Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The number of moles of hydroxide ion in 0.300 L of \(0.0050 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is (a) \(0.0015 ;(\mathrm{b}) 0.0030 ;(\mathrm{c}) 0.0050\) (d) 0.010.

Short Answer

Expert verified
So, the number of moles of hydroxide ion in 0.300 L of 0.0050 M Ba(OH)2 solution is 0.0030 moles (option b).

Step by step solution

01

Identifying the given variables

In our problem, the concentration [C] of barium hydroxide is 0.0050 M and the volume [V] given is 0.300L.
02

Formulate the relationship

Ba(OH)2 \[ \rightarrow \] Ba2+ + 2OH-, from one formula unit of Ba(OH)2 we get 2 hydroxide ions. Thus, the concentration of hydroxide ions is twice the concentration of the barium hydroxide.
03

Calculate the moles of Ba(OH)2

To calculate the moles (n) of Ba(OH)2, use the formula n = C x V. Substituting the given values, number of moles of Ba(OH)2 = 0.0050 M x 0.300L = 0.0015 moles.
04

Resolve for the concentration of OH⁻

Since each mole of Ba(OH)2 produces 2 moles of OH⁻, the number of moles of OH⁻ will be 2 times the moles of Ba(OH)2. The moles of OH⁻ is therefore = 2 x 0.0015 moles = 0.0030 moles.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Barium hydroxide
Barium hydroxide, with the chemical formula \( ext{Ba(OH)}_2\), is an inorganic compound used in various applications such as chemistry and industry. It is often utilized in laboratory settings for titration and analytical procedures due to its reliable reaction properties.

When dissolved in water, barium hydroxide dissociates completely to produce barium ions (\( ext{Ba}^{2+}\)) and hydroxide ions (\( ext{OH}^-\)). This ability to produce hydroxide ions makes it particularly useful for pH adjustment and chemical reactions that require a strong base. Understanding how barium hydroxide dissociates in solution is crucial when determining the chemical concentration and behavior in reactions. By knowing the concentration of barium hydroxide, one can calculate important parameters such as the pH of the solution and the number of moles of hydroxide ions produced.
Hydroxide ion
The hydroxide ion (\( ext{OH}^-\)) plays a vital role in many chemical reactions and is known for its basic properties. It is composed of one oxygen atom and one hydrogen atom, carrying a negative charge.

In aqueous solutions, the hydroxide ion is a strong base because it can accept protons from acids. This property allows it to neutralize acids in various chemical processes.

  • One of the most notable features of the hydroxide ion is its ability to affect the pH level of solutions, increasing the alkalinity when present in higher concentrations.
  • Neutralization reactions involving hydroxide ions are essential in titrations, where the unknown concentration of an acidic solution is determined by reacting it with a base like a hydroxide ion.
Understanding the behavior and calculation of the number of hydroxide ions in a solution is essential for any chemistry student, especially when dealing with compounds like barium hydroxide.
Chemical concentration
Chemical concentration refers to the amount of a substance present in a given volume of solution. Expressed often in terms of molarity (M), which is moles per liter (mol/L), concentration is a crucial parameter in chemistry as it dictates the reactivity and properties of a solution.

In our context, calculating the concentration is essential for understanding how much barium hydroxide is present in a given volume of solution. If you know the concentration, you can predict the behavior of the solution in chemical reactions.

  • Higher concentration indicates more dissolved particles, which typically increases the reactivity of the solution.
  • Concentration plays a key role in stoichiometry calculations, where balanced chemical equations are used to predict products and reactants.
By grasping the concept of concentration, students can better understand the dynamics of chemical reactions as well as accurately carry out calculations necessary for laboratory settings.
Stoichiometry
Stoichiometry is the branch of chemistry concerned with the quantities of reactants and products in chemical reactions. It allows chemists to predict the outcomes of reactions quantitatively. This field involves using balanced chemical equations to relate the amounts of reactants used to the amounts of products formed.

In the provided problem, stoichiometry comes into play when we calculate the number of hydroxide ions produced from barium hydroxide. One formula unit of \( ext{Ba(OH)}_2\) dissociates to form two hydroxide ions, doubling the concentration of produced \( ext{OH}^-\) ions compared to the original barium hydroxide concentration.

  • Understanding stoichiometry is crucial for solving problems involving moles, mass, volume, and concentration.
  • It helps in planning practical applications, like ensuring sufficient reactant quantities are available for desired reactions.
Stoichiometry not only guides chemists in practical laboratory tasks but also deepens their understanding of how reactions occur at the molecular level, allowing for more accurate and efficient chemical experimentation and application.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When treated with dilute \(\mathrm{HCl}(\mathrm{aq}),\) the solid that reacts to produce a gas is (a) \(\mathrm{BaSO}_{3} ;\) (b) \(\mathrm{ZnO};\) (c) \(\mathrm{NaBr} ;\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\).

Following are some laboratory methods occasionally used for the preparation of small quantities of chemicals. Write a balanced equation for each. (a) preparation of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{FeS}(\mathrm{s})\) (b) preparation of \(\mathrm{Cl}_{2}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{MnO}_{2}(\mathrm{s}) ; \mathrm{MnCl}_{2}(\mathrm{aq})\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are other products (c) preparation of \(\mathrm{N}_{2}: \mathrm{Br}_{2}\) and \(\mathrm{NH}_{3}\) react in aqueous solution; \(\mathrm{NH}_{4} \mathrm{Br}\) is another product (d) preparation of chlorous acid: an aqueous suspension of solid barium chlorite is treated with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\)

A \(7.55 \mathrm{g}\) sample of \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is added to \(125 \mathrm{mL}\) of a vinegar that is \(0.762 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} .\) Will the resulting solution still be acidic? Explain.

An unknown white solid consists of two compounds, each containing a different cation. As suggested in the illustration, the unknown is partially soluble in water. The solution is treated with \(\mathrm{NaOH}(\mathrm{aq})\) and yields a white precipitate. The part of the original solid that is insoluble in water dissolves in \(\mathrm{HCl}(\mathrm{aq})\) with the evolution of a gas. The resulting solution is then treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})\) and yields a white precipitate. (a) Is it possible that any of the cations \(M g^{2+}, C u^{2+}\) \(\mathrm{Ba}^{2+}, \mathrm{Na}^{+},\) or \(\mathrm{NH}_{4}^{+}\) were present in the original unknown? Explain your reasoning. (b) What compounds could be in the unknown mixture (that is, what anions might be present)?

Consider the following redox reaction: $$\begin{array}{r}4 \mathrm{NO}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 4 \mathrm{NO}_{3}^{-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq})\end{array} $$ (a) Which species is oxidized? (b) Which species is reduced? (c) Which species is the oxidizing agent? (d) Which species is the reducing agent? (e) Which species gains electrons? (f) Which species loses electrons?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free