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What volume of a 0.246 M \( \text{HNO}_3 \) solution is required to react completely with 38.6 mL of a 0.515 M \( \text{Ba(OH)}_2 \) solution?

Answer :

Final answer:

To react completely with 38.6 mL of 0.515 M Ba(OH)2, 161.540 mL of 0.246 M HNO3 solution is required, based on the stoichiometric relationship outlined in the balanced chemical equation.

Explanation:

The question asks for the volume of 0.246 M HNO3 solution needed to react completely with 38.6 mL of 0.515 M Ba(OH)2. To solve this, we must use stoichiometry and the balanced chemical equation. First, we need to write the balanced chemical equation for the reaction between nitric acid (HNO3) and barium hydroxide (Ba(OH)2):

2 HNO3 (aq) + Ba(OH)2 (aq)
ightarrow Ba(NO3)2 (aq) + 2 H2O (l)

From the equation, we see that 2 moles of HNO3 react with 1 mole of Ba(OH)2. Now, we can calculate the number of moles of Ba(OH)2 that we have:

Moles of Ba(OH)2 = 0.515 M * 0.0386 L = 0.019879 moles

Using the stoichiometry of the balanced equation to find the moles of HNO3 required:

Moles of HNO3 needed = 2 * Moles of Ba(OH)2 = 2 * 0.019879 moles = 0.039758 moles

Finally, we can find the volume of HNO3 solution needed:

Volume of HNO3 = Moles of HNO3 / Molarity of HNO3 = 0.039758 moles / 0.246 M = 0.161540 L or 161.540 mL

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