Chapter 16: Problem 7
Which is more likely to dissolve in an acidic solution, silver sulfide or silver chloride? Why?
Chapter 16: Problem 7
Which is more likely to dissolve in an acidic solution, silver sulfide or silver chloride? Why?
All the tools & learning materials you need for study success - in one app.
Get started for freeThe overall formation constant for \(\mathrm{HgI}_{4}{ }^{2-}\) is \(1.0 \times 10^{30} .\) That is, $$1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[1^{-}\right]^{4}}$$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{~mL}\) of a solution that was originally \(0.010 \mathrm{M} \mathrm{Hg}^{2+}\) and \(0.78 \mathrm{M} \mathrm{I}^{-}\) ? The reaction is $$\mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}{ }^{2-}(a q)$$
a. Calculate the molar solubility of \(\mathrm{Sr} \mathrm{F}_{2}\) in water, ignoring the basic properties of \(\mathrm{F}^{-} .\) (For \(\mathrm{SrF}_{2}, K_{\mathrm{se}}=7.9 \times 10^{-10} .\) ) 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 \(7.2 \times 10^{-4} .\) )
A mixture contains \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Mn}^{2+}\) and is saturated with \(0.10 M \mathrm{H}_{2} \mathrm{~S}\). Determine a \(\mathrm{pH}\) where CuS precipitates but MnS does not precipitate. \(K_{\text {op }}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and \(K_{\mathrm{s}}\) for \(\mathrm{MnS}=2.3 \times 10^{-1.3}\)
Calculate the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=5.9 \times 10^{-11}\).
A solution is \(1 \times 10^{-4} \mathrm{M}\) 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 \(K_{\text {sp }}\) values are \(K_{\text {sp }}\left(\mathrm{PbF}_{2}\right)=\) \(4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29}\), and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=1 \times 10^{-54}\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.