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Chapter 16: Problem 60

Consider the following solubility data for calcium oxalate \(\left(\mathrm{Ca} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) : $$ \begin{aligned} &K_{s p} \text { at } 25^{\circ} \mathrm{C}=4 \times 10^{-9} \\ &K_{s p} \text { at } 95^{\circ} \mathrm{C}=1 \times 10^{-8} \end{aligned} $$ Five hundred mL of a saturated solution is prepared at \(95^{\circ} \mathrm{C}\). How many milligrams of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) will precipitate when the solution is cooled to \(25^{\circ} \mathrm{C}\) ? (Assume that supersaturation does not take place.)

Short Answer

Expert verified
Answer: In order to find the amount of CaC2O4 that will precipitate when cooled down to 25°C, we need to follow the steps which involve calculating molar solubility at both temperatures, finding the amount dissolved at each temperature, and then determining the difference between the two. Once we find the difference in moles, we can convert it to milligrams by using the molar mass of CaC2O4.

Step by step solution

01

Calculate molar solubility at 95°C

First, we need to calculate the molar solubility of CaC2O4 at 95°C. The solubility product constant, Ksp, at this temperature, is 1 × 10⁻⁸. Since the balanced dissolution reaction of CaC2O4 is: $$ \mathrm{CaC}_{2} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(aq) $$ Let the molar solubility of CaC2O4 at 95°C be S mol/L. Then, the Ksp expression becomes: $$ K_{s p}=[\mathrm{Ca}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]=S^{2} $$ Now, solve for S: $$ S=\sqrt{1 \times 10^{-8}} $$
02

Calculate the amount of CaC2O4 dissolved at 95°C

Using the molar solubility calculated in step 1, we can now find the amount of CaC2O4 that is dissolved in 500 mL of saturated solution at 95°C: $$ \text { Amount dissolved at } 95^{\circ} \mathrm{C}=\text { Molar solubility }\times \text { Volume of solution } $$ $$ =S \times 0.5\,\text{L} $$
03

Calculate molar solubility at 25°C

Next, we need to calculate the molar solubility of CaC2O4 at 25°C, using the given value of Ksp (4 × 10⁻⁹) in the same manner as in Step 1: $$ S=\sqrt{4 \times 10^{-9}} $$
04

Calculate the amount of CaC2O4 dissolved at 25°C

Using the molar solubility calculated in step 3, we can now find the amount of CaC2O4 that will be dissolved in 500 mL of solution at 25°C: $$ \text { Amount dissolved at } 25^{\circ} \mathrm{C}=\text { Molar solubility }\times \text { Volume of solution } $$ $$ =S \times 0.5\,\text{L} $$
05

Calculate the amount of CaC2O4 precipitated

To find the amount of CaC2O4 that will precipitate when the solution is cooled from 95°C to 25°C, subtract the amount dissolved at 25°C from the amount dissolved at 95°C: $$ \text { Amount precipitated }=\text { Amount dissolved at } 95^{\circ} \mathrm{C}-\text { Amount dissolved at } 25^{\circ} \mathrm{C} $$
06

Convert the amount precipitated to milligrams

The amount precipitated calculated in step 5 is in moles. We need to convert it to milligrams. To do this, multiply the amount precipitated by the molar mass of CaC2O4 and then by 1000 to convert grams to milligrams: $$ \text { Amount precipitated in mg }=\text { Amount precipitated in moles }\times \text { Molar mass of CaC }_{2} \mathrm{O}_{4}\times 1000 $$ Finally, the amount precipitated in milligrams is the answer we are looking for.

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Most popular questions from this chapter

\(\mathrm{K}_{\mathrm{sp}}\) for silver acetate \(\left(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) at \(80^{\circ} \mathrm{C}\) is estimated to be \(2 \times 10^{-2}\). Ten grams of silver acetate are added to \(1.0 \mathrm{~L}\) of water at \(25^{\circ} \mathrm{C}\). (a) Will all the silver acetate dissolve at \(25^{\circ} \mathrm{C} ?\) (b) If the solution (assume the volume to be \(1.0 \mathrm{~L}\) ) is heated to \(80^{\circ} \mathrm{C}\), will all the silver acetate dissolve?

Calculate the molar solubility of gold(I) chloride \(\left(K_{s p}=2.0 \times 10^{-13}\right)\) in \(0.10 \mathrm{M} \mathrm{NaCN}\). The complex ion formed is \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\) with \(K_{\mathrm{f}}=2 \times 10^{38}\). Ignore any other competing equilibrium systems.

Shown below is a representation of the ionic solid \(\mathrm{MX}\), where \(\mathrm{M}\) cations are represented by squares and \(X\) anions are represented by circles. Fill in the box after the arrow to represent what happens to the solid after it has been completely dissolved in water. For simplicity, do not represent the water molecules.

Write net ionic equations for the reaction of \(\mathrm{H}^{+}\) with (a) \(\mathrm{Cu}_{2} \mathrm{~S}\) (b) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{SrCO}_{3}\) (d) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}\) (e) \(\mathrm{Ca}(\mathrm{OH})_{2}\)

The concentrations of various cations in seawater, in moles per liter, are $$ \begin{array}{llllll} \hline \text { Ion } & \mathrm{Na}^{+} & \mathrm{Mg}^{2+} & \mathrm{Ca}^{2+} & \mathrm{A}^{3+} & \mathrm{Fe}^{3+} \\ \text { Molarity }(M) & 0.46 & 0.056 & 0.01 & 4 \times 10^{-7} & 2 \times 10^{-7} \\ \hline \end{array} $$ (a) At what \(\left[\mathrm{OH}^{-}\right]\) does \(\mathrm{Mg}(\mathrm{OH})_{2}\) start to precipitate? (b) At this concentration, will any of the other ions precipitate? (c) If enough \(\mathrm{OH}^{-}\) is added to precipitate \(50 \%\) of the \(\mathrm{Mg}^{2+}\), what percentage of each of the other ions will precipitate? (d) Under the conditions in (c), what mass of precipitate will be ob- tained from one liter of seawater?
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