/*! 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 33 Calculate the concentration of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 33

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 \(\mathrm{g} \mathrm{K}_{3} \mathrm{PO}_{4}\) in 250.0 \(\mathrm{mL}\) of solution

Short Answer

Expert verified
In summary, the concentrations of the ions in each solution are as follows: a. Ca虏鈦�: 1.00 M, NO鈧冣伝: 2.00 M b. Na鈦�: 4.00 M, SO鈧劼测伝: 2.00 M c. NH鈧勨伜: 0.187 M, Cl鈦�: 0.187 M d. K鈦�: 0.0564 M, PO鈧劼斥伝: 0.0188 M

Step by step solution

01

(a) 0.100 mole of Ca(NO3)鈧� in 100.0 mL of solution

1. Calculate the moles of ions produced: One mole of Ca(NO3)鈧� produces one mole of Ca虏鈦� and two moles of NO鈧冣伝 ions when it dissolves. Therefore, we have 0.100 moles of Ca虏鈦� and 0.100 * 2 = 0.200 moles of NO鈧冣伝 ions. 2. Calculate the molarity of each ion: Since the solution has a volume of 100.0 mL (0.100 L), we can compute the concentration of the ions. Molarity of Ca虏鈦� = Moles of Ca虏鈦� / Volume of solution in Liters = 0.100 moles / 0.100 L = 1.00 M Molarity of NO鈧冣伝 = Moles of NO鈧冣伝 / Volume of solution in Liters = 0.200 moles / 0.100 L = 2.00 M
02

(b) 2.5 moles of Na鈧係O鈧� in 1.25 L of solution

1. Calculate the moles of ions produced: One mole of Na鈧係O鈧� produces two moles of Na鈦� and one mole of SO鈧劼测伝 ions when it dissolves. So, we have 2.5 * 2 = 5.0 moles of Na鈦� and 2.5 moles of SO鈧劼测伝 ions. 2. Calculate the molarity of each ion: Molarity of Na鈦� = Moles of Na鈦� / Volume of solution in Liters = 5.0 moles / 1.25 L = 4.00 M Molarity of SO鈧劼测伝 = Moles of SO鈧劼测伝 / Volume of solution in Liters = 2.5 moles / 1.25 L = 2.00 M
03

(c) 5.00 g of NH鈧凜l in 500.0 mL of solution

1. Convert grams to moles: To convert grams to moles, we need to use the molar mass of NH鈧凜l, which is approximately 53.49 g/mol. Moles of NH鈧凜l = 5.00 g / 53.49 g/mol 鈮� 0.0935 moles 2. Calculate the moles of ions produced: One mole of NH鈧凜l produces one mole of NH鈧勨伜 and one mole of Cl鈦� ions when it dissolves. So, we have 0.0935 moles of NH鈧勨伜 and 0.0935 moles of Cl鈦� ions. 3. Calculate the molarity of each ion: Molarity of NH鈧勨伜 = Moles of NH鈧勨伜 / Volume of solution in Liters = 0.0935 moles / 0.500 L = 0.187 M Molarity of Cl鈦� = Moles of Cl鈦� / Volume of solution in Liters = 0.0935 moles / 0.500 L = 0.187 M
04

(d) 1.00 g K鈧働O鈧� in 250.0 mL of solution

1. Convert grams to moles: To convert grams to moles, we need to use the molar mass of K鈧働O鈧�, which is approximately 212.27 g/mol. Moles of K鈧働O鈧� = 1.00 g / 212.27 g/mol 鈮� 0.00471 moles 2. Calculate the moles of ions produced: One mole of K鈧働O鈧� produces three moles of K鈦� and one mole of PO鈧劼斥伝 ions when it dissolves. So, we have 0.00471 * 3 鈮� 0.0141 moles of K鈦� and 0.00471 moles of PO鈧劼斥伝 ions. 3. Calculate the molarity of each ion: Molarity of K鈦� = Moles of K鈦� / Volume of solution in Liters = 0.0141 moles / 0.250 L = 0.0564 M Molarity of PO鈧劼斥伝 = Moles of PO鈧劼斥伝 / Volume of solution in Liters = 0.00471 moles / 0.250 L = 0.0188 M

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 91影视!

