Question

To calculate the concentration of a solution using acid–base titration data.

In an acid–base titration, an acid (or base) of known concentration is added to a base (or acid) of unknown concentration until the number of moles of H⁺ and OH^- are equal, a condition called the equivalence point. Since you know the number of moles of H⁺ (or OH^- ) that you added, you can determine the number of moles of OH^- (or H⁺) in the unknown solution.

For example, a solution containing 1 mol of H₂SO₄ contains 2 mol of ionizable hydrogen atoms, and would therefore require 2 mol of NaOH for neutralization.

A chemist needs to determine the concentration of a solution of nitric acid, HNO₃. She puts 715 mLof the acid in a flask along with a few drops of indicator. She then slowly adds 0.600 M Ba(OH)2 to the flask until the solution turns pink, indicating the equivalence point of the titration. She notes that 135 mL of Ba(OH)₂was needed to reach the equivalence point.

Solution map

In this titration, the concentration of base is known and can be used to calculate the unknown acid concentration:

concentration of base ⟶ moles of base ⟶ moles of acid ⟶ concentration of acid

b. What is the concentration of HNO₃ if 0.162 mol are present in 715 mL of the solution?

Express your answer with the appropriate units.

Hint 2. Convert the volume to liters

Answer 1

As 1 mole of HNO₃ contains, 1 mole of ionizable proton (H⁺), and 1 mole of Ba(OH)₂ contains 2 mole of OH^-, therfore in the titration reaction, 1 moles of HNO₃ would be neutralized by half or 0.5 mole of Ba(OH)₂,

Or, if 1 mole of Ba(OH)₂ is used up, then 2 moles of HNO₃ was present.

mmole of Ba(OH)₂ used up = volume of Ba(OH)₂ taken in ml x Molarity (i.e. moles per litre) of Ba(OH)₂ solution.

                                              = 135 x 0.600 M

                                             = 81.0

Hence, mmole of HNO₃ present = 81.0 x 2 = 162

or mol of HNO₃ present = 0.162

Now,

Volume of HNO₃ solution taken = 715 ml = 0.715 litre                   ....since 1 litre = 1000 ml

$\Rightarrow$ concentration of HNO₃ in M (i.e. mol per litre) = 0.162/0.715 = 0.226 M

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