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Chapter 5: Problem 4

Which compound or compounds in each of the following groups is (are) expected to be soluble in water? (a) \(\mathrm{BaSO}_{4}, \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{BaCO}_{3}\) (b) \(\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{NaClO}_{4}, \mathrm{NaCH}_{3} \mathrm{CO}_{2}\) (c) \(\mathrm{AgBr}, \mathrm{KBr}, \mathrm{Al}_{2} \mathrm{Br}_{6}\)

Short Answer

Expert verified
(a) Ba(NO鈧�)鈧�, (b) All compounds, (c) KBr, Al鈧侭r鈧� are soluble.

Step by step solution

01

Identify Solubility Rules

To determine the solubility of the compounds, we first need to recall some general solubility rules: - Most nitrate (NO鈧冣伝) salts are soluble. - Salts containing alkali metal ions (like Na鈦�) are generally soluble. - Most sulfate (SO鈧劼测伝) salts are soluble except for barium sulfate (BaSO鈧�), and a few others like calcium sulfate (CaSO鈧�). - Carbonates (CO鈧兟测伝) are generally insoluble except for those of alkali metals. - Most chlorates (ClO鈧勨伝) are soluble. - AgBr (and other silver halides) are generally insoluble. - Most bromides (Br鈦�) are soluble, except for some such as AgBr.
02

Evaluate Group (a) Compounds

In group (a): - BaSO鈧�: Barium sulfate is generally insoluble in water. - Ba(NO鈧�)鈧�: Following the rule that most nitrates are soluble, Barium nitrate is soluble in water. - BaCO鈧�: Barium carbonate is generally insoluble in water. Thus, only Ba(NO鈧�)鈧� is expected to be soluble in water.
03

Evaluate Group (b) Compounds

In group (b): - Na鈧係O鈧�: Sodium sulfate should be soluble as sulfates are generally soluble and it contains the alkali metal Na鈦�. - NaClO鈧�: Sodium perchlorate is soluble as chlorates are generally soluble and it also contains Na鈦�. - NaCH鈧僀O鈧�: Sodium acetate is soluble due to the presence of Na鈦�. Thus, all compounds in group (b) are expected to be soluble in water.
04

Evaluate Group (c) Compounds

In group (c): - AgBr: Silver bromide is generally insoluble. - KBr: Potassium bromide is soluble; potassium salts are usually soluble. - Al鈧侭r鈧�: Aluminum bromide is soluble (although less commonly addressed by direct rules, bromides are typically soluble unless paired with metals like Ag). Thus, KBr and Al鈧侭r鈧� are expected to be soluble in water.

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Key Concepts

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

Water Solubility
Understanding water solubility is crucial for predicting how compounds will behave when dissolved in water. This concept revolves around the ability of a substance to dissolve in water, forming a homogeneous solution.
Water is known as a "universal solvent" because it can dissolve many substances due to its polar nature. However, not all substances dissolve in water, and this depends on several factors.
General solubility rules help us predict which compounds are water-soluble:
  • Compounds containing alkali metal ions (Li鈦�, Na鈦�, K鈦�, etc.) are usually soluble.
  • Nitrate ( ext{NO}_3^{-}) compounds are generally soluble.
  • Most chlorides, bromides, and iodides are soluble, except when paired with silver, lead, or mercury.

Solubility can be affected by temperature, ionic strength, and the presence of other substances in the solution. Understanding the specific rules and exceptions is key to making accurate predictions.
Nitrate Solubility
Nitrate solubility is a specific concept explaining why most nitrate compounds ( ext{NO}_3^{-}) are soluble in water. Nitrate salts, regardless of the metal they are paired with, exhibit high water solubility.
This is because the nitrate ion is very stable and forms multiple hydrogen bonds with water molecules.
Key facts about nitrate solubility include:
  • All common inorganic nitrates are soluble.
  • They dissolve by forming strong interactions with water molecules, overcoming lattice energy.
  • Nitrate solubility is utilized in various applications like fertilizers and explosives because of their ready dissolution.

