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Chapter 4: Problem 135

A \(0.500-\mathrm{L}\) sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0-mL aliquot and adding \(50.0 \mathrm{~mL}\) of \(0.213 \mathrm{M} \mathrm{NaOH}\). After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required \(13.21 \mathrm{~mL}\) of \(0.103 M\) \(\mathrm{HCl}\) for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Sulfuric acid has two acidic hydrogens.

Short Answer

Expert verified
The molarity of the original sample of H2SO4 is approximately 0.232 M.

Step by step solution

01

Calculate the moles of excess OH⁻ ions

First, we determine the moles of excess OH⁻ ions present in the solution by using the molarity and volume of the HCl that neutralizes the excess: Moles of excess OH⁻ = (Volume of HCl) × (Molarity of HCl) Moles of excess OH⁻ = (13.21 mL) × (0.103 mol/L) Convert mL to L: Moles of excess OH⁻ = (0.01321 L) × (0.103 mol/L) = 0.00136013 mol
02

Calculate the moles of NaOH initially added

Next, we determine the moles of NaOH initially added to the aliquot: Moles of NaOH = (Volume of NaOH) × (Molarity of NaOH) Moles of NaOH = (50.0 mL) × (0.213 mol/L) Convert mL to L: Moles of NaOH = (0.050 L) × (0.213 mol/L) = 0.01065 mol
03

Determine the moles of NaOH that reacted with the H2SO4

Since we now know the moles of excess OH⁻ ions and the initial moles of NaOH, we can subtract the excess from the initial amount to determine the moles of NaOH that reacted with the H2SO4: Moles of NaOH reacted = (Initial moles of NaOH) - (Moles of excess OH⁻) Moles of NaOH reacted = 0.01065 mol - 0.00136013 mol = 0.00928987 mol
04

Calculate the moles of H2SO4 in the aliquot

Using the stoichiometry of the reaction between H2SO4 and NaOH, we can now determine the moles of H2SO4 in the aliquot, as H2SO4 has two acidic hydrogens, and it reacts with NaOH in a 1:2 ratio. Moles of H2SO4 = (Moles of reacted NaOH) / 2 = 0.00928987 mol / 2 = 0.004644935 mol
05

Calculate the molarity of the original H2SO4 sample

Finally, we can determine the molarity of the original H2SO4 sample by dividing the moles of H2SO4 in the aliquot by the volume of the aliquot, then multiplying by 5 since the aliquot represents only 1/5 (100 mL out of 500 mL) of the original volume: Molarity of H2SO4 = (Moles of H2SO4 in aliquot) / (Volume of aliquot) × 5 Molarity of H2SO4 = (0.004644935 mol) / (100 mL) × 5 Convert mL to L: Molarity of H2SO4 = (0.004644935 mol) / (0.1 L) × 5 = 0.23224675 M The molarity of the original sample of H2SO4 is approximately 0.232 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. \(0.0200\) mole of sodium phosphate in \(10.0 \mathrm{~mL}\) of solution b. \(0.300\) mole 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

A \(450.0-\mathrm{mL}\) sample of a \(0.257-M\) solution of silver nitrate is mixed with \(400.0 \mathrm{~mL}\) of \(0.200 M\) calcium chloride. What is the concentration of \(\mathrm{Cl}^{-}\) in solution after the reaction is complete?

A \(10.00-\mathrm{mL}\) sample of vinegar, an aqueous solution of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\), is titrated with \(0.5062 \mathrm{M} \mathrm{NaOH}\), and \(16.58 \mathrm{~mL}\) is required to reach the equivalence point. a. What is the molarity of the acetic acid? b. If the density of the vinegar is \(1.006 \mathrm{~g} / \mathrm{cm}^{3}\), what is the mass percent of acetic acid in the vinegar?

Separate samples of a solution of an unknown soluble ionic compound are treated with \(\mathrm{KCl}, \mathrm{Na}_{2} \mathrm{SO}_{4}\), and \(\mathrm{NaOH}\). A precipitate forms only when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added. Which cations could be present in the unknown soluble ionic compound?

In most of its ionic compounds, cobalt is either \(\mathrm{Co}(\mathrm{II})\) or \(\mathrm{Co}(\mathrm{III})\). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A \(0.256-\mathrm{g}\) sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of \(0.308 \mathrm{~g}\). A second sample of \(0.416 \mathrm{~g}\) of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was \(0.145 \mathrm{~g}\). a. What is the percent composition, by mass, of the compound? b. Assuming the compound contains one cobalt ion per formula unit, what is the formula? c. Write balanced equations for the three reactions described.
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