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When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{MgI}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{AlBr}_{3}(a q)\)

Short Answer

Expert verified
In summary, a precipitate will form in reactions a and c: a. \(Hg_{2}(NO_{3})_{2} + CuSO_{4}\) will form \(Hg_{2}SO_{4}\) as a precipitate. c. \(K_{2}CO_{3} + MgI_{2}\) will form \(MgCO_{3}\) as a precipitate.

Step by step solution

01

a. Hg₂(NO₃)₂(aq) + CuSO₄(aq)

In this case, let's consider the possible double replacement reaction products: - Hg₂SO₄ and Cu(NO₃)₂ Now, let's apply the solubility rules: - Hg₂SO₄ is insoluble (rule: most sulfates are soluble, except for BaSO₄, PbSO₄, Hg₂SO₄, and CaSO₄) - Cu(NO₃)₂ is soluble (rule: all nitrates are soluble) Since Hg₂SO₄ is insoluble, a precipitate will form in this case.
02

b. Ni(NO₃)₂(aq) + CaCl₂(aq)

The possible double replacement reaction products are: - NiCl₂ and Ca(NO₃)₂ Let's use the solubility rules: - NiCl₂ is soluble (rule: all chlorides are soluble, except for AgCl, PbCl₂, and Hg₂Cl₂) - Ca(NO₃)₂ is soluble (rule: all nitrates are soluble) Since both products are soluble, no precipitate will form in this case.
03

c. K₂CO₃(aq) + MgI₂(aq)

The possible double replacement reaction products are: - KI and MgCO₃ Let's apply the solubility rules: - KI is soluble (rule: all alkali metal salts are soluble) - MgCO₃ is insoluble (rule: most carbonates are insoluble, except for those of alkali metals and ammonium) Since MgCO₃ is insoluble, a precipitate will form in this case.
04

d. Na₂CrO₄(aq) + AlBr₃(aq)

The possible double replacement reaction products are: - NaBr and Al₂(CrO₄)₃ Let's apply solubility rules: - NaBr is soluble (rule: all alkali metal salts are soluble) - Al₂(CrO₄)₃ is soluble (rule: most chromates are insoluble, except for those of alkali metals and ammonium) Since both products are soluble, no precipitate will form in this case. In summary, a precipitate will form in reactions a, and c: a. Hg₂(NO₃)₂ + CuSO₄ will form Hg₂SO₄ as a precipitate. c. K₂CO₃ + MgI₂ will form MgCO₃ as a precipitate.

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Key Concepts

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

Solubility Rules
Solubility rules are guidelines that help us predict whether a compound will dissolve in water or form a precipitate. They are essential in determining the outcomes of chemical reactions, particularly when solutions are mixed together. Here are some of the key solubility rules:
  • All nitrates (\(\mathrm{NO}_3^-\)) are soluble. This means compounds containing the nitrate ion will not form a precipitate.
  • Chlorides (\(\mathrm{Cl}^-\)) are generally soluble, except those combined with silver, lead, or mercury (\(\mathrm{AgCl, \; PbCl}_2, \; \text{and} \, Hg_2Cl_2\)).
  • Sulfates (\(\mathrm{SO}_4^{2-}\)) are mostly soluble, except for barium sulfate (\(\mathrm{BaSO}_4\)), lead sulfate (\(\mathrm{PbSO}_4\)), mercury sulfate (\(\mathrm{Hg}_2SO_4\)), and calcium sulfate (\(\mathrm{CaSO}_4\)).
  • Carbonates (\(\mathrm{CO}_3^{2-}\)) are insoluble except when paired with alkali metals (e.g., sodium or potassium) or ammonium.
  • Alkali metal salts (like those containing \(\mathrm{Li}^+, \; Na^+, \; ext{and} \, K^+\)) and ammonium (\(\mathrm{NH}_4^+\)) are soluble.
Understanding these rules helps predict which compounds will remain dissolved or form solids as precipitates when solutions are mixed.
Double Replacement Reactions
Double replacement reactions involve the exchange of ions between two compounds in aqueous solution. These reactions can lead to the formation of a precipitate, gas, or water. They are generally represented by the equation:\[\mathrm{AB}\,(aq) + \mathrm{CD}\,(aq) \rightarrow \mathrm{AD} + \mathrm{CB}\]where \(\mathrm{A}\) and \(\mathrm{C}\) are cations and \(\mathrm{B}\) and \(\mathrm{D}\) are anions. In a double replacement reaction:
  • The cation from one compound combines with the anion from the other, potentially forming an insoluble precipitate.
  • These reactions often involve ionic compounds dissociating in water, making the ions available for exchange.
  • The driving force for these reactions includes the formation of a precipitate, gas, or a stable molecule like water.
For instance, when potassium carbonate (\(\mathrm{K}_2\mathrm{CO}_3\)) and magnesium iodide (\(\mathrm{MgI}_2\)) react, they can form magnesium carbonate (\(\mathrm{MgCO}_3\)), which is insoluble in water and precipitates out of the solution. This satisfies the condition for a double replacement reaction leading to a precipitate.
Insoluble Compounds
Insoluble compounds are those that do not dissolve significantly in water. In the context of precipitation reactions, they are crucial as they form the solid precipitate observed during a reaction. Precipitation occurs when the product of a reaction is insoluble in the solvent, usually water, and separates from the rest of the mixture as a solid. Key characteristics of insoluble compounds include:
  • They tend to have low solubility product constants, indicating their low tendency to dissolve in water.
  • In solubility rules, many insoluble compounds are exceptions to the general trends. For example, most carbonates are insoluble except when paired with alkali metals.
  • In reactions, such compounds drive the formation of the observable precipitate. As seen with Hg₂SO₄ or MgCO₃, their low solubility makes them precipitate out of the solution.
The formation of an insoluble compound is an important factor in predicting and analyzing chemical reactions. By understanding solubility rules and recognizing potential insoluble products, one can determine if a precipitate will form.

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Most popular questions from this chapter

The concentration of a certain sodium hydroxide solution was determined by using the solution to titrate a sample of potassium hydrogen phthalate (abbreviated as KHP). KHP is an acid with one acidic hydrogen and a molar mass of 204.22 g/ mol. In the titration, 34.67 mL of the sodium hydroxide solution was required to react with 0.1082 g KHP. Calculate the molarity of the sodium hydroxide.

A 6.50-g sample of a diprotic acid requires 137.5 mL of a 0.750 M NaOH solution for complete neutralization. Determine the molar mass of the acid.

Many plants are poisonous because their stems and leaves contain oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) or sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) . When ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-},\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride, \(\mathrm{CaCl}_{2},\) in aqueous solution.

A 10.00-mL sample of sulfuric acid from an automobile battery requires 35.08 mL of 2.12 M sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid? Sulfuric acid contains two acidic hydrogens

What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from 75.0 mL of a 0.100-M solution of \(\mathrm{AgNO}_{3} ?\)

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