/*! 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 96 Lithium salts, such as lithium c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 96

Lithium salts, such as lithium carbonate, are used to treat manic-depressives. What is the solubility product for lithium carbonate if 1.36 grams of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) dissolve in \(100 \mathrm{~mL}\) of water?

Short Answer

Expert verified
Ksp for \( \text{Li}_2\text{CO}_3 \) is approximately 0.0249.

Step by step solution

01

Calculate Molar Mass

First, calculate the molar mass of lithium carbonate \( \text{Li}_2\text{CO}_3 \). The molar masses are: Li = 6.94 g/mol, C = 12.01 g/mol, O = 16.00 g/mol. Therefore, \( \text{Li}_2\text{CO}_3 \) has a molar mass of \( 2(6.94) + 12.01 + 3(16.00) = 73.89 \text{ g/mol} \).
02

Calculate Moles of Li2CO3 Dissolved

Convert the dissolved \( \text{Li}_2\text{CO}_3 \) into moles. Given 1.36 g of \( \text{Li}_2\text{CO}_3 \): \( \frac{1.36 \, \text{g}}{73.89 \, \text{g/mol}} = 0.0184 \, \text{mol} \).
03

Calculate Concentration of Li2CO3

Calculate the molarity of the lithium carbonate solution. Since it dissolves in 100 mL (or 0.1 L) of water, the concentration is \( \frac{0.0184 \, \text{mol}}{0.1 \, \text{L}} = 0.184 \, \text{M} \).
04

Write Dissociation Equation

The dissociation equation for lithium carbonate is \( \text{Li}_2\text{CO}_3(s) \rightleftharpoons 2\text{Li}^+(aq) + \text{CO}_3^{2-}(aq) \). Each formula unit produces 2 lithium ions and 1 carbonate ion.
05

Set up Ksp Expression

Calculate the solubility product (Ksp). The formula is \( K_{sp} = [\text{Li}^+]^2 \times [\text{CO}_3^{2-}] \). From the dissociation, \([\text{Li}^+] = 2 \times 0.184 = 0.368 \, \text{M}\) and \([\text{CO}_3^{2-}] = 0.184 \, \text{M}\).
06

Calculate Ksp

Substitute the concentrations into the Ksp expression: \( K_{sp} = (0.368)^2 \times (0.184) = 0.024876 \). Thus, the solubility product is approximately 0.0249.

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.

Lithium Carbonate
Lithium carbonate, denoted as \( \text{Li}_2\text{CO}_3 \), is a vital compound in the pharmaceutical industry, particularly in the treatment of bipolar disorder.This compound appears as a white, powdery substance and stands out due to its limited solubility in water.When dissolved, it dissociates into lithium ions (\( \text{Li}^+ \) and carbonate ions (\( \text{CO}_3^{2-} \).
  • Applications: It is predominantly used in the medical field to stabilize mood swings in manic-depressive patients.
  • Solubility: Despite its therapeutic benefits, lithium carbonate is sparingly soluble, making its precise solubility critical for medical dosages.
The solubility of lithium carbonate is represented through the solubility product constant, \( K_{sp} \), which is a measure of how much \( \text{Li}_2\text{CO}_3 \) can dissolve in water before the solution becomes saturated.
Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction's forward and reverse processes happen at the same rate, resulting in a stable state.In the context of lithium carbonate, this dynamic is crucial as it informs how the compound dissolves and maintains balance in solution.When \( \text{Li}_2\text{CO}_3 \) is dissolved in water, it achieves equilibrium between its solid form and the ions in the solution.
  • Dissociation Equation: The dissociation of lithium carbonate is represented as \( \text{Li}_2\text{CO}_3(s) \rightleftharpoons 2\text{Li}^+(aq) + \text{CO}_3^{2-}(aq) \). Each mole of \( \text{Li}_2\text{CO}_3 \) produces 2 moles of \( \text{Li}^+ \) and 1 mole of \( \text{CO}_3^{2-} \).
  • Equilibrium Considerations: The degree of dissolution and the concentrations of ions are dictated by the solubility product, \( K_{sp} \).
This equilibrium state is essential for predicting how and to what extent \( \text{Li}_2\text{CO}_3 \) can be dissolved in varying conditions, such as temperature and the presence of other ions.
Molar Mass Calculation
Understanding molar mass calculation is fundamental in determining the amount of a substance in a chemical reaction.In this exercise, calculating the molar mass of lithium carbonate (\( \text{Li}_2\text{CO}_3 \)) is a critical step.Each element in the compound contributes to its total mass based on its atomic weight.- Lithium (Li) has an atomic mass of 6.94 g/mol.- Carbon (C) has an atomic mass of 12.01 g/mol.- Oxygen (O) comes in at 16.00 g/mol.The formula for lithium carbonate consists of two lithium atoms, one carbon atom, and three oxygen atoms.Thus, the molar mass is:\[ 2(6.94) + 12.01 + 3(16.00) = 73.89 \text{ g/mol} \]Calculating this value is crucial when converting grams of a substance to moles, which is necessary for further calculations such as determining the concentration or the \( K_{sp} \). Knowing how to calculate molar mass allows students to confidently handle and predict outcomes in chemical reactions.

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

How many grams of silver sulfide will dissolve in 500 \(\mathrm{mL}\) of a \(0.050 \mathrm{M} \mathrm{S}^{2-}\) solution? \(\left(\mathrm{Ag}_{2} \mathrm{~S}: K_{\mathrm{sp}}=6.3 \times\right.$$\left.10^{-50}\right)\)

What is a rate constant? How does it differ from the rate law?

Calculate the equilibrium concentrations of \(\mathrm{CO}, \mathrm{H}_{2} \mathrm{O},\) \(\mathrm{CO}_{2}\), and \(\mathrm{H}_{2}\) present in the water-gas shift reaction at \(800^{\circ} \mathrm{C}\) if the initial concentrations of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) are $$ \begin{array}{l} 1.00 \mathrm{M} . \\ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \\ K_{\mathrm{c}}=0.72\left(\text { at } 800^{\circ} \mathrm{C}\right) \end{array} $$

When equilibrium is reached for the dissolving of solid calcium sulfate in water at \(25^{\circ} \mathrm{C},\) it is found that \(\left[\mathrm{Ca}^{2+}\right]=\left[\mathrm{SO}_{4}^{2-}\right]=4.9 \times 10^{-3} \mathrm{M}\) $$ \mathrm{CaSO}_{4}(s) \rightleftarrows \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) $$ (a) Calculate the equilibrium constant \(\left(K_{\mathrm{sp}}\right)\) for the above reaction. (b) A solution contains \(\mathrm{Ca}^{2+}(a q)\) at a concentration of \(3.6 \times 10^{-3} M\) and \(\mathrm{SO}_{4}^{2-}(a q)\) at a concentration of \(8.0 \times 10^{-3} M\). Will solid \(\mathrm{CaSO}_{4}\) be formed? Show your calculations. (c) \(\mathrm{CaSO}_{4}\) is allowed to dissolve in \(1.0 \mathrm{~L}\) of water at \(25^{\circ} \mathrm{C}\) until equilibrium is reached. Then the water is allowed to evaporate to half its original volume. What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) in this solution? Explain.

What is the solubility product for strontium fluoride if the solubility of \(\mathrm{SrF}_{2}\) in water is 0.107 gram per liter?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

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.