The solubility of copper(II) hydroxide in water can be increased by adding either the base \(\mathrm{NH}_{3}\) or the acid \(\mathrm{HNO}_{3}\) . Explain. Would added \(\mathrm{NH}_{3}\) or \(\mathrm{HNO}_{3}\) have the same effect on the solubility of silver acetate or silver chloride? Explain.

Short Answer

Expert verified
Adding \(\mathrm{NH}_{3}\) increases the solubility of copper(II) hydroxide, silver acetate, and silver chloride by forming soluble complexes with their respective metal ions, while adding \(\mathrm{HNO}_{3}\) only increases the solubility of copper(II) hydroxide by reacting with \(\mathrm{OH^{-}}\) ions. The solubility of silver acetate and silver chloride remains unaffected by the addition of an acid like \(\mathrm{HNO}_{3}\) since their anions are not basic.

Step by step solution

01

1. Solubility of Copper(II) Hydroxide and Effect of NH₃ and HNO₃

Copper(II) hydroxide (\(\mathrm{Cu(OH)_{2}}\)) is only slightly soluble in water and the equilibrium can be represented as: \[\mathrm{Cu(OH)_{2}(s) \rightleftharpoons Cu^{2+}(aq) + 2\,OH^{-}(aq)}\] When we add \(\mathrm{NH}_{3}\) (ammonia, a base) to the solution, it reacts with \(\mathrm{Cu^{2+}}\) ions to form the soluble copper-ammonia complex ion, \(\mathrm{[Cu(NH_{3})_{4}]^{2+}}\). The reaction is: \[\mathrm{Cu^{2+}(aq) + 4\,NH_{3}(aq) \rightleftharpoons [Cu(NH_{3})_{4}]^{2+}(aq)}\] This reaction reduces the concentration of \(\mathrm{Cu^{2+}}\) ions in the solution, causing the equilibrium of \(\mathrm{Cu(OH)_{2}}\) dissolution to shift to the right, and making it more soluble. Now, when we add \(\mathrm{HNO}_{3}\) (nitric acid) to the solution, it reacts with the \(\mathrm{OH^{-}}\) ions as follows: \[\mathrm{H^{+}(aq) + OH^{-}(aq) \rightleftharpoons H_{2}O(l)}\] In this case, the addition of acid neutralizes the \(\mathrm{OH^{-}}\) ions and decreases their concentration, causing the \(\mathrm{Cu(OH)_{2}}\) solubility equilibrium to shift to the right.
02

2. Comparing the Solubility of Silver Acetate and Silver Chloride with Copper(II) Hydroxide

Now, let's consider the solubility of silver acetate (\(\mathrm{AgC_{2}H_{3}O_{2}}\)) and silver chloride (\(\mathrm{AgCl}\)) in water, represented by the following equilibria: \[\mathrm{AgC_{2}H_{3}O_{2}(s) \rightleftharpoons Ag^{+}(aq) + C_{2}H_{3}O_{2}^{-}(aq)}\] \[\mathrm{AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)}\] When we add ammonia to these equilibria, \(\mathrm{Ag^{+}}\) ions form soluble complexes – \(\mathrm{[Ag(NH_{3})_{2}]^{+}}\) – similar to the copper case, increasing the solubility of both silver acetate and silver chloride. \[\mathrm{Ag^{+}(aq) + 2\,NH_{3}(aq) \rightleftharpoons [Ag(NH_{3})_{2}]^{+}(aq)}\] However, if we add an acid like \(\mathrm{HNO}_{3}\) to these solutions, there will be no significant reaction with the anions of silver acetate (\(\mathrm{C_{2}H_{3}O_{2}^{-}}\)) and silver chloride (\(\mathrm{Cl^{-}}\)), as they are not basic anions. Hence, the solubility of silver acetate and silver chloride will not change significantly when an acid is added. In conclusion, adding ammonia increases the solubility of all three compounds, while adding nitric acid only has a significant effect on the solubility of copper(II) hydroxide in water and not on the solubility of silver acetate or silver chloride.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A solution contains 0.018 mole each of \(\mathrm{I}^{-}, \mathrm{Br}^{-},\) and \(\mathrm{Cl}^{-}.\) When the solution is mixed with \(200 . \mathrm{mL}\) of \(0.24 M\) \(\mathrm{AgNO}_{3}\) what mass of \(\mathrm{AgCl}(s)\) precipitates out, and what is \(\left[\mathrm{Ag}^{+}\right] ?\) Assume no volume change. $$\mathrm{AgI} : K_{\mathrm{sp}}=1.5 \times 10^{-16}$$ $$\operatorname{AgBr} ; K_{\mathrm{sp}}=5,0 \times 10^{-13}$$ $$\mathrm{AgCl} : K_{\mathrm{sp}}=1.6 \times 10^{-10}$$

a. Calculate the molar solubility of \(\mathrm{AgBr}\) in pure water. \(K_{\mathrm{sp}}\) for \(\mathrm{AgBr}\) is \(5.0 \times 10^{-13}\) . b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of \(\mathrm{AgBr}\) will dissolve in \(250.0 \mathrm{mL}\) of $3.0 M\( \)\mathrm{NH}_{3}?$ e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

A solution contains \(1.0 \times 10^{-5} M \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What concentrations of \(\mathrm{A} \mathrm{g} \mathrm{NO}_{3}\) will cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\) \(\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)

Calculate the mass of manganese hydroxide that dissolves to form 1300 mL of a saturated manganese hydroxide solution. For $\mathrm{Mn}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=2.0 \times 10^{-13}.$

Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free