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For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), (e) \(\mathrm{CuBr}\).

Short Answer

Expert verified
The net ionic equations for the reaction of slightly soluble salts with acid are: a) \( MnS(s) + 2H鈦(aq) \rightarrow Mn虏鈦(aq) + H鈧係(g) \) b) \( PbF鈧(s) + 2Cl鈦(aq) \rightarrow PbCl鈧(s) + 2F鈦(aq) \) c) No Reaction d) \( Hg鈧侰鈧侽鈧(s) + 2H鈦(aq) \rightarrow Hg鈧侰l鈧(s) + 2HOC鈧侽鈧(aq) \) e) No Reaction

Step by step solution

01

Identify the acid and the ions present in the salt

Generally, acids that are commonly used are hydrochloric acid (HCl), sulfuric acid (H鈧係O鈧), and nitric acid (HNO鈧). For this exercise, we will use hydrochloric acid (HCl) as the acid. Let's identify the ions present in each salt: a) MnS - Mn虏鈦 (Manganese ion), S虏鈦 (Sulfide ion) b) PbF鈧 - Pb虏鈦 (Lead ion), F鈦 (Fluoride ion) c) AuCl鈧 - Au鲁鈦 (Gold ion), Cl鈦 (Chloride ion) d) Hg鈧侰鈧侽鈧 - Hg鈧偮测伜 (Mercury ion), C鈧侽鈧劼测伝 (Oxalate ion) e) CuBr - Cu鈦 (Copper ion), Br鈦 (Bromide ion)
02

Predict the possible products of the reaction

In this step, we will predict the possible products of the reaction between the salt and hydrochloric acid (HCl). Usually, slightly soluble salts will form a solid (precipitate) in the presence of an acid. a) MnS + 2HCl 鈫 MnCl鈧 + H鈧係 b) PbF鈧 + 2HCl 鈫 PbCl鈧 + 2HF c) AuCl鈧 + 3HCl 鈫 No Reaction (AuCl鈧 is soluble in HCl and forms chloroauric acid, HAuCl鈧) d) Hg鈧侰鈧侽鈧 + 2HCl 鈫 Hg鈧侰l鈧 + 2HOC鈧侽鈧 e) CuBr + HCl 鈫 No Reaction (CuBr is not soluble in HCl)
03

Write the balanced chemical equations

Now, we will write the balanced chemical equations for the reactions that took place: a) MnS(s) + 2HCl(aq) 鈫 MnCl鈧(aq) + H鈧係(g) b) PbF鈧(s) + 2HCl(aq) 鈫 PbCl鈧(s) + 2HF(aq) d) Hg鈧侰鈧侽鈧(s) + 2HCl(aq) 鈫 Hg鈧侰l鈧(s) + 2HOC鈧侽鈧(aq)
04

Write the net ionic equations by eliminating spectator ions

Finally, we will write the net ionic equations for the reactions by eliminating the spectator ions: a) MnS(s) + 2H鈦(aq) 鈫 Mn虏鈦(aq) + H鈧係(g) b) PbF鈧(s) + 2H鈦(aq) + 2Cl鈦(aq) 鈫 PbCl鈧(s) + 2F鈦(aq) + 2H鈦(aq) d) Hg鈧侰鈧侽鈧(s) + 2H鈦(aq) 鈫 Hg鈧侰l鈧(s) + 2HOC鈧侽鈧(aq) Note that in (b), the 2H鈦 ions are present on both sides of the equation. So, they are spectator ions and can be eliminated: b) PbF鈧(s) + 2Cl鈦(aq) 鈫 PbCl鈧(s) + 2F鈦(aq)

