/*! 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 67 Write the ion-product expression... [FREE SOLUTION] | 91影视

91影视

Write the ion-product expressions for (a) silver carbonate; (b) barium fluoride; (c) copper(II) sulfide.

Short Answer

Expert verified
(a) K_{sp} = [Ag鈦篯虏[CO鈧兟测伝](b) K_{sp} = [Ba虏鈦篯[F鈦籡虏(c) K_{sp} = [Cu虏鈦篯[S虏鈦籡

Step by step solution

01

Determine the chemical formula of each compound

First, identify the chemical formula for each compound.(a) Silver carbonate: Ag鈧侰O鈧(b) Barium fluoride: BaF鈧(c) Copper(II) sulfide: CuS
02

Write the dissociation equation for each compound

Next, write the dissociation equation for each compound in water.(a) Ag鈧侰O鈧 鈫 2Ag鈦 + CO鈧兟测伝(b) BaF鈧 鈫 Ba虏鈦 + 2F鈦(c) CuS 鈫 Cu虏鈦 + S虏鈦
03

Write the ion-product expression for each compound

The ion-product expression (Ksp) is derived from the dissociation equations. It is the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients.(a) For Ag鈧侰O鈧: K_{sp} = [Ag鈦篯虏[CO鈧兟测伝](b) For BaF鈧: K_{sp} = [Ba虏鈦篯[F鈦籡虏(c) For CuS: K_{sp} = [Cu虏鈦篯[S虏鈦籡

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.

chemical formula
A chemical formula shows the types and numbers of atoms in a compound. It uses element symbols from the periodic table and numerical subscripts. For example, Silver Carbonate's formula is written as \( \text{Ag}_{2}\text{CO}_{3} \). This formula means it has two silver (Ag) atoms and one carbonate (CO鈧) group.

For Barium Fluoride, the formula is \( \text{BaF}_{2} \). It means one barium (Ba) atom and two fluoride (F) atoms. For Copper(II) Sulfide, we write \( \text{CuS} \), meaning one copper (Cu) atom and one sulfide (S) atom.

Understanding chemical formulas helps us predict the behavior and interaction of compounds. It is essential for writing dissociation equations and ion-product expressions.
dissociation equation
A dissociation equation shows how a compound splits into its ions when dissolved in water. For example, Silver Carbonate (\( \text{Ag}_{2}\text{CO}_{3} \)) dissociates into two silver ions (Ag鈦) and one carbonate ion (CO鈧兟测伝). We write it as follows:

\( \text{Ag}_{2}\text{CO}_{3} \rightarrow 2\text{Ag}^{+} + \text{CO}_{3}^{2-} \).

For Barium Fluoride (\( \text{BaF}_{2} \)), it dissociates into one barium ion (Ba虏鈦) and two fluoride ions (F鈦):
\( \text{BaF}_{2} \rightarrow \text{Ba}^{2+} + 2\text{F}^{-} \).

For Copper(II) Sulfide (\( \text{CuS} \)), it dissociates into one copper ion (Cu虏鈦) and one sulfide ion (S虏鈦):
\( \text{CuS} \rightarrow \text{Cu}^{2+} + \text{S}^{2-} \).

By writing dissociation equations, we can then derive the ion-product expressions, which are crucial in understanding solubility.
solubility product constant (Ksp)
The solubility product constant, or \( \text{Ksp} \), tells us how much of a compound can dissolve in water. It is derived from the dissociation equations. It is the product of ion concentrations, each raised to their stoichiometric coefficients. For Silver Carbonate (\( \text{Ag}_{2}\text{CO}_{3} \)), we get the following ion-product expression:

\( \text{Ksp} = [\text{Ag}^{+}]^{2}[\text{CO}_{3}^{2-}] \).

For Barium Fluoride (\( \text{BaF}_{2} \)), it is:
\( \text{Ksp} = [\text{Ba}^{2+}][\text{F}^{-}]^{2} \).

For Copper(II) Sulfide (\( \text{CuS} \)), we have:
\( \text{Ksp} = [\text{Cu}^{2+}][\text{S}^{2-}] \).

The \( \text{Ksp} \) helps us understand and calculate the extent to which these compounds will dissolve in water. Knowing the \( \text{Ksp} \) allows chemists to predict and manipulate the solubility of different ionic compounds, critical in various scientific and industrial processes.

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

Find the molar solubility of \(\mathrm{SrCO}_{3}\left(K_{\mathrm{sp}}=5.4 \times 10^{-10}\right)\) in (a) pure water and (b) \(0.13 M \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}\).

Gout is caused by an error in metabolism that leads to a buildup of uric acid in body fluids, which is deposited as slightly soluble sodium urate \(\left(\mathrm{C}_{5} \mathrm{H}_{3} \mathrm{~N}_{4} \mathrm{O}_{3} \mathrm{Na}\right)\) in the joints. If the extracellular \(\left[\mathrm{Na}^{+}\right]\) is \(0.15 \mathrm{M}\) and the solubility of sodium urate is \(0.085 \mathrm{~g} / 100 . \mathrm{mL}\) what is the minimum urate ion concentration (abbreviated [Ur-]) that will cause a deposit of sodium urate?

Find the \(\mathrm{pH}\) of a buffer that consists of \(1.3 \mathrm{M}\) sodium phenolate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{ONa}\right)\) and \(1.2 M\) phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\left(\mathrm{p} K_{\mathrm{a}}\right.\) of phenol \(\left.=10.00\right)\)

Explain how strong acid-strong base, weak acid-strong base, and weak base- strong acid titrations using the same concentrations differ in terms of (a) the initial pH and (b) the pH at the equivalence point. (The component in italics is in the flask.)

Find the solubility of \(\mathrm{Cr}(\mathrm{OH})_{3}\) in a buffer of \(\mathrm{pH} 13.0\left[K_{\mathrm{sp}}\right.\) of \(\mathrm{Cr}(\mathrm{OH})_{3}=6.3 \times 10^{-31} ; K_{\mathrm{f}}\) of \(\left.\mathrm{Cr}(\mathrm{OH})_{4}^{-}=8.0 \times 10^{29}\right]\)

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.