Because of these reliable rules, predicting whether a nitrate compound will dissolve in water is usually straightforward, making it a dependable rule in chemistry.
Sulfate Solubility
The concept of sulfate solubility involves understanding which sulfate ( ext{SO}_4^{2-}) compounds dissolve in water. Most sulfates are soluble, but there are notable exceptions due to specific interactions that affect their solubility.
General sulfate solubility rules include:
  • Most sulfates dissolve in water, including those of metals like magnesium and sodium.
  • Exceptions include barium sulfate ( ext{BaSO}_4), lead sulfate ( ext{PbSO}_4), and calcium sulfate ( ext{CaSO}_4), which are not soluble or only slightly soluble.
  • The solubility depends on the cation. Larger and more positively charged cations tend to form less soluble sulfates.

Understanding these exceptions is critical, especially in fields like medicine and environmental science, where the behavior of sulfate compounds affects health and ecosystems.
Carbonates Insolubility
Carbonates ( ext{CO}_3^{2-}) are generally insoluble in water, with few exceptions. The insolubility arises from the large, complex nature of carbonate ions and how they interact with different cations.
Important points about carbonate solubility are:
  • Most carbonate compounds, such as metal carbonates, do not dissolve in water.
  • Exceptions are carbonates of alkali metals (such as sodium carbonate, ext{Na}_2 ext{CO}_3) and ammonium carbonate, which are soluble.
  • The insolubility is due to the strong ionic bonds in the lattice structure that water cannot easily dissociate.

Understanding the insolubility of carbonates is crucial for applications like water treatment and geology, where these compounds play a significant role.

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Most popular questions from this chapter

What volume of \(0.123 \mathrm{M} \mathrm{NaOH},\) in milliliters, contains \(25.0 \mathrm{g}\) of \(\mathrm{NaOH} ?\)

A The cancer chemotherapy drug cisplatin, \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) can be made by reacting (NH_) \(_{2} \mathrm{PtCl}_{4}\) with ammonia in aqueous solution. Besides cisplatin, the other product is \(\mathrm{NH}_{1} \mathrm{Cl}\) (a) Write a balanced equation for this reaction. (b) To obtain \(12.50 \mathrm{g}\) of cisplatin, what mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtCl}_{4}\) is required? What volume of \(0.125 \mathrm{M}\) \(\mathrm{NH}_{3}\) is required? (c) Cisplatin can react with the organic compound pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N},\) to form a new compound. $$\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{aq})+x \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(\mathrm{aq}) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{x}(\mathrm{s})$$ Suppose you treat \(0.150 \mathrm{g}\) of cisplatin with what you believe is an excess of liquid pyridine \((1.50 \mathrm{mL}\) \(d=0.979 \mathrm{g} / \mathrm{mL}) .\) When the reaction is complete, you can find out how much pyridine was not used by titrating the solution with standardized HCl. If 37.0 mL. of \(0.475 \mathrm{M} \mathrm{HCl}\) is required to titrate the excess pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) what is the formula of the unknown compound \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{x} ?\)

Which of the following methods would you use to prepare \(300 . \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\) (a) Add \(30.0 \mathrm{mL}\) of \(1.50 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to \(270 .\) mL of water. (b) Dilute \(250 .\) mL of \(0.600 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a volume of \(300 . \mathrm{mL}\).

A compound has been isolated that can have either of two possible formulas: (a) \(\mathrm{K}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) or (b) \(\mathrm{K}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right] .\) To find which is correct, you dissolve a weighed sample of the compound in acid and then titrate the oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}^{2}\right)\) that comes from the compound with potassium permanganate, \(\mathrm{KMnO}_{4}\) (the source of the \(\mathrm{MnO}_{4}^{-}\) ion). The balanced, net ionic equation for the titration is $$\begin{array}{rl}5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(\mathrm{aq})+2 \mathrm{MnO}_{4}^{-}(\mathrm{aq})+16 \mathrm{H}^{+}(\mathrm{aq}) & \longrightarrow \\\2 \mathrm{Mn}^{2+}(\mathrm{aq})+10 \mathrm{CO}_{2}(\mathrm{g})+8 & \mathrm{H}_{2} \mathrm{O}(\ell) \end{array}$$ Titration of \(1.356 \mathrm{g}\) of the compound requires \(34.50 \mathrm{mL}\) of \(0.108 \mathrm{M} \mathrm{KMnO}_{4} .\) Which is the correct formula of the iron-containing compound: (a) or (b)?

Which contains the greater mass of solute: 1 I. of \(0.1 \mathrm{M}\) NaCl or 1 L of \(0.06 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3} ?\)
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