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What volume of \(0.155 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is required to titrate \(0.293 \mathrm{~g}\) of \(90.0 \%\) pure \(\mathrm{LiOH} ?\)

Short Answer

Expert verified
35.5 mL of 0.155 M Hâ‚‚SOâ‚„ is required.

Step by step solution

01

Calculate the Moles of Lithium Hydroxide (LiOH)

First, determine the moles of pure LiOH in the sample. The molecular weight of LiOH is approximately 23.95 g/mol. The mass of pure LiOH is given by \( 0.293 \text{ g} \times 0.9 \). Divide this mass by the molecular weight to find the moles of LiOH:\[\text{Moles of LiOH} = \frac{0.2637 \text{ g}}{23.95 \text{ g/mol}} \approx 0.01101 \text{ mol}\]
02

Determine the Moles of Sulfuric Acid (Hâ‚‚SOâ‚„) Needed

In the titration of LiOH with Hâ‚‚SOâ‚„, the balanced equation is:\[2 \text{LiOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Li}_2\text{SO}_4 + 2 \text{H}_2\text{O}\]This shows that 2 moles of LiOH react with 1 mole of Hâ‚‚SOâ‚„. Hence, the moles of Hâ‚‚SOâ‚„ needed are half the moles of LiOH:\[\text{Moles of H}_2\text{SO}_4 = \frac{0.01101}{2} \approx 0.005505 \text{ mol}\]
03

Calculate the Volume of Hâ‚‚SOâ‚„ Solution Required

Given the molarity of Hâ‚‚SOâ‚„ is 0.155 M, use the formula for molarity to find the volume required:\[\text{Molarity} = \frac{\text{Moles}}{\text{Volume (L)}}\]Rearrange to solve for volume:\[\text{Volume (L)} = \frac{0.005505 \text{ mol}}{0.155 \text{ M}} \approx 0.0355 \text{ L}\]Convert this volume into milliliters by multiplying by 1000:\[0.0355 \text{ L} \times 1000 = 35.5 \text{ mL}\]

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Key Concepts

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

Titration
Titration is an analytical chemical technique used to determine the concentration of a solute in a solution. It involves the gradual addition of a titrant, which is a solution of known concentration, to a sample until a reaction reaches completion. In the context of our exercise, sulfuric acid (Hâ‚‚SOâ‚„) is the titrant, while lithium hydroxide (LiOH) is the substance being analyzed.

As the titrant is added, it reacts with the substance in the sample. The completion of this reaction is typically indicated by a color change due to an indicator, or by reaching an endpoint detected with a pH meter. The volume of titrant used at this endpoint is then used to calculate the concentration of the unknown solution.
  • Titration allows for precise determination of concentration.
  • It requires a balanced chemical equation to relate the stoichiometry of the reactants.
  • The endpoint indicates when the reaction is stoichiometrically complete.
Molarity
Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution, often represented by the symbol \( ext{M}\). In the exercise, the molarity of sulfuric acid (Hâ‚‚SOâ‚„) is given as 0.155 M.

Calculating molarity involves dividing the moles of solute by the total volume of the solution in liters. In the context of titration, knowing the molarity of the titrant (the acid in this case) allows us to calculate how much of it is needed to react with the given amount of the substance being titrated.
  • Molarity is critical for understanding the quantitative aspect of chemical reactions.
  • It allows chemists to prepare solutions of exact concentrations.
  • Different units of concentration may exist, but molarity is most common in many laboratory settings.
Lithium Hydroxide (LiOH)
Lithium Hydroxide, represented chemically as LiOH, is a strong base used in various applications such as the chemical industry and battery manufacturing. In titration, LiOH is often used as a reagent due to its well-understood chemical properties.

In our exercise, the LiOH sample is 90% pure, meaning only 90% of the weight is true LiOH, while the remaining 10% may consist of impurities. To calculate the moles of pure LiOH, it's important to account for its purity quotient when considering the weight.
  • LiOH in titration reacts with acids like Hâ‚‚SOâ‚„ to form lithium sulfate and water.
  • Its high solubility in water makes it practical for various scalable chemical applications.
  • Considering purity is crucial for accurate stoichiometric calculations.
Sulfuric Acid (Hâ‚‚SOâ‚„)
Sulfuric Acid is a strong mineral acid with the chemical formula Hâ‚‚SOâ‚„. It's widely used in various industries, including manufacturing metabolites, fertilizers, and in chemical analysis lab settings.

In the context of titration, Hâ‚‚SOâ‚„ serves as an excellent titrant because it reliably reacts with bases such as LiOH to form neutral products. The balanced equation of the reaction involved here is \[2 ext{LiOH} + ext{H}_2 ext{SO}_4 ightarrow ext{Li}_2 ext{SO}_4 + 2 ext{H}_2 ext{O}\]. This means one mole of sulfuric acid reacts with two moles of lithium hydroxide.
  • Hâ‚‚SOâ‚„'s strong acidic properties make it effective for reacting with a variety of bases.
  • Understanding the stoichiometry helps accurately determine the endpoint of titration.
  • Safety measures must be considered, as sulfuric acid is highly corrosive.

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Most popular questions from this chapter

Sodium hydroxide and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) will titrate together to a phenolphthalein end point \(\left(\mathrm{OH}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O} ; \mathrm{CO}_{3}{ }^{2-} \rightarrow \mathrm{HCO}_{3}^{-}\right) .\) A mixture of \(\mathrm{NaOH}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is titrated with \(0.250 \mathrm{M} \mathrm{HCl}\), requiring \(26.2 \mathrm{~mL}\) for the phenolphthalein end point and an additional \(15.2 \mathrm{~mL}\) to reach the modified methyl orange end point. How many milligrams \(\mathrm{NaOH}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) are in the mixture?

Calculate the \(\mathrm{pH}\) at \(0,25.0,50.0,75.0,100,\) and \(125 \%\) titration in the titration of both protons of the diprotic acid \(\mathrm{H}_{2} \mathrm{~A}\) with \(0.100 \mathrm{M} \mathrm{NaOH}\), starting with \(100 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2}\) A. \(K_{a 1}=1.0 \times 10^{-3}, K_{a 2}=1.0 \times 10^{-7}\).

Explain why boiling the solution near the end point in the titration of sodium carbonate increases the sharpness of the end point.

A sample of \(\mathrm{P}_{2} \mathrm{O}_{5}\) contains some \(\mathrm{H}_{3} \mathrm{PO}_{4}\) impurity. A \(0.405-\mathrm{g}\) sample is reacted with water \(\left(\mathrm{P}_{2} \mathrm{O}_{5}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{H}_{3} \mathrm{PO}_{4}\right),\) and the resulting solution is titrated with \(0.250 \mathrm{MNaOH}\left(\mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow \mathrm{Na}_{2} \mathrm{HPO}_{4}\right)\). If \(42.5 \mathrm{~mL}\) is required for the titration, what is the percent of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) impurity?

A hydrochloric acid solution is standardized by titrating \(0.2329 \mathrm{~g}\) of primary standard sodium carbonate to a methyl red end point by boiling the carbonate solution near the end point to remove carbon dioxide. If \(42.87 \mathrm{~mL}\) acid is required for the titration, what is its molarity?

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