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Key Concepts

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

Net Ionic Equations
A net ionic equation is a way to express a chemical reaction that focuses only on the ions that participate directly in the formation of the products. In such equations, spectator ions, which do not change during the reaction, are excluded. This simplifies the equation and highlights the chemistry occurring in the solution.
For instance, in the reaction of MnS with HCl, the full chemical equation is:
  • MnS(s) + 2HCl(aq) 鈫 MnCl鈧(aq) + H鈧係(g)
To create a net ionic equation, we remove the spectator ions (in this case, Cl鈦 ions). The net ionic equation, which shows only the reacting species, becomes:
  • MnS(s) + 2H鈦(aq) 鈫 Mn虏鈦(aq) + H鈧係(g)
Net ionic equations help clarify the essence of the reaction by excluding parts that don鈥檛 directly participate. They are particularly useful in understanding precipitation reactions, redox reactions, and acid-base reactions.
Slightly Soluble Salts
Slightly soluble salts are compounds that do not dissolve well in water, meaning only a small amount will break apart into ions. This "limited solubility" often results in the formation of a solid or precipitate in solutions. Understanding the behavior of these salts is crucial when predicting reactions, especially in aqueous environments.
Taking lead(II) fluoride (PbF鈧) as an example, this salt is slightly soluble in water. When it reacts with hydrochloric acid (HCl), it forms lead(II) chloride (PbCl鈧), which is less soluble than PbF鈧 and precipitates out:
  • PbF鈧(s) + 2HCl(aq) 鈫 PbCl鈧(s) + 2HF(aq)
Writing the net ionic equation involves removing spectator ions. Here, ions like H鈦 do not appear in the net equation as they do not contribute directly to the formation of the precipitate. Recognition and understanding of slightly soluble salts help in predicting reactions and their products in various chemical settings.
Reaction with Acids
Reactions between slightly soluble salts and acids are fascinating as they typically involve the exchange of ions that form either new acids, dissolved species, or precipitates. Commonly used acids include hydrochloric acid (HCl), sulfuric acid (H鈧係O鈧), and nitric acid (HNO鈧), with HCl being most frequently used in educational settings for ease of understanding.
These reactions often dissolve the original salt or alter it to produce a different compound. For instance, Hg鈧侰鈧侽鈧 (mercury(II) oxalate) reacts with HCl as follows:
  • Hg鈧侰鈧侽鈧(s) + 2HCl(aq) 鈫 Hg鈧侰l鈧(s) + 2HOC鈧侽鈧(aq)
This example illustrates the transformation of a slightly soluble salt into another compound (Hg鈧侰l鈧), which also precipitates. The ability to recognize and predict such transformations provides deep insights into the behavior of ions in solutions.
Spectator Ions
Spectator ions are ions that do not participate directly in the chemical reaction. They appear on both sides of a chemical equation, unchanged, and their role is essentially to "watch" the reaction unfold, hence the term "spectator."
To illustrate, consider the reaction involving PbF鈧 and HCl mentioned earlier. The spectator ions are Cl鈦, present on both sides of the equation:
  • PbF鈧(s) + 2Cl鈦(aq) 鈫 PbCl鈧(s) + 2F鈦(aq)
In this reaction, the Cl鈦 ions do not partake in the formation of the precipitate (PbCl鈧), which means they are spectators. Their presence in ionic reactions does not affect the product outcome directly. Understanding and identifying these ions allows chemists to simplify complex reactions into more manageable net ionic equations, providing clearer insights into the chemical processes at play.

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Most popular questions from this chapter

(a) Precipitation of the group 4 cations (Figure 17.22) requires a basic medium. Why is this so? (b) What is the most significant difference between the sulfides precipitated in group 2 and those precipitated in group \(3 ?\) (c) Suggest a procedure that would serve to redissolve the group 3 cations following their precipitation.

Calculate the \(\mathrm{pH}\) at the equivalence point in titrating \(0.100 \mathrm{M}\) solutions of each of the following with \(0.080 \mathrm{M}\) \(\mathrm{NaOH}:\) (a) hydrobromic acid \((\mathrm{HBr})\), (b) lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right]\), (c) sodium hydrogen chromate \(\left(\mathrm{NaHCrO}_{4}\right)\)

(a) What is the common-ion effect? (b) Give an example of a salt that can decrease the ionization of \(\mathrm{HNO}_{2}\) in solution.

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C}\) ?

A biochemist needs \(750 \mathrm{~mL}\) of an acetic acid-sodium ?cetate buffer with pH 4.50. Solid sodium acetat猫 \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) and glacial acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) are available Glacial acetic acid is \(99 \% \mathrm{CH}_{3} \mathrm{COOH}\) by mass and has a density of \(1.05 \mathrm{~g} / \mathrm{mL}\). If the buffer is to be \(0.15 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{COOH}\), how many grams of \(\mathrm{CH}_{3} \mathrm{COONa}\) and how many milliliters of glacial acetic acid must be used?

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