Chapter 16: Problem 4
A friend tells you: "The constant \(K_{
Chapter 16: Problem 4
A friend tells you: "The constant \(K_{
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Get started for freeCalcium 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 \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\).
When aqueous \(\mathrm{KI}\) 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 \(K_{\text {sp }}\) values for \(\mathrm{AgCl}, \mathrm{AgBr}\), and \(\mathrm{AgI}\) ?
a. Calculate the molar solubility of \(\mathrm{Sr} \mathrm{F}_{2}\) in water, ignoring the basic properties of \(\mathrm{F}^{-} .\left(\right.\) For \(\mathrm{SrF}_{2}, K_{\text {sp }}=7.9 \times 10^{-10}\).) 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 \(\left.7.2 \times 10^{-4} .\right)\)
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 \(3.0 \mathrm{M} \mathrm{NH}_{3} .\) Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text {owendl }}\) 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 $$ \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) $$
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