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Chapter 4: Problem 41

A solution is prepared by dissolving \(0.5842 \mathrm{~g}\) oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) in enough water to make \(100.0 \mathrm{~mL}\) of solution. A \(10.00-\mathrm{mL}\) aliquot (portion) of this solution is then diluted to a final volume of \(250.0\) \(\mathrm{mL}\). What is the final molarity of the oxalic acid solution?

Short Answer

Expert verified
The final molarity of the oxalic acid solution is approximately \(0.00260\,\mathrm{M}\).

Step by step solution

01

Calculate initial moles of oxalic acid

The molar mass of oxalic acid can be calculated as: \(\mathrm{M_{H_{2}C_{2}O_{4}} = 2\times M_{H} + 2\times M_{C} + 4\times M_{O}}\) \(\mathrm{M_{H_{2}C_{2}O_{4}} = 2\times1.01 + 2\times 12.01 + 4\times 16.00}\) \(\mathrm{M_{H_{2}C_{2}O_{4}} \approx 90.04 \mathrm{~g/mol}}\) Now, we can calculate the initial moles of oxalic acid, using the given mass \(0.5842 \mathrm{~g}\) as follows: \(\mathrm{n_{initial} = \frac{mass}{molar\,mass}}\) \(\mathrm{n_{initial} = \frac{0.5842\,g}{90.04\,g/mol}}\) \(\mathrm{n_{initial} \approx 0.00649\,mol}\)
02

Calculate the initial molarity

The initial molarity can be found using the initial moles and the volume of \(100.0 \mathrm{~mL}\): \(\mathrm{M_{initial} = \frac{n_{initial}}{V_{initial}}}\) \(\mathrm{M_{initial} = \frac{0.00649\,mol}{0.100\,L}}\) \(\mathrm{M_{initial} \approx 0.0649\,M}\)
03

Find the new moles of oxalic acid after the aliquot

After taking \(10.00 \mathrm{~mL}\) aliquot, we need to find the new moles of oxalic acid in that portion. \(\mathrm{n_{new} = M_{initial} \times V_{aliquot}}\) \(\mathrm{n_{new} = 0.0649\,M \times 0.0100\,L}\) \(\mathrm{n_{new} \approx 0.000649 \,mol}\)
04

Calculate the final molarity

Finally, calculate the final molarity of the oxalic acid, using the aliquot moles and the final volume of \(250.0 \mathrm{~mL}\): \(\mathrm{M_{final} = \frac{n_{new}}{V_{final}}}\) \(\mathrm{M_{final} = \frac{0.000649\,mol}{0.250\,L}}\) \(\mathrm{M_{final} \approx 0.00260\,M}\) So, the final molarity of the oxalic acid solution is approximately \(0.00260\,\mathrm{M}\).

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Most popular questions from this chapter

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.0200 \mathrm{~mol}\) of sodium phosphate in \(10.0 \mathrm{~mL}\) of solution b. \(0.300 \mathrm{~mol}\) of barium nitrate in \(600.0 \mathrm{~mL}\) of solution c. \(1.00 \mathrm{~g}\) of potassium chloride in \(0.500 \mathrm{~L}\) of solution d. \(132 \mathrm{~g}\) of ammonium sulfate in \(1.50 \mathrm{~L}\) of solution

Many plants are poisonous because their stems and leaves contain oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) or sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\); when ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) in aqueous solution.

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

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The balanced equation is \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow\) $$ 4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) $$ This reaction is an oxidation-reduction reaction. What species is reduced, and what species is oxidized? How many electrons are transferred in the balanced equation above?

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of 1.0 ppm is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{~mL}\), which equals 1.0 g solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10\) ppm DDT \(\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)
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