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Chapter 16: Problem 7

Which is more likely to dissolve in an acidic solution, silver sulfide or silver chloride? Why?

Short Answer

Expert verified
Silver sulfide is more likely to dissolve in an acidic solution compared to silver chloride. This is because silver sulfide reacts with protons in acidic conditions to form hydrogen sulfide gas, which shifts the equilibrium and increases its solubility. On the other hand, the solubility of silver chloride is not significantly affected by acidic conditions.

Step by step solution

01

Recall the solubility rules

Recall that solubility rules help us predict which compounds will be soluble or insoluble in water. Some general rules are: 1. Most nitrates, acetates, and perchlorates are soluble. 2. Most alkali metal salts and ammonium salts are soluble. 3. Most chloride, bromide, and iodide salts are soluble, except for those formed with silver, lead, and mercury. 4. Most sulfate salts are soluble, except for those formed with barium, calcium, strontium, and lead. 5. Most carbonate, phosphate, and hydroxide salts are insoluble, except for those formed with alkali metals and ammonium.
02

Compare the Ksp values of silver sulfide and silver chloride

The solubility product constant (Ksp) is an equilibrium constant that quantitatively describes the extent to which an ionic compound dissolves in water. A higher Ksp value indicates a more soluble compound. The Ksp values for silver sulfide and silver chloride are: - Silver sulfide (Ag2S): \( K_{sp} = 6.3 \times 10^{-51} \) - Silver chloride (AgCl): \( K_{sp} = 1.8 \times 10^{-10} \) Comparing these values, we can see that silver chloride has a much higher Ksp, indicating that it is more soluble in water than silver sulfide.
03

Discuss the effects of acidic conditions on solubility

Under acidic conditions, H+ ions (protons) in the solution can react with anions from the insoluble salts. This can cause a shift in the equilibrium and increase the solubility of some ionic compounds. In the case of silver chloride, the chloride anion (Cl-) does not react significantly with protons. Therefore, the solubility of silver chloride is not significantly affected by the presence of an acidic solution. Silver sulfide, on the other hand, reacts with protons in acidic conditions to form hydrogen sulfide (H2S) gas. This reaction shifts the equilibrium and leads to an increase in solubility of silver sulfide in an acidic solution.
04

Compare the solubility of silver sulfide and silver chloride in acidic conditions

Considering both the Ksp values and the effects of acidic conditions on solubility, we can conclude that silver sulfide is more likely to dissolve in an acidic solution. Although silver chloride has a higher Ksp, its solubility is not significantly affected by acidic conditions, while the solubility of silver sulfide increases due to the formation of hydrogen sulfide gas.

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Most popular questions from this chapter

The concentration of \(\mathrm{Ag}^{+}\) in a solution saturated with \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\) is \(2.2 \times 10^{-4}\) M. Calculate \(K_{\text {sp }}\) for \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\).

A solution is prepared by adding \(0.10 \mathrm{~mol} \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to \(0.50 \mathrm{~L}\) of \(3.0 \mathrm{M} \mathrm{NH}_{3} .\) Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text {owerall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{\mathrm{x}}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

The concentration of \(\mathrm{Pb}^{2+}\) in a solution saturated with \(\mathrm{PbBr}_{2}(s)\) is \(2.14 \times 10^{-2} M .\) Calculate \(K_{\text {sp }}\) for \(\mathrm{PbBr}_{2}\).

a. Calculate the molar solubility of \(\mathrm{Sr} \mathrm{F}_{2}\) in water, ignoring the basic properties of \(\mathrm{F}^{-} .\) (For \(\mathrm{SrF}_{2}, K_{\mathrm{se}}=7.9 \times 10^{-10} .\) ) b. Would the measured molar solubility of \(\mathrm{Sr} \mathrm{F}_{2}\) be greater than or less than the value calculated in part a? Explain. c. Calculate the molar solubility of \(\mathrm{SrF}_{2}\) in a solution buffered at \(\mathrm{pH}=2.00 .\left(K_{\mathrm{a}}\right.\) for \(\mathrm{HF}\) is \(7.2 \times 10^{-4} .\) )

The overall formation constant for \(\mathrm{HgI}_{4}{ }^{2-}\) is \(1.0 \times 10^{30} .\) That is, $$1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[1^{-}\right]^{4}}$$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{~mL}\) of a solution that was originally \(0.010 \mathrm{M} \mathrm{Hg}^{2+}\) and \(0.78 \mathrm{M} \mathrm{I}^{-}\) ? The reaction is $$\mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}{ }^{2-}(a q)$$
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