Key Concepts

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

Strong Electrolytes
In the realm of chemistry, strong electrolytes play a vital role in conducting electricity when dissolved in water. This conductance is because they dissociate completely into ions.
A compound like Ca(NO鈧�)鈧� is a classic example. In solution, it breaks down entirely into Ca虏鈦� and NO鈧冣伝 ions, leading to a mixture capable of conducting an electrical current efficiently.
  • Complete dissociation: Strong electrolytes dissociate fully, leaving no intact molecules in the solution.
  • Examples include most salts, strong acids (e.g., HCl), and strong bases (e.g., NaOH).
Understanding the behavior of strong electrolytes is critical, especially in processes like electrolysis, where ions are the carriers of electricity.
Remember, their complete dissociation not only affects conductivity but also the solution's chemical reactivity.
Molarity Calculation
Molarity is a key concept in understanding the concentration of solutions. It measures the number of moles of a solute per liter of solution and is a standard unit in chemistry.
Calculating the molarity involves dividing the number of solute moles by the solution's volume in liters. It's a straightforward method that provides a clear picture of the solution's concentration.
For example, if you dissolve 0.100 moles of Ca(NO鈧�)鈧� in 0.100 L of water, the solution's molarity for Ca虏鈦� would be 1.00 M, indicating a strong concentration.
  • Formula: Molarity (M) = Moles of solute / Liters of solution
  • Allows comparison of concentrations in various solutions.
It's essential to ensure precise measurements since molarity influences reaction rates and equilibrium in solutions.
Chemical Solutions
Chemical solutions are homogeneous mixtures of two or more substances. The solute is dissolved in the solvent, creating a solution.
Understanding the nature of solutions is crucial for tasks like dilutions and reaction predictions.
When dealing with ionic compounds, such as Na鈧係O鈧� or NH鈧凜l, they dissociate into ions that remain evenly distributed throughout the solvent. This uniformity is what defines them as homogenous.
  • Solvent: The primary component in which the solute is dissolved.
  • Solute: The substance that is dissolved in the solvent.
By adjusting the concentration, you can tailor the solution's properties to fit specific experimental needs. Mixing solutions properly ensures uniform composition, predicting chemical behavior more accurately.
Mole Conversion
Converting between grams and moles is a fundamental skill in chemistry. It involves using the molar mass of a compound, which is the mass of one mole of that substance.
This conversion is pivotal when you have a substance's mass but need to understand it in terms of number of moles, and vice versa.
For instance, to determine moles from a given mass of NH鈧凜l, divide the mass (5.00 g) by the molar mass (approximately 53.49 g/mol). This yields around 0.0935 moles of NH鈧凜l.
  • Conversion steps: Use molar mass for converting between grams and moles.
  • Key formula: Moles = Mass in grams / Molar mass
This unit conversion is crucial for calculating the concentration of ions, as it directly influences the molarity and overall chemical analysis.

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

What acid and what base would react in aqueous solution so that the following salts appear as products in the formula equation? Write the balanced formula equation for each reaction. a. potassium perchlorate b. cesium nitrate c. calcium iodide

One of the classic methods for determining the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing the periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.

Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: $$\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)$$ Triodide ion concentration is determined by titration with a sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right)\) solution. The products are iodide ion and tetrathionate ion \(\left(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\right)\) a. Balance the equation for the reaction of \(\mathrm{IO}_{3}^{-}\) with \(\mathrm{I}^{-}\) ions. b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HCl required to convert all of the 1 \(\mathrm{O}_{3}^{-}\) ions to \(\mathrm{I}_{3}^{-}\) ions? c. Write and balance the equation for the reaction of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with \(\mathrm{I}_{3}-\) in acidic solution. d. A 25.00 -mL sample of a 0.0100\(M\) solution of \(\mathrm{KIO}_{3}\) is reacted with an excess of \(\mathrm{KL}\) . It requires 32.04 \(\mathrm{mL}\) of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution to titrate the \(\mathrm{I}_{3}^{-}\) ions present. What is the molarity of the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution? e. How would you prepare 500.0 \(\mathrm{mL}\) of the KIO\(_{3}\)solution in part d using solid \(\mathrm{KIO}_{3} ?\)

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid鈥揵ase reactions. a. \(\mathrm{HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow\) b. \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow\) c. \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow\)

A 500.0-mL sample of 0.200 M sodium phosphate is mixed with 400.0 mL of 0.289 M barium chloride. What is the mass of the solid produced?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation
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.