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A \(30.0-\mathrm{mL}\) sample of an unknown strong base is neutralized after the addition of \(12.0 \mathrm{~mL}\) of a \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) solution. If the unknown base concentration is \(0.0300 M\), give some possible identities for the unknown base.

Short Answer

Expert verified
Possible identities for the unknown strong base are Sodium hydroxide (NaOH), Potassium hydroxide (KOH), Calcium hydroxide (Ca(OH)鈧), and Barium hydroxide (Ba(OH)鈧). These bases can neutralize with HNO鈧, and they have the same moles (0.00180 mol) and concentration (0.0300 M) as given in the problem.

Step by step solution

01

Calculate the moles of the strong acid HNO鈧

To calculate the moles of the acid, we can use the formula moles = molarity x volume. In this case, the molarity of HNO鈧 is given as 0.150 M, and the volume is given as 12.0 mL. Moles of HNO鈧 = (0.150 mol/L) x (12.0 mL x 10鈦宦 L/mL) = 0.00180 mol
02

Calculate the moles of the unknown base

Since the moles of acid = moles of base in a neutralization reaction, we have: Moles of unknown base = moles of HNO鈧 = 0.00180 mol
03

Calculate the volume of the unknown base

We are given the concentration of the unknown base as 0.0300 M and now we know the moles of the base as 0.00180 mol, we can calculate the volume of the unknown base using the formula volume = moles/concentration. Volume of unknown base = 0.00180 mol / 0.0300 mol/L = 0.0600 L
04

Identify possible strong bases

We can now identify some strong bases based on their possible reactions with HNO鈧. Some common strong bases are: 1. Sodium hydroxide (NaOH) 2. Potassium hydroxide (KOH) 3. Calcium hydroxide (Ca(OH)鈧) 4. Barium hydroxide (Ba(OH)鈧) These are just a few examples of possible strong bases. More options could be present, as long as they neutralize with HNO鈧 and have the same moles and concentration as given in the problem.

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Key Concepts

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

Strong Base
A strong base is a chemical compound that can completely dissociate in water, releasing hydroxide ions ( OH鈦 ). This complete dissociation makes strong bases highly effective at neutralizing acids, such as nitric acid (HNO鈧). When a strong base reacts with a strong acid, they combine to form water and a salt, effectively reducing the acidity or alkalinity of a solution. Some common examples of strong bases include:
  • Sodium hydroxide ( NaOH )
  • Potassium hydroxide ( KOH )
  • Calcium hydroxide ( Ca(OH)鈧 )
  • Barium hydroxide ( Ba(OH)鈧 )
These bases are known for their robust reactivity due to their ability to release hydroxide ions so readily. This property is important in numerous chemical processes, including environmental, industrial, and laboratory settings. Understanding the behavior of strong bases is crucial for predicting their reactions with acids and determining their potential applications.
HNO鈧 (Nitric Acid)
Nitric acid, with the chemical formula HNO鈧 , is a strong acid widely used in various chemical industries. As a strong acid, it fully ionizes in water to produce hydrogen ions ( H鈦 ) and nitrate ions ( NO鈧冣伝 ). This complete ionization is what makes nitric acid effective in neutralization reactions, particularly with strong bases. In a neutralization reaction with a strong base, HNO鈧 reacts to form water and a corresponding nitrate salt. For example, when reacting with sodium hydroxide ( NaOH ), the products would be water ( H鈧侽 ) and sodium nitrate ( NaNO鈧 ). Nitric acid is not only used for neutralization but also plays a significant role in the production of fertilizers, explosives, and dyes. It's also employed in metallurgy and chemicals for cleaning purposes due to its oxidizing properties. Understanding HNO鈧 is essential in controlling and predicting how it will interact with other substances, especially in processes that require neutralization.
Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict how much of a substance is needed or produced in a reaction. The central concept of stoichiometry is based on the conservation of mass and the mole concept. In the context of neutralization reactions, stoichiometry helps in calculating the amount of base required to neutralize a given amount of acid. This involves:
  • Determining the moles of acid or base using their respective molarities and volumes.
  • Using the balanced chemical equation to find the mole ratio between reactants and products.
  • Calculating the required or produced amount using these relationships.
For instance, if we know the moles of HNO鈧 used, stoichiometry allows us to deduce the moles and volume of the unknown strong base needed for complete neutralization. This calculation is essential for accurately understanding the changes in a reaction and is widely used in practical applications, such as producing chemicals, pharmaceuticals, and industrial solutions.
Molarity
Molarity is a measure of the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution (mol/L). Molarity is a fundamental concept in chemistry because it allows for the quantification of a substance's concentration, which is critical for various calculations and reactions.To calculate molarity, you use the formula:\[\text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}}\]In practical applications, knowing the molarity of a solution is essential for understanding how much of a chemical is present in a given volume. This knowledge allows chemists to prepare solutions with precise concentrations required for specific reactions.In a neutralization reaction, such as the one featuring HNO鈧 and a strong base, molarity is used to determine the number of moles of each reactant involved. Accurate measurements of molarity are vital for ensuring the reaction goes to completion and that all acid and base are fully reacted without excess. Understanding molarity and practicing its calculation is a cornerstone for anyone delving into chemistry, from students to professional chemists.

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Most popular questions from this chapter

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 \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(a q) \longrightarrow\) \(4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(t)\) 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?

Consider the reaction between sodium metal and fluorine \(\left(\mathrm{F}_{2}\right)\) gas to form sodium fluoride. Using oxidation states, how many electrons would each sodium atom lose, and how many electrons would each fluorine atom gain? How many sodium atoms are needed to react with one fluorine molecule? Write a balanced equation for this reaction.

Suppose \(50.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CoCl}_{2}\) solution is added to \(25.0 \mathrm{~mL}\) of \(0.350 \mathrm{M} \mathrm{NiCl}_{2}\) solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing.

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.100\) mole of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(100.0 \mathrm{~mL}\) of solution b. \(2.5\) moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in \(1.25 \mathrm{~L}\) of solution c. \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(500.0 \mathrm{~mL}\) of solution d. \(1.00 \mathrm{~g} \mathrm{~K}_{2} \mathrm{PO}_{4}\) in \(250.0 \mathrm{~mL}\) of solution

A sample may contain any or all of the following ions: \(\mathrm{Hg}_{2}{ }^{2+}\), \(\mathrm{Ba}^{2+}\), and \(\mathrm{Mn}^{2+}\). a. No precipitate formed when an aqueous solution of \(\mathrm{NaCl}\) was added to the sample solution. b. No precipitate formed when an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) was added to the sample solution. c. A precipitate formed when the sample solution was made basic with \(\mathrm{NaOH}\). Which ion or ions are present in the sample solution?

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