Question

Consider the following balanced equation for the complete neutralization of nitric acid of unknown concentration with aqueous 0.100 M potassium hydroxide in a titration

reaction.
HNO3(aq) + KOH(aq)
KNO3(aq) + H₂O(1)
If 73.4 mL of the base are required to neutralize 38.5 mL of the acid, what is the molarity of the acid? Enter only the numeric value for your answer (no units).

Answer 1

The molarity of the nitric acid is 0.19.

Step-by-step explanation:

To determine the molarity of the acid from the given balanced equation and titration data, we can use the concept of stoichiometry. From the equation, we see that 1 mole of nitric acid (HNO3) reacts with 1 mole of potassium hydroxide (KOH), so the molar ratio is 1:1.

Given that 73.4 mL of the base (KOH) are required to neutralize 38.5 mL of the acid (HNO3), we can set up the following equation:

Moles of acid (HNO3) = Moles of base (KOH)

Molarity of acid (HNO3) × Volume of acid (HNO3) = Molarity of base (KOH) × Volume of base (KOH)

Plugging in the values, we have:

Molarity of acid (HNO3) × 38.5 mL = 0.100 M × 73.4 mL

Solving for the molarity of the acid:

Molarity of acid (HNO3) = (0.100 M × 73.4 mL) ÷ 38.5 mL

= 0.19 M