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Chapter 16: Problem 42

Calculate the solubility of $\mathrm{Co}(\mathrm{OH})_{2}(s)\left(K_{\mathrm{sp}}=2.5 \times 10^{-16}\right)\( in a buffered solution with a \)\mathrm{pH}$ of 11.00.

Short Answer

Expert verified
The solubility of Co(OH)₂ in a buffered solution with a pH of 11.00 is calculated using the given Ksp value and the concentration of OH⁻ ions. Using the dissolution of Co(OH)₂ in water, the balanced chemical equation, and the Ksp expression, we find the concentration of Co²⁺ ions. After converting the given pH to the concentration of OH⁻ ions and substituting the given Ksp and [OH⁻] into the Ksp expression, we solve for the concentration of Co²⁺ ions. The solubility of Co(OH)₂ in the buffered solution is found to be 2.5 × 10⁻¹⁰ M.

Step by step solution

01

Write the balanced chemical equation

The dissolution of Co(OH)2 in water can be represented by the following balanced chemical equation: Co(OH)2 (s) ↔ Co2+ (aq) + 2 OH- (aq)
02

Write the expression for Ksp

The solubility product constant (Ksp) for the dissolution of Co(OH)2 is given by: Ksp = [Co²⁺] × [OH⁻]² where [Co²⁺] and [OH⁻] are the equilibrium concentrations of Co2+ and OH- ions, respectively. We are given that Ksp = 2.5 × 10⁻¹⁶.
03

Convert the given pH to [OH⁻] concentration

We are given the pH of the buffered solution as 11.00. To find the concentration of OH- ions, we will first find the pOH: pOH = 14 - pH = 14 - 11 = 3 Now, we can calculate the concentration of OH- ions using the pOH: [OH⁻] = 10^(-pOH) = 10⁻³ M
04

Substitute the given Ksp and [OH⁻] into the Ksp expression

Now that we have the concentration of OH- ions, we can substitute it into the Ksp expression: 2.5 × 10⁻¹⁶ = [Co²⁺] × (10⁻³)² 2.5 × 10⁻¹⁶ = [Co²⁺] × 10⁻⁶
05

Solve for the concentration of Co²⁺ ions

Now we will solve for the equilibrium concentration of Co²⁺ ions: [Co²⁺] = (2.5 × 10⁻¹⁶) / 10⁻⁶ [Co²⁺] = 2.5 × 10⁻¹⁰ M
06

Calculate the solubility of Co(OH)2

The solubility of Co(OH)2 in the given buffered solution is equal to the equilibrium concentration of Co²⁺ ions: Solubility of Co(OH)2 = 2.5 × 10⁻¹⁰ M

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Most popular questions from this chapter

A solution contains $3.0 \times 10^{-3} M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2} .$ What concentrations of \(\mathrm{KF}\) will cause precipitation of solid \(\mathrm{MgF}_{2}\left(K_{\mathrm{sp}}=6.4 \times 10^{-9}\right) ?\)

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On a hot day, a 200.0 -mL sample of a saturated solution of \(\mathrm{PbI}_{2}\) was allowed to evaporate until dry. If 240 mg of solid \(\mathrm{PbI}_{2}\) was collected after evaporation was complete, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{PbI}_{2}\) on this hot day.

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s),\) but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-}.\) In terms of solubility, All \((\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the pH dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3},\) as a function of \(\left[\mathrm{H}^{+}\right],\) obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is 40.0 and \(K_{\mathrm{sp}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\) Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the ph range \(4-12.\)

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