Chapter 16: Problem 14
When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?
Chapter 16: Problem 14
When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?
All the tools & learning materials you need for study success - in one app.
Get started for freeCalculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties. a. \(A g_{3} P O_{4}, K_{s p}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+} \right.$ is the cation in is the cation in solution.\()\)
For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{FeC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.1 \times 10^{-7},\) or $\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-7}$ b. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\mathrm{sp}}=8.1 \times 10^{-12},\) or \(\mathrm{Mn}(\mathrm{OH})_{2},\) \(K_{\mathrm{sp}}=2 \times 10^{-13}\)
A solution contains $1.0 \times 10^{-6} M \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\( and \)5.0 \times 10^{-7} M$ \(\mathrm{K}_{3} \mathrm{PO}_{4} .\) Will \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) precipitate? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}=1.0 \times 10^{-31} . ] \right.$
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. $$\mathrm{AgI} : K_{\mathrm{sp}}=1.5 \times 10^{-16}$$ $$\operatorname{AgBr} ; K_{\mathrm{sp}}=5,0 \times 10^{-13}$$ $$\mathrm{AgCl} : K_{\mathrm{sp}}=1.6 \times 10^{-10}$$
Two different compounds have about the same molar solubility. Do they also have about the same \(K_{\text {sp}}\) value?
What do you think about this solution?
We value your feedback to improve our textbook solutions.