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Chapter 17: Problem 64

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

Short Answer

Expert verified
The short answer for the net ionic equations for each slightly soluble salt reacting with a strong acid would be: (a) MnS (s) + 2 H⁺ (aq) → Mn²⁺ (aq) + H₂S (g) (b) PbF₂ (s) + 2 H⁺ (aq) → Pb²⁺ (aq) + 2 HF (aq) (c) No net ionic equation for AuCl₃ with strong acid. (d) Hg₂C₂O₄ (s) + 2 H⁺ (aq) → Hg₂²⁺ (aq) + H₂C₂O₄ (aq) (e) CuBr (s) + H⁺ (aq) → Cu²⁺ (aq) + HBr (aq)

Step by step solution

01

(a) MnS with strong acid

MnS is a slightly soluble salt, so when it reacts with the strong acid HCl, we expect to produce Mn²⁺ ions and hydrogen sulfide (H₂S) gas. The balanced complete ionic equation is: MnS (s) + 2 HCl (aq) → Mn²⁺ (aq) + 2 Cl⁻ (aq) + H₂S (g) Removing the spectator ions (Cl⁻) yields the net ionic equation: MnS (s) + 2 H⁺ (aq) → Mn²⁺ (aq) + H₂S (g)
02

(b) PbF₂ with strong acid

PbF₂ is a slightly soluble salt, so when it reacts with the strong acid HCl, we expect to produce Pb²⁺ ions and HF. The balanced complete ionic equation is: PbF₂ (s) + 2 HCl (aq) → Pb²⁺ (aq) + 2 Cl⁻ (aq) + 2 HF (aq) Removing the spectator ions (Cl⁻) yields the net ionic equation: PbF₂ (s) + 2 H⁺ (aq) → Pb²⁺ (aq) + 2 HF (aq)
03

(c) AuCl₃ with strong acid

AuCl₃ is soluble in water, so it won't react with strong acids. Therefore, there is no net ionic equation for this reaction.
04

(d) Hg₂C₂O₄ with strong acid

Hg₂C₂O₄ is a slightly soluble salt, so when it reacts with the strong acid HCl, we expect to produce Hg₂²⁺ ions and oxalic acid (H₂C₂O₄). The balanced complete ionic equation is: Hg₂C₂O₄ (s) + 2 HCl (aq) → Hg₂²⁺ (aq) + 2 Cl⁻ (aq) + H₂C₂O₄ (aq) Removing the spectator ions (Cl⁻) yields the net ionic equation: Hg₂C₂O₄ (s) + 2 H⁺ (aq) → Hg₂²⁺ (aq) + H₂C₂O₄ (aq)
05

(e) CuBr with strong acid

CuBr is a slightly soluble salt, so when it reacts with the strong acid HCl, we expect to produce Cu²⁺ ions and HBr. The balanced complete ionic equation is: CuBr (s) + HCl (aq) → Cu²⁺ (aq) + Cl⁻ (aq) + HBr (aq) Removing the spectator ions (Cl⁻) yields the net ionic equation: CuBr (s) + H⁺ (aq) → Cu²⁺ (aq) + HBr (aq)

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Most popular questions from this chapter

What is the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of water saturated with \(\mathrm{CO}_{2}\) at a partial pressure of \(111.5 \mathrm{kPa}\) ? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is $3.1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}-\mathrm{kPa}$.

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