Chapter 16

A solution contains \(3.0 \times 10^{-3} M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2} .\) What concentrations of \(\mathrm{KF}\) will cause precipitation of solid \(\mathrm{MgF}_{2}\left(K_{\mathrm{sp}}=6.4 \times 10^{-9}\right) ?\)

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}\), \(\mathrm{Na}_{2} \mathrm{S},\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are \(K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},\) and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=1 \times 10^{-54}.\)

A solution contains 0.25\(M \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.25\(M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) Can the metal ions be separated by slowly adding \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) Assume that for successful separation 99\(\%\) of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3}.\)

Write equations for the step wise formation of each of the following complex ions. a. \(N i(C N)_{4}^{2-}\) b. \(V\left(C_{2} O_{4}\right)_{3}^{3-}\)

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-} .\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are \(8.5 \times 10^{-40} \mathrm{M}\) and \(1.5 \times 10^{-3} M,\) respectively, in a \(0.11-M\) KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$

In the presence of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\) forms the complex ion \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .\) If the equilibrium concentrations of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) are \(1.8 \times 10^{-17} \mathrm{M}\) and \(1.0 \times 10^{-3} \mathrm{M},\) respectively, in a \(1.5-M \mathrm{NH}_{3}\) solution, calculate the value for the overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}.\) $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \qquad K_{\mathrm{overall}}=?$$

When aqueous KI is added gradually to mercury (II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{HgI}_{4}^{2-}.)\)

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr},\) and \(\mathrm{AgI?}\)

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

A solution is formed by mixing \(50.0 \mathrm{mL}\) of \(10.0 \mathrm{M}\) \(\mathrm{NaX}\) with \(50.0 \mathrm{mL}\) of \(2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3} .\) Assume that \(\mathrm{Cu}^{+}\) forms complex ions with \(\mathrm{X}^{-}\) as follows: $$\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) \quad K_{1}=1.0 \times 10^{2}$$ $$\operatorname{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) \qquad K_{2}=1.0 \times 10^{4}$$ $$\operatorname{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K_{3}=1.0 \times 10^{3}$$ with an overall reaction $$\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \qquad K=1.0 \times 10^{9}$$ Calculate the following concentrations at equilibrium a. \(\mathrm{CuX}_{3}^{2-} \quad\) b. \(\mathrm{CuX}_{2}^{-} \quad\) c. \(\mathrm{Cu}^{+}\)

A solution is prepared by mixing 100.0 \(\mathrm{mL}\) of \(1.0 \times 10^{-4} M\) \(\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{2}\) and 100.0 \(\mathrm{mL}\) of \(8.0 M\) \(\mathrm{NaF}.\) $$\mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}^{+}(a q) \quad K_{1}=7.9 \times 10^{4}$$ $$\operatorname{Be} \mathrm{F}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{2}(a q) \quad K_{2}=5.8 \times 10^{3}$$ $$\operatorname{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{3}^{-}(a q) \quad K_{3}=6.1 \times 10^{2}$$ $$\operatorname{Be} \mathrm{F}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{4}^{2-}(a q) \qquad K_{4}=2.7 \times 10^{1}$$ Calculate the equilibrium concentrations of \(\mathrm{F}^{-}, \mathrm{Be}^{2+}, \mathrm{BeF}^{+},\) \(\mathrm{BeF}_{2}, \mathrm{BeF}_{3}^{-},\) and \(\mathrm{BeF}_{4}^{2-}\) in this solution.

Solutions of sodium thiosulfate are used to dissolve unexposed \(\operatorname{AgBr}\left(K_{\mathrm{sp}}=5.0 \times 10^{-13}\right)\) in the developing process for black-and-white film. What mass of \(\mathrm{AgBr}\) can dissolve in \(1.00 \mathrm{L}\) of \(0.500 M\) \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} ? \mathrm{Ag}^{+}\) reacts with \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to form a complex ion: $$\begin{aligned} \mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) & \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \\ K &=2.9 \times 10^{13} \end{aligned}$$

\(K_{\mathrm{f}}\) for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7} . K_{\mathrm{sp}}\) for \(\mathrm{AgCl}\) is \(1.6 \times 10^{-10} .\) Calculate the molar solubility of \(\mathrm{AgCl}\) in \(1.0M\) \(\mathrm{NH}_{3}.\)

The copper(I) ion forms a chloride salt that has \(K_{\mathrm{sp}}= 1.2 \times 10^{-6} .\) Copper (I) also forms a complex ion with \(\mathrm{Cl}^{-} :\) $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \qquad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in 0.10\(M\) \(\mathrm{NaCl}\).

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.

