Chapter 15: Problem 7
Which is more likely to dissolve in an acidic solution, silver sulfide or silver chloride? Why?
Chapter 15: Problem 7
Which is more likely to dissolve in an acidic solution, silver sulfide or silver chloride? Why?
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Get started for freeCalculate the final concentrations of \(\mathrm{K}^{+}(a q), \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)\),\(\mathrm{Ba}^{2+}(a q),\) and \(\mathrm{Br}^{-}(a q)\) in a solution prepared by adding \(0.100 \mathrm{L}\) of \(0.200 M \mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to \(0.150 \mathrm{L}\) of \(0.250 M \mathrm{BaBr}_{2}\). (For \(\left.\mathrm{BaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.3 \times 10^{-8} .\right)\)
A solution contains \(2.0 \times 10^{-3} M \mathrm{Ce}^{3+}\) and \(1.0 \times 10^{-2} M\) IO \(_{3}^{3-}\) Will \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}(s)\) \(\left[K_{\mathrm{sp}} \text { for } \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} \text { is } 3.2 \times 10^{-10} .\right]\)
Consider \(1.0 \mathrm{L}\) 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 \(\mathrm{pH},\) the concentration of barium ions in the final solution, and the mass of solid formed.
A solution is formed by mixing \(50.0 \mathrm{mL}\) of \(10.0 \mathrm{M}\) 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 \(X^{-}\) as follows: $$\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) \quad K_{1}=1.0 \times 10^{2}$$ $$\mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) \quad K_{2}=1.0 \times 10^{4}$$ $$\mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K_{3}=1.0 \times 10^{3}$$ with an overall reaction $$\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K=1.0 \times 10^{9}$$ Calculate the following concentrations at equilibrium. a. \(\mathrm{CuX}_{3}^{2-}\) b. \(\mathrm{CuX}_{2}^{-}\) c. \(\mathrm{Cu}^{+}\)
The overall formation constant for \(\mathrm{HgI}_{4}^{2-}\) is \(1.0 \times 10^{30}\) That is, $$ 1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{4}} $$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{mL}\) of a solution that was originally \(0.010\) \(M\) \(\mathrm{Hg}^{2+}\) and \(0.78\) \(M\) \(\mathrm{I}^{-} ?\) The reaction is $$\mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}^{2-}(a q)$$
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