/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 127 A 0.500-L sample of \(\mathrm{H}... [FREE SOLUTION] | 91影视

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A 0.500-L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a \(100.0-\mathrm{mL}\) aliquot and adding \(50.0 \mathrm{~mL}\) of \(0.213 M \mathrm{NaOH}\). After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required \(13.21 \mathrm{~mL}\) of \(0.103 \mathrm{M} \mathrm{HCl}\) 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 H鈧係O鈧 is approximately 0.0465 M.

Step by step solution

01

Calculate the moles of OH鈦 ions from HCl neutralization

We are given that after the reaction and with the excess base, it took 13.21 mL of 0.103 M HCl to neutralize the OH鈦 ions. To find the moles of OH鈦 ions, we use the formula: Moles = Molarity (of HCl) 脳 Volume (of HCl) Moles of OH鈦 = 0.103 mol/L 脳 13.21 mL 脳 (1L/1000 mL) Moles of OH鈦 = 0.00135923 mol
02

Calculate the moles of OH鈦 ions reacted with H鈧係O鈧

We are given that 50.0 mL of 0.213 M NaOH was added, so we calculate the total moles of OH鈦 ions: Total moles of OH鈦 = Molarity (of NaOH) 脳 Volume (of NaOH) Total moles of OH鈦 = 0.213 mol/L 脳 50 mL 脳 (1L/1000 mL) Total moles of OH鈦 = 0.010650 mol Now, subtract the moles of OH鈦 ions from HCl neutralization to find the moles of OH鈦 ions that reacted with H鈧係O鈧: Moles of OH鈦 (reacted with H鈧係O鈧) = 0.010650 mol - 0.00135923 mol Moles of OH鈦 (reacted with H鈧係O鈧) = 0.00929077 mol
03

Calculate moles of H鈧係O鈧 and find the molarity

Since H鈧係O鈧 has two acidic hydrogens, it reacts with two OH鈦 ions per molecule. To find the moles of H鈧係O鈧 that reacted, we divide the moles of OH鈦 ions by 2: Moles of H鈧係O鈧 = Moles of OH鈦 (reacted with H鈧係O鈧) / 2 Moles of H鈧係O鈧 = 0.00929077 mol / 2 Moles of H鈧係O鈧 = 0.004645385 mol Recall that the H鈧係O鈧 solution was originally a 0.500-L sample, and 100 mL aliquot was taken for the reaction. We can now find the concentration (molarity) of the original H鈧係O鈧 solution: Molarity = Moles of H鈧係O鈧 / Volume of aliquot (in liters) Molarity (of H鈧係O鈧) = 0.004645385 mol / 0.1 L Molarity (of H鈧係O鈧) = 0.04645385 mol/L Therefore, the molarity of the original sample of H鈧係O鈧 is approximately 0.0465 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Titration Analysis
Titration analysis is a critical laboratory procedure used to determine the concentration of an unknown solution. Imagine you're on a treasure hunt, but instead of a map, you have to rely on a reaction to guide you to the 'X marks the spot'鈥攖he molarity of a sample. In the context of titration, the treasure is the point where an acid and base neutralize each other, known as the equivalence point.

To reach this, a known concentration of titrant (in this case, NaOH and then HCl) is gradually added to the analyte (H2SO4 solution). Observing the volume of titrant used until neutralization indicates how much of the analyte is present. But, finding out the molarity doesn't end here. What's crucial is the stoichiometry of the reaction鈥攗nderstanding how the acid and base interact on a mole-to-mole basis. Improving your grasp of this concept will make titration a powerful tool in your chemistry toolkit.
Stoichiometry
Stoichiometry bridges the gap between the molecular world and measurable quantities. It is like a baking recipe that tells you how much of each ingredient you need to mix together. In chemistry, stoichiometry uses balanced chemical equations to relate the amounts of reactants and products. It focuses on mole ratios, which are the coefficients of the chemicals involved in a reaction.

For example, in the neutralization of H2SO4 by NaOH, we must consider that sulfuric acid can donate two H+ ions, meaning it reacts with two OH- ions, a detail that's pivotal when computing the exact moles of acid in the original sample. If we miss this, we would incorrectly estimate the concentration of our 'recipe'. Dissecting the stoichiometric relationships in the provided problem is fundamental for students. Breaking down complex recipes into simpler, digestible parts will ensure you don't just memorize; you understand and apply.
Acid-Base Neutralization
Acid-base neutralization is a reaction where an acid like H2SO4 interacts with a base like NaOH, to produce a salt and water. It's a chemical dance where the acid's H+ ions pair with the base's OH- ions to create H2O. The beauty of this reaction is its predictability. With acids releasing H+ and bases releasing OH-, when they meet, they always aim to form water, reaching a balance point, or neutrality.

To improve the understanding of neutralization, visualize it as a seesaw where both acid and base are trying to balance each other out. Too much of one, and we'd need to add the other to level things. This is essentially what's happening when excess hydroxide ions from NaOH are neutralized by HCl. The exercise showcases how much acid is needed to bring the seesaw to a level position after introducing a known amount of base, highlighting the symmetry inherent in acid-base reactions. Always remember, the goal is to achieve that perfect harmony, also known as the equivalence point.

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Most popular questions from this chapter

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?

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 \(\mathrm{H}_{2} \mathrm{O}\)

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.

What mass of barium sulfate can be produced when \(100.0 \mathrm{~mL}\) of a \(0.100 M\) solution of barium chloride is mixed with \(100.0\) \(\mathrm{mL}\) of a \(0.100 \mathrm{M}\) solution of iron(III) sulfate?

Citric acid, which can be obtained from lemon juice, has the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} .\) A \(0.250-\mathrm{g}\) sample of citric acid dissolved in \(25.0 \mathrm{~mL}\) of water requires \(37.2 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{NaOH}\) for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?

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