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. $$\mathrm{AgI} : K_{\mathrm{sp}}=1.5 \times 10^{-16}$$ $$\operatorname{AgBr} ; K_{\mathrm{sp}}=5,0 \times 10^{-13}$$ $$\mathrm{AgCl} : K_{\mathrm{sp}}=1.6 \times 10^{-10}$$

Magnesium hydroxide, \(\operatorname{Mg}(\mathrm{OH})_{2},\) is the active ingredient in the antacid TUMS and has a \(K_{\mathrm{sp}}\) value of \(8.9 \times 10^{-12}\) . If a 10.0 -g sample of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is placed in \(500.0 \mathrm{mL}\) of solution, calculate the moles of OH -ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is much less than \(1,\) not a lot solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acid.

Tooth enamel is composed of the mineral hydroxyapatite. The \(K_{\mathrm{sp}}\) of hydroxyapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) is \(6.8 \times 10^{-37}\) . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\) , forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times 10^{-60}\) . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is 1 \(\mathrm{mg} \mathrm{F}^{-}\) per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{CaF}_{2}=4.0 \times 10^{-11}\right)\)

The \(K_{\mathrm{sp}}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32} .\) At what pH will a \(0.2-M\) \(\mathrm{Al}^{3+}\) solution begin to show precipitation of \(\mathrm{Al}(\mathrm{OH})_{3} ?\)

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

On a hot day, a 200.0 -mL sample of a saturated solution of \(\mathrm{PbI}_{2}\) was allowed to evaporate until dry. If 240 mg of solid \(\mathrm{PbI}_{2}\) was collected after evaporation was complete, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{PbI}_{2}\) on this hot day.

The active ingredient of Pepto-Bismol is the compound bismuth subsalicylate, which undergoes the following dissociation when added to water: $$\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}^{2-}(a q) +\mathrm{Bi}^{3+}(a q)+\mathrm{OH}^{-}(a q) \qquad K=?$$ If the maximum amount of bismuth subsalicylate that reacts by this reaction is \(3.2 \times 10^{-19} \mathrm{mol} / \mathrm{L}\) , calculate the equilibrium constant for the preceding reaction.

Silver cyanide \((\mathrm{AgCN})\) is an insoluble salt with \(K_{\mathrm{sp}}=2.2 \times 10^{-12}\) . Compare the effects on the solubility of silver cyanide by addition of \(\mathrm{HNO}_{3}(a q)\) or by addition of \(\mathrm{NH}_{3}(a q).\)

Nanotechnology has become an important field, with applications ranging from high-density data storage to the design of “nano machines.” One common building block of nano structured architectures is manganese oxide nano particles. The particles can be formed from manganese oxalate nano rods, the formation of which can be described as follows: $$\mathrm{Mn}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MnC}, \mathrm{O}_{4}(a q) \quad K_{1}=7.9 \times 10^{3}$$ $$\mathrm{MnC}_{2} \mathrm{O}_{4}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}(a q) \quad K_{2}=7.9 \times 10^{1}$$ Calculate the value for the overall formation constant for \(\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-} :\) $$K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}}$$

The equilibrium constant for the following reaction is \(1.0 \times 10^{23} :\) $$\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q) \rightleftharpoons \mathrm{CrEDTA}^{-}(a q)+2 \mathrm{H}^{+}(a q)$$ EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt \(\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}\) , are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 M\) in \(\mathrm{Cr}^{3+}\) and \(0.050 M\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00.\)

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}\) b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{ED} \mathrm{TA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+} \mathrm{is}\) $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(\mathrm{M}\) Na_thion. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

Will a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) form if \(1.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(1.0 \mathrm{L}\) of \(5.0 \mathrm{MNH}_{3} ?\) $$\mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{7}$$ $$\mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=5.9 \times 10^{-15}$$

a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\) calculate the value for the equilibrium constant for the following reaction: $$\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$$ b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0M\) \(\mathrm{NH}_{3} .\) In \(5.0M\) \(\mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\) .

Describe how you could separate the ions in each of the following groups by selective precipitation. a. \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\) c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\)

Nitrate salts are generally considered to be soluble salts. One of the least soluble nitrate salts is barium nitrate. Approximately 15 g of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) will dissolve per liter of solution. Calculate the \(K_{\mathrm{sp}}\) value for barium nitrate.

Assuming that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is \(1.6 \times 10^{-7} \mathrm{mol} / \mathrm{L}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{sp}}\) for this salt. Ignore any potential reactions of the ions with water.

