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Chapter 16: Problem 32

Calculate the molar solubility of \(\mathrm{Co}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2.5 \times 10^{-43}\).

Short Answer

Expert verified
The molar solubility of Co(OH)₃ is approximately \( 1.25 \times 10^{-11}\: \text{mol/L} \).

Step by step solution

01

1. Write the balanced chemical equation of Co(OH)₃ dissolving in water

First, we need to know how Co(OH)₃ behaves in the water when it dissolves. The balanced reaction for the dissolution of cobalt(III) hydroxide is: Co(OH)₃ (s) ⇌ Co³⁺ (aq) + 3OH⁻ (aq)
02

2. Express the solubility in terms of x

Now, let's assume the molar solubility of Co(OH)₃ is x mol/L. Then at equilibrium: - [Co³⁺] = x mol/L - [OH⁻] = 3x mol/L
03

3. Write the expression for Ksp

Now, let's write the expression for Ksp for this reaction using the molar concentrations at equilibrium: Ksp = [Co³⁺] [OH⁻]³
04

4. Substitute the given Ksp value and molar concentrations in the Ksp expression

Since the question provides the Ksp value of 2.5 × 10⁻⁴³, now we can substitute the values in our expression: \( 2.5 \times 10^{-43} \) = (x) (3x)³
05

5. Solve for x

To find the molar solubility x, we need to simplify and solve the equation for x: \( 2.5 \times 10^{-43} \) = 27x⁴ Divide both sides by 27: \( x^{4} = \frac{2.5 \times 10^{-43}}{27} \) Now, take the fourth root of both sides: \( x = \left(\frac{2.5 \times 10^{-43}}{27}\right)^{\frac{1}{4}} \)
06

6. Calculate the molar solubility

Finally, we can plug in the given values into our calculator to find the molar solubility: \( x = \left(\frac{2.5 \times 10^{-43}}{27}\right)^{\frac{1}{4}} \approx 1.25 \times 10^{-11}\: \text{mol/L} \) So, the molar solubility of Co(OH)₃ is approximately \( 1.25 \times 10^{-11}\: \text{mol/L} \).

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Most popular questions from this chapter

A solution contains \(2.0 \times 10^{-3} \mathrm{M} \mathrm{Ce}^{3+}\) and \(1.0 \times 10^{-2} \mathrm{M} \mathrm{IO}_{3}^{3-}\). Will \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}(s)\) precipitate? \(\left[K_{\text {sp }}\right.\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) is \(\left.3.2 \times 10^{-10} .\right]\)

Calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) is relatively insoluble in water \(\left(K_{\mathrm{xp}}=\right.\) \(\left.2 \times 10^{-9}\right)\). However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in \(0.10 \mathrm{M}\) \(\mathrm{H}^{+}\) than in pure water? In pure water, ignore the basic properties of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)

You have two salts, \(\mathrm{Ag} \mathrm{X}\) and \(\mathrm{AgY}\), with very similar \(K_{\mathrm{xp}}\) values. You know that \(K_{a}\) for \(\mathrm{HX}\) is much greater than \(K_{n}\) for HY. Which salt is more soluble in acidic solution? Explain.

Consider a solution made by mixing \(500.0 \mathrm{~mL}\) of \(4.0 \mathrm{M} \mathrm{NH}_{3}\) and \(500.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{AgNO}_{3} \cdot \mathrm{Ag}^{+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}:\) \(\mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \quad K_{1}=2.1 \times 10^{3}\) \(\mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K_{2}=8.2 \times 10^{3}\)

A solution contains \(1.0 \times 10^{-5} \mathrm{M} \mathrm{Ag}^{+}\) and \(2.0 \times 10^{-6} \mathrm{M} \mathrm{CN}^{-}\). Will \(\mathrm{AgCN}(s)\) precipitate? \(\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{AgCN}(s)\) is \(2.2 \times 10^{-12}\).)
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