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Chapter 15: Problem 36

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 \(\mathrm{Mn}(\mathrm{OH})_{2}\) \(K_{\mathrm{sp}}=2 \times 10^{-13}\)

Short Answer

Expert verified
For the given pairs of solids, FeC2O4 has a smaller molar solubility than Cu(IO₄)₂, and Mn(OH)₂ has a smaller molar solubility than Ag₂CO₃. Therefore, the solids with the smallest molar solubilities are FeC2O4 and Mn(OH)₂.

Step by step solution

01

Balanced chemical equation for FeC2O4

$$\mathrm{FeC}_{2}\mathrm{O}_{4(s)} \rightleftharpoons \mathrm{Fe^{2+}_{(aq)}} + 2 \mathrm{C}_{2}\mathrm{O}_{4^{2-}_{(aq)}}$$ Step 2: Set up the Ksp expression
02

Ksp expression for FeC2O4

Let the molar solubility of FeC2O4 be 's' Then, we have Ksp = [Fe²⁺][C₂O₄²⁻]² = (s)(2s)² Step 3: Calculate molar solubility 's'
03

Calculate molar solubility of FeC2O4

Given, Ksp = 2.1 x 10⁻⁷ (2.1 x 10⁻⁷) = s*(2s)² Solve for 's' which is approximately 4.91 x 10⁻³ M 2) Cu(IO₄)₂ Step 1: Write the balanced chemical equation for the dissolution
04

Balanced chemical equation for Cu(IO₄)₂

$$\mathrm{Cu}(\mathrm{IO}_{4})_{ 2(s)} \rightleftharpoons \mathrm{Cu^{2+}_{(aq)}} + 2 \mathrm{IO}_{4^{-}_{(aq)}}$$ Step 2: Set up the Ksp expression
05

Ksp expression for Cu(IO₄)₂

Let the molar solubility of Cu(IO₄)₂ be 's' Then, we have Ksp = [Cu²⁺][IO₄⁻]² = (s)(2s)² Step 3: Calculate molar solubility 's'
06

Calculate molar solubility of Cu(IO₄)₂

Given, Ksp = 1.4 x 10⁻⁷ (1.4 x 10⁻⁷) = s*(2s)² Solve for 's' which is approximately 5.04 x 10⁻³ M
07

Compare the molar solubility of FeC2O4 and Cu(IO₄)₂

Molar solubility of FeC2O4: 4.91 x 10⁻³ M Molar solubility of Cu(IO₄)₂: 5.04 x 10⁻³ M FeC2O4 has a smaller molar solubility than Cu(IO₄)₂. b. We have two solids: 1) Ag₂CO₃ Step 1: Write the balanced chemical equation for the dissolution
08

Balanced chemical equation for Ag₂CO₃

$$\mathrm{Ag}_{2}\mathrm{CO}_{3(s)} \rightleftharpoons 2 \mathrm{Ag^{+}_{(aq)}} + \mathrm{CO}_{3^{2-}_{(aq)}}$$ Step 2: Set up the Ksp expression
09

Ksp expression for Ag₂CO₃

Let the molar solubility of Ag₂CO₃ be 's' Then, we have Ksp = [Ag⁺]²[CO₃²⁻] = (2s)²(s) Step 3: Calculate molar solubility 's'
10

Calculate molar solubility of Ag₂CO₃

Given, Ksp = 8.1 x 10⁻¹² (8.1 x 10⁻¹²) = (2s)²(s) Solve for 's' which is approximately 9.45 x 10⁻⁵ M 2) Mn(OH)₂ Step 1: Write the balanced chemical equation for the dissolution
11

Balanced chemical equation for Mn(OH)₂

$$\mathrm{Mn}(\mathrm{OH})_{2(s)} \rightleftharpoons \mathrm{Mn^{2+}_{(aq)}} + 2 \mathrm{OH^{-}_{(aq)}}$$ Step 2: Set up the Ksp expression
12

Ksp expression for Mn(OH)₂

Let the molar solubility of Mn(OH)₂ be 's' Then, we have Ksp = [Mn²⁺][OH⁻]² = (s)(2s)² Step 3: Calculate molar solubility 's'
13

Calculate molar solubility of Mn(OH)₂

Given, Ksp = 2 x 10⁻¹³ (2 x 10⁻¹³) = s*(2s)² Solve for 's' which is approximately 6.48 x 10⁻⁵ M
14

Compare the molar solubility of Ag₂CO₃ and Mn(OH)₂

Molar solubility of Ag₂CO₃: 9.45 x 10⁻⁵ M Molar solubility of Mn(OH)₂: 6.48 x 10⁻⁵ M Mn(OH)₂ has a smaller molar solubility than Ag₂CO₃. So the solids with smaller molar solubilities are FeC2O4 and Mn(OH)₂.

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Most popular questions from this chapter

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?

The \(\mathrm{Hg}^{2+}\) ion forms complex ions with \(\mathrm{I}^{-}\) as follows: $$\begin{aligned} \mathrm{Hg}^{2+}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}^{+}(a q) & & K_{1}=1.0 \times 10^{8} \\ \mathrm{HgI}^{+}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{2}(a q) & & K_{2}=1.0 \times 10^{5} \\ \mathrm{HgI}_{2}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{3}^{-}(a q) & & K_{3}=1.0 \times 10^{9} \\ \mathrm{HgI}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{4}^{2-}(a q) & & K_{4}=1.0 \times 10^{8} \end{aligned}$$ A solution is prepared by dissolving 0.088 mole of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) and 5.00 moles of NaI in enough water to make 1.0 L of solution. a. Calculate the equilibrium concentration of \(\left[\mathrm{HgI}_{4}^{2-}\right] .\) b. Calculate the equilibrium concentration of \(\left[\mathrm{I}^{-}\right] .\) c. Calculate the equilibrium concentration of \(\left[\mathrm{Hg}^{2+}\right]\).

$$K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}}$$

A solution contains \(1.0 \times 10^{-5} M \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What is the minimum concentration of \(\mathrm{AgNO}_{3}\) that would cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{PbI}_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-8}\) b. \(\mathrm{CdCO}_{3}, K_{\mathrm{sp}}=5.2 \times 10^{-12}\) c. \(\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\mathrm{sp}}=1 \times 10^{-31}\)
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