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Chapter 4: Problem 44

A solution was prepared by mixing 50.00 \(\mathrm{mL}\) of 0.100 \(\mathrm{M}\) \(\mathrm{HNO}_{3}\) and 100.00 \(\mathrm{mL}\) of 0.200 $\mathrm{M} \mathrm{HNO}_{3}$ . Calculate the molarity of the final solution of nitric acid.

Short Answer

Expert verified
The molarity of the final solution of nitric acid is \(0.167 \ \mathrm{M}\).

Step by step solution

01

Calculate moles of Nitric Acid from both solutions

First, we will calculate the moles of nitric acid in each solution using the given volumes and molarities. From Solution 1 (50.00 mL of 0.100 M HNO3): Moles of \(\mathrm{HNO}_{3}\) = Molarity * Volume Moles of \(\mathrm{HNO}_{3}\) = 0.100 M * 0.050 L (Note: We converted mL to L by dividing by 1000) Moles of \(\mathrm{HNO}_{3}\) (1) = 0.005 mol From Solution 2 (100.00 mL of 0.200 M HNO3): Moles of \(\mathrm{HNO}_{3}\) = Molarity * Volume Moles of \(\mathrm{HNO}_{3}\) = 0.200 M * 0.100 L (Note: We converted mL to L by dividing by 1000) Moles of \(\mathrm{HNO}_{3}\) (2) = 0.020 mol
02

Calculate the Total Moles of Nitric Acid

Now, we will combine the moles of \(\mathrm{HNO}_{3}\) from both solutions to find the total moles of nitric acid in the final solution. Total moles of \(\mathrm{HNO}_{3}\) = Moles of \(\mathrm{HNO}_{3}\) (1) + Moles of \(\mathrm{HNO}_{3}\) (2) Total moles of \(\mathrm{HNO}_{3}\) = 0.005 mol + 0.020 mol = 0.025 mol
03

Calculate Final Volume of the Solution

We need the volume of the final solution. Since the volumes are additive in this case, we simply add the volumes of both initial solutions to find the total volume. Total Volume = Volume of Solution 1 + Volume of Solution 2 Total Volume = 0.050 L + 0.100 L = 0.150 L
04

Calculate the Molarity of the Final Solution

Now that we have the total moles of \(\mathrm{HNO}_{3}\) and the total volume of the final solution, we can calculate the molarity. Final Molarity = Total moles of \(\mathrm{HNO}_{3}\)/Total Volume Final Molarity = 0.025 mol / 0.150 L Final Molarity = 0.167 M So, the molarity of the final solution of nitric acid is \(0.167 \ \mathrm{M}\).

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Most popular questions from this chapter

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) A 0.205-g sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The BaSO\(_{4}\) that is formed is filtered, dried, and weighed. Its mass is 0.298 g. What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

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