Order the following solids (a–d) from least soluble to most soluble. Ignore any potential reactions of the ions with water. a. \(\mathrm{AgCl} \quad K_{s p}=1.6 \times 10^{-10}\) b. \(\mathrm{Ag}_{2} \mathrm{S} \quad K_{\mathrm{sp}}=1.6 \times 10^{-49}\) c. \(\mathrm{CaF}_{2} \quad K_{\mathrm{sp}}=4.0 \times 10^{-11}\) d. \(\mathrm{CuS} \quad K_{\mathrm{sp}}=8.5 \times 10^{-45}\)

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(7.2 \times 10^{-2}-M\) \(\mathrm{KIO}_{3}\) solution is \(6.0 \times 10^{-9} \mathrm{mol} / \mathrm{L}\) . Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\)

The copper(l) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: $$\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ a. Calculate the solubility of \(\mathrm{CuBr}(s)\left(K_{\mathrm{sp}}=1.0 \times 10^{-5}\right)\) in \(1.0 \mathrm{L}\) of \(1.0 \mathrm{M} \mathrm{NaCN}\) . b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.

Consider a solution made by mixing \(500.0 \mathrm{mL}\) of \(4.0 \mathrm{M} \mathrm{NH}_{3}\) and \(500.0 \mathrm{mL}\) of \(0.40 \mathrm{M} \mathrm{AgNO}_{3} . \mathrm{Ag}^{+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}:\) $$\mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \qquad K_{1}=2.1 \times 10^{3}$$ $$\operatorname{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K_{2}=8.2 \times 10^{3}$$ Determine the concentration of all species in solution.

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?

Calculate the equilibrium concentrations of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+},\) and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) in a solution prepared by mixing \(500.0 \mathrm{mL}\) of \(3.00M\) \(\mathrm{NH}_{3}\) with \(500.0 \mathrm{mL}\) of \(2.00 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) . The step wise equilibria are $$\mathrm{Cu}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{CuNH}_{3}^{2+}(a q) \quad K_{1}=1.86 \times 10^{4}$$ $$\mathrm{CuNH}_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q) \quad K_{2}=3.88 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{2}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{2}\right)_{3}^{2+}(a q) \quad K_{3}=1.00 \times 10^{3}$$ $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K_{4}=1.55 \times 10^{2}$$

Calculate the solubility of \(\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\) in a solution containing \(1.0 M\) \(\mathrm{H}^{+} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN} \text { is } 6.2 \times 10^{-10} .\right)\)

Calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) is relatively insoluble in water \(\left(K_{\mathrm{sp}}=2 \times 10^{-9}\right) .\) However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in 0.10\(M \mathrm{H}^{+}\) than in pure water? In pure water, ignore the basic properties of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}.\)

What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10 M\) \(\mathrm{H}_{2} \mathrm{S}\) and maintained at \(\mathrm{pH} 3.0\) with \(\mathrm{HCl} ?\)

A mixture contains \(1.0 \times 10^{-3} M \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-3} M\) \(\mathrm{Mn}^{2+}\) and is saturated with 0.10\(M \mathrm{H}_{2} \mathrm{S} .\) Determine a pH where CuS precipitates but MnS does not precipitate. \(K_{\mathrm{sp}}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and \(K_{\mathrm{sp}} \mathrm{for} \mathrm{MnS}=2.3 \times 10^{-13} .\)

The salt MX has a solubility of \(3.17 \times 10^{-8} \mathrm{mol} / \mathrm{L}\) in a solution with \(\mathrm{pH}=0.000 .\) If \(K_{\mathrm{a}}\) for \(\mathrm{HX}\) is \(1.00 \times 10^{-15}\) , calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{MX}\) .

Consider 1.0 L of an aqueous solution that contains 0.10 M sulfuric acid to which 0.30 mole of barium nitrate is added. Assuming no change in volume of the solution, determine the pH, the concentration of barium ions in the final solution, and the mass of solid formed.

What mass of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to 1.0 \(\mathrm{L}\) of a \(1.0-M \mathrm{HF}\) solution to begin precipitation of \(\mathrm{CaF}_{2}(s) ?\) For \(\mathrm{CaF}_{2}, K_{\mathrm{sp}}= 4.0 \times 10^{-11}\) and \(K_{\mathrm{a}}\) for \(\mathrm{HF}=7.2 \times 10^{-4}\) . Assume no volume change on addition of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s).\)

The \(K_{\mathrm{sp}}\) for \(Q,\) a slightly soluble ionic compound composed of \(\mathrm{M}_{2}^{2+}\) and \(\mathrm{X}^{-}\) ions, is \(4.5 \times 10^{-29} .\) The electron configuration of \(\mathrm{M}^{+}\) is \([\mathrm{Xe}] 6 s^{1} 4 f^{4} 5 d^{10} .\) The \(\mathrm{X}^{-}\) anion has 54 electrons. What is the molar solubility of \(Q\) in a solution of \(\mathrm{NaX}\) prepared by dissolving \(1.98 \mathrm{g}\) \(\mathrm{NaX}\) in \(150 .\) mL solution?

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s),\) but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-}.\) In terms of solubility, All \((\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the pH dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3},\) as a function of \(\left[\mathrm{H}^{+}\right],\) obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is 40.0 and \(K_{\mathrm{sp}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\) Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the ph range \(4-12.\)