A 725 mL sample of a saturated aqueous solution of calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4},\) at \(95^{\circ} \mathrm{C}\) is cooled to \(13^{\circ} \mathrm{C}\). How many milligrams of calcium oxalate will precipitate? For \(\mathrm{CaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=1.2 \times 10^{-8}\) at \(95^{\circ} \mathrm{C}\) and \(2.7 \times 10^{-9}\) at \(13^{\circ} \mathrm{C}\).

Short Answer

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To find the short answer, follow the calculations set out in the steps. The final result will be the amount in milligrams of calcium oxalate that precipitates when cooled to \(13^{\circ} C\).

Step by step solution

01

Calculate the molar solubility at \(95^{\circ} C\)

The solubility product constant, \(K_{sp}\), at \(95^{\circ} C\) is given as \(1.2 \times 10^{-8}\). Since the equation for calcium oxalate, \(CaC_{2}O_{4}\), can be written as \(CaC_{2}O_{4} ⇌ Ca^{2+} + C_{2}O_{4}^{2-}\), we know that the \(K_{sp}\) expression is \(K_{sp} = [Ca^{2+}][C_{2}O_{4}^{2-}]\). Because calcium oxalate is a 1:1 salt, the molar solubility at \(95^{\circ} C\) can be represented by \(s\) and so \(K_{sp}\) can be expressed as \(s^{2}\). Solving for \(s\), we get \(s = \sqrt{K_{sp}} = \sqrt{1.2 \times 10^{-8}}\).
02

Calculate the molar solubility at \(13^{\circ} C\)

We repeat the step for temperature \(13^{\circ} C\) where \(K_{sp}\) is given as \(2.7 \times 10^{-9}\). Thus, \(s = \sqrt{K_{sp}} = \sqrt{2.7 \times 10^{-9}}\).
03

Calculate the difference in molar solubility

The difference in molar solubility will give us the amount of \(CaC_{2}O_{4}\) that precipitates out when cooled to \(13^{\circ} C\). It's calculated as: \(\Delta s = s_{95} - s_{13}\).
04

Convert molar solubility to mass

Calculate the milligrams of \(CaC_{2}O_{4}\) that precipitates out by converting the molar solubility difference (\(\Delta s\)) into grams using its molar weight (one mole of \(CaC_{2}O_{4}\) has a weight of \(128.1g\)) and then into milligrams. Multiply this with the volume of the solution in litres to get the total milligrams that precipitate.

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Most popular questions from this chapter

Which of the following has the highest molar solubility? (a) \(\mathrm{MgF}_{2}, K_{\mathrm{sp}}=3.7 \times 10^{-8}\) \(\mathrm{MgCO}_{3}\), \(K_{\mathrm{sp}}=3.5 \times 10^{-8} ;(\mathrm{c}) \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\mathrm{sp}}=1 \times 10^{-25}\); (d) \(\mathrm{Li}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=3.2 \times 10^{-9}\).

The slightly soluble solute \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is most soluble in (a) pure water; (b) \(0.10 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4} ;\) (c) \(0.25 \mathrm{M} \mathrm{KNO}_{3}\); (d) \(0.40 \mathrm{M} \mathrm{AgNO}_{3}\).

All but two of the following solutions yield a precipitate when the solution is also made \(2.00 \mathrm{M}\) in \(\mathrm{NH}_{3}\). Those two are (a) \(\mathrm{MgCl}_{2}(\mathrm{aq}) ;\) (b) \(\mathrm{FeCl}_{3}(\mathrm{aq})\); (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq}) ;(\mathrm{d}) \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\); (e) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq})\).

Write net ionic equations for each of the following observations. (a) When concentrated \(\mathrm{CaCl}_{2}(\mathrm{aq})\) is added to \(\mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq}),\) a white precipitate forms that is \(38.7 \%\) Ca by mass. (b) When a piece of dry ice, \(\mathrm{CO}_{2}(\mathrm{s}),\) is placed in a clear dilute solution of limewater \(\left[\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})\right]\), bubbles of gas evolve. At first, a white precipitate forms, but then it redissolves.

A solution is \(0.010 \mathrm{M}\) in both \(\mathrm{CrO}_{4}^{2-}\) and \(\mathrm{SO}_{4}^{2-}\). To this solution, \(0.50 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\text { aq })\) is slowly added. (a) Which anion will precipitate first from solution? (b) What is \(\left[\mathrm{Pb}^{2+}\right]\) at the point at which the second anion begins to precipitate? (c) Are the two anions effectively separated by this fractional precipitation?

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