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

Short Answer

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Reacting MnS with Strong Acid: \(MnS + 2H^+ \rightarrow Mn^{2+} + H_2S\)

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01

Reacting MnS with Strong Acid

\(MnS + 2H^+ \rightarrow Mn^{2+} + H_2S\) In the case of PbI鈧, the net ionic equation is not formed because HI is a strong acid and will not react with H鈦 ions. In the case of AuCl鈧, the net ionic equation is not formed because Cl鈦 ion is not reacting with H鈦 ions. In the case of Hg鈧侰鈧侽鈧, the anions don't react with H鈦, so no net ionic equation is formed. In the case of CuBr, the net ionic equation is not formed because HBr is a strong acid that cannot react with H鈦 ions. So, the only net ionic equation in this exercise is for MnS reacting with a strong acid to form Mn虏鈦 and H鈧係.

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Key Concepts

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

Slightly Soluble Salts
Slightly soluble salts are compounds that only dissolve to a small extent in water. These salts have low solubility, meaning that only a small amount of the salt can dissolve in the solution at any given time. In practical terms, when these salts are added to water, they tend to form a solid precipitate rather than dissolving completely. Examples include substances like calcium fluoride (\( \text{CaF}_2 \) ) and magnesium carbonate (\( \text{MgCO}_3 \)).Understanding slightly soluble salts is crucial because their solubility can be affected by different factors, such as changes in pH or the presence of other ions in the solution. Chemists often deal with these salts to predict and analyze the output of various chemical reactions in aqueous (water-based) environments. When dealing with chemical equations, slightly soluble salts commonly appear on the reactant side, as they are generally present in the system as solid forms.
Strong Acid Reactions
Strong acids play a significant role in chemical reactions, particularly when dealing with ionic compounds. Known for their complete ionization in water, strong acids like hydrochloric acid (\( \text{HCl} \)) and sulfuric acid (\( \text{H}_2\text{SO}_4 \)) are often used to facilitate reactions with slightly soluble salts.During a reaction with a strong acid, the donation of protons (\( \text{H}^+ \)) can change the solubility of the salt and cause some ions previously locked in solid form to dissolve.For example, in the case of \( \text{MnS} \), the addition of a strong acid such as hydrochloric acid results in an ionic reaction where \( \text{MnS} \) reacts with the \( \text{H}^+ \) ions to form \( \text{Mn}^{2+} \) and hydrogen sulfide gas (\( \text{H}_2\text{S} \)).Thus, strong acids are valuable tools in manipulating the solubility and reactivity of slightly soluble salts in various chemical evaluations.
MnS Reaction
The reaction between manganese sulfide (\( \text{MnS} \)) and strong acids is a classic example of how slightly soluble salts interact in solution to form new products.In this reaction, \( \text{MnS} \), which is only slightly soluble in water, reacts with strong acids to form manganese ions (\( \text{Mn}^{2+} \)) and hydrogen sulfide gas (\( \text{H}_2\text{S} \)).The net ionic equation is as follows:\[ \text{MnS} (s) + 2\text{H}^+ (aq) \rightarrow \text{Mn}^{2+} (aq) + \text{H}_2\text{S} (g) \]This equation signifies that the strong acid supplies the \( \text{H}^+ \) ions needed to dissolve the otherwise slightly soluble \( \text{MnS} \), resulting in the formation of sulfur gas and soluble manganese ions.Such reactions are vital in the realm of chemistry for extracting or dissolving specific components from mixtures.
Solubility Rules
Solubility rules are guidelines that help predict the solubility of various ionic compounds in water. These rules assist chemists in determining which salts will dissolve and which will form precipitates.Here are a few basic solubility rules to consider:
  • Most nitrate (\( \text{NO}_3^- \)) salts are soluble in water.
  • Salts containing alkali metal ions and ammonium (\( \text{NH}_4^+ \)) are generally soluble.
  • Chloride (\( \text{Cl}^- \)), bromide (\( \text{Br}^- \)), and iodide (\( \text{I}^- \)) salts are soluble, except those of silver, lead (\( \text{Pb}^{2+} \)), and mercury (\( \text{Hg}_2^{2+} \)).
  • Sulfate (\( \text{SO}_4^{2-} \)) salts are soluble, with exceptions like barium sulfate (\( \text{BaSO}_4 \)) and calcium sulfate (\( \text{CaSO}_4 \)).
  • Most sulfide (\( \text{S}^{2-} \)), carbonate (\( \text{CO}_3^{2-} \)), and phosphate (\( \text{PO}_4^{3-} \)) salts are insoluble, except those containing alkali metals and \( \text{NH}_4^+ \).
Applying these solubility rules aids in anticipating the outcomes of reactions involving ionic compounds and in understanding which products will dissolve or precipitate in water.

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Most popular questions from this chapter

Equal quantities of 0.010\(M\) solutions of an acid \(\mathrm{HA}\) and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is \(9.2 .\) (a) Write the chemical equation and equilibrium-constant expression for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\) . (b) If \(K_{a}\) for \(\mathrm{HA}\) is \(8.0 \times 10^{-5}\) , what is the value of the equilibrium constant for the reaction between \(\mathrm{HA}\) and \(\mathrm{B} ?\) (c) What is the value of \(K_{b}\) for \(\mathrm{B}\) ?

A 20.0 -mL sample of 0.200 \(\mathrm{M}\) HBr solution is titrated with 0.200 \(\mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) of the solution after the following volumes of base have been added: (a) 15.0 \(\mathrm{mL}\) \((\mathbf{b}) 19.9 \mathrm{mL},(\mathbf{c}) 20.0 \mathrm{mL},(\mathbf{d}) 20.1 \mathrm{mL},(\mathbf{e}) 35.0 \mathrm{mL}\)

(a) Calculate the \(\mathrm{pH}\) of a buffer that is 0.105 \(\mathrm{M}\) in \(\mathrm{NaHCO}_{3}\) and 0.125 \(\mathrm{M}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) . (b) Calculate the pH of a solution formed by mixing 65 \(\mathrm{mL}\) of 0.20 \(\mathrm{M} \mathrm{NaHCO}_{3}\) with 75 \(\mathrm{mL}\) of 0.15 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3} .\)

The beaker on the right contains 0.1 Macetic acid solution with methyl orange as an indicator. The beaker on the left contains a mixture of 0.1\(M\) acetic acid and 0.1\(M\) sodium acetate with methyl orange. (a) Using Figures 16.8 and 16.9, which solution has a higher pH? (b) Which solution is better able to maintain its pH when small amounts of NaOH are added? Explain. [Sections 17.1 and 17.2]

A sample of 7.5 \(\mathrm{L}\) of \(\mathrm{NH}_{3}\) gas at \(22^{\circ} \mathrm{C}\) and 735 torr is bubbled into a 0.50 -L solution of 0.40 \(\mathrm{M}\) HCl. Assuming that all the \(\mathrm{NH}_{3}\) dissolves and that the volume of the solution remains \(0.50 \mathrm{L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

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