(a) True or false: "solubility" and "solubility-product constant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Short Answer

Expert verified
(a) The statement is false, as solubility and solubility-product constant are not the same number for a given compound. (b) The solubility-product constant expressions are as follows: \(K_{sp}(\mathrm{MnCO}_{3}) = [\mathrm{Mn}^{2+}][\mathrm{CO}_{3}^{2-}]\), \(K_{sp}(\mathrm{Hg}(\mathrm{OH})_{2}) = [\mathrm{Hg}^{2+}][\mathrm{OH}^-]^2\), and \(K_{sp}(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}) = [\mathrm{Cu}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2\).

Step by step solution

01

Part (a) - Solubility vs Solubility-Product Constant

Solubility is a measure of how much solute can dissolve in a given solvent, typically measured in moles per liter (M). The solubility of a compound depends on temperature, pressure, and the nature of the solvent. The solubility-product constant (Ksp) is a value that represents the equilibrium between a solid ionic compound and its dissolved ions. The Ksp of a compound is a constant value for a given temperature. Therefore, the solubility and solubility-product constant are not the same number for a given compound. So, the statement is false.
02

Part (b) - Expression for Solubility-Product Constant for MnCO3

The dissolution of MnCO3 in water can be represented as: \[\mathrm{MnCO}_{3}(s) \rightleftharpoons \mathrm{Mn}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\] The solubility-product constant expression (Ksp) for MnCO3 is: \[K_{sp} = [\mathrm{Mn}^{2+}][\mathrm{CO}_{3}^{2-}]\]
03

Part (b) - Expression for Solubility-Product Constant for Hg(OH)2

The dissolution of Hg(OH)2 in water can be represented as: \[\mathrm{Hg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Hg}^{2+}(aq) + 2\mathrm{OH}^{-}(aq)\] The solubility-product constant expression (Ksp) for Hg(OH)2 is: \[K_{sp} = [\mathrm{Hg}^{2+}][\mathrm{OH}^-]^2\]
04

Part (b) - Expression for Solubility-Product Constant for Cu3(PO4)2

The dissolution of Cu3(PO4)2 in water can be represented as: \[\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Cu}^{2+}(aq) + 2\mathrm{PO}_{4}^{3-}(aq)\] The solubility-product constant expression (Ksp) for Cu3(PO4)2 is: \[K_{sp} = [\mathrm{Cu}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2\]

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!

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

Baking soda (sodium bicarbonate, \(\mathrm{NaHCO}_{3}\) ) reacts with acids in foods to form carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right),\) which in turn decomposes to water and carbon dioxide gas. In a cake batter, the \(\mathrm{CO}_{2}(g)\) forms bubbles and causes the cake to rise. \((\mathbf{a})\) A rule of thumb in baking is that \(1 / 2\) teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\). Write the chemical equation for this neutralization reaction. (b) The density of baking soda is \(2.16 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of mol/L. (One cup \(=236.6 \mathrm{~mL}=48\) teaspoons \() .(\mathbf{c})\) If \(1 / 2\) teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at a pressure of \(101.3 \mathrm{kPa}\), in an oven set to \(177^{\circ} \mathrm{C}\).

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is $2.5 \times 10^{-14} .(\mathbf{a})$ What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?(\mathbf{b})\) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to $1.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?$

Which of these statements about the common-ion effect is most correct? (a) The solubility of a salt MA is decreased in a solution that already contains either \(\mathrm{M}^{+}\) or \(\mathrm{A}^{-} .\) (b) Common ions alter the equilibrium constant for the reaction of an ionic solid with water. \((\mathbf{c})\) The common-ion effect does not apply to unusual ions like \(\mathrm{SO}_{3}^{2-} .(\mathbf{d})\) The solubility of a salt MA is affected equally by the addition of either \(\mathrm{A}^{-}\) or a noncommon ion.

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

A 10.0-mL sample of \(0.250 \mathrm{MHNO}_{3}\) solution is titrated with $0.100 M$ KOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) \(20.0 \mathrm{~mL}\), (b) \(24.9 \mathrm{~mL}\) (c) \(25.0 \mathrm{~mL}\) (d) \(25.1 \mathrm{~mL}\), (e) \(30.0 \mathrm{~mL}\).

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