titrate it to an end point with 7.47 mL of 0.25 M KOH. What is the molar concentration of nitric acid in the sample? a. 0.505 M b. 0.258 M c. 0.187 M d. 0.383 M

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Answer:

c. 0.187

Explanation:

The strategy here is for us to utilze the fact that at the equivalence point all of the acid would have been consumed by the neutralization reaction of KOH and HNO₃, so

KOH + HNO₃ ⇒ KNO₃ + H₂O

moles of KOH consumed = moles of HNO₃ present

But molarity = number of moles/V(L), so

moles KOH = 0.25 mol/L x 7.47 x 10⁻³ L = 1.87 x 10⁻³ mol KOH

mol HNO₃ = 1.87 x 10⁻³ mol

M HNO₃ = mol HNO₃ / V (L) = 1.87 x 10⁻³ / 1.0 x 10⁻² L = 0.187 M

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