Chapter 16: Problem 11
\(\mathrm{Ag}_{2} \mathrm{S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.
Chapter 16: Problem 11
\(\mathrm{Ag}_{2} \mathrm{S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.
All the tools & learning materials you need for study success - in one app.
Get started for freeNitrate 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.
What happens to the \(K_{\mathrm{sp}}\) value of a solid as the temperature of the solution changes? Consider both increasing and decreasing temperatures, and explain your answer.
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. \(\mathrm{Cu}^{+}\)
Approximately 0.14 g nickel(II) hydroxide, Ni(OH) \(_{2}(s),\) dissolves per liter of water at \(20^{\circ} \mathrm{C}\) . Calculate \(K_{\text { sp }}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.
A solution is prepared by mixing \(50.0 \mathrm{mL}\) of \(0.10M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(50.0 \mathrm{mL}\) of $1.0 \mathrm{M}\( \)\mathrm{KCl}$ . Calculate the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) at equilibrium. $\left[K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_{2}(s) \text { is } 1.6 \times 10^{-5}.\right]$
What do you think about this solution?
We value your feedback to improve our textbook solutions.