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Chapter 16: Problem 27

Approximately 0.14 g nickel(II) hydroxide, Ni(OH) \(_{2}(s),\) dissolves per liter of water at \(20^{\circ} \mathrm{C}\) . Calculate \(K_{\text { sp }}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

Short Answer

Expert verified
The solubility product constant, \(K_{sp}\), for Ni(OH)₂(s) at 20°C is approximately \(4.37 * 10^{-8}\).

Step by step solution

01

Write the balanced chemical equation for the dissolution of Ni(OH)₂(s) in water

The balanced chemical equation for the dissolution of nickel(II) hydroxide in water is: Ni(OH)₂(s) ⇌ Ni²⁺(aq) + 2OH⁻(aq)
02

Write the equilibrium expression for this reaction

The equilibrium expression, Ksp, for the dissolution of Ni(OH)₂(s) in water is given by: Ksp = [Ni²⁺] * [OH⁻]² where [Ni²⁺] and [OH⁻] represent the equilibrium concentrations of the Ni²⁺ and OH⁻ ions, respectively.
03

Calculate molar solubility, s, of Ni(OH)₂ in water

We are given that approximately 0.14 g of Ni(OH)₂ dissolves per liter of water at 20°C. To convert this mass to moles, we need the molar mass of Ni(OH)₂: Molar mass of Ni(OH)₂ = 58.69 g/mol (Ni) + 2 * (15.999 g/mol + 1.0079 g/mol) = 92.708 g/mol for Ni(OH)₂ Now, we can find the molar solubility, s: s = (0.14 g Ni(OH)₂) / (92.708 g/mol Ni(OH)₂) = 1.51 * 10⁻³ mol/L
04

Determine the equilibrium concentrations of Ni²⁺ and OH⁻ ions in the saturated solution

From the balanced chemical equation, we can see that for every mole of Ni(OH)₂ that dissolves, one mole of Ni²⁺ and two moles of OH⁻ ions are produced. Thus, we can use the molar solubility, s, to determine the equilibrium concentrations of Ni²⁺ and OH⁻ ions in the saturated solution: [Ni²⁺] = 1.51 * 10⁻³ mol/L [OH⁻] = 2 * 1.51 * 10⁻³ mol/L = 3.02 * 10⁻³ mol/L
05

Calculate Ksp for Ni(OH)₂(s)

Finally, we can use the equilibrium concentrations of Ni²⁺ and OH⁻ ions in the saturated solution to calculate Ksp for Ni(OH)₂(s): Ksp = [Ni²⁺] * [OH⁻]² Ksp = (1.51 * 10⁻³)(3.02 * 10⁻³)² Ksp = 4.37 * 10⁻⁸ So, the solubility product constant, Ksp, for Ni(OH)₂(s) at 20°C is approximately 4.37 * 10⁻⁸.

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Most popular questions from this chapter

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

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.

Two different compounds have about the same molar solubility. Do they also have about the same \(K_{\text {sp}}\) value?

Tooth enamel is composed of the mineral hydroxyapatite. The \(K_{\mathrm{sp}}\) of hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\( is \)6.8 \times 10^{-37}$ . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\) , forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times 10^{-60}\) . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

What mass of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to 1.0 \(\mathrm{L}\) of a \(1.0-M \mathrm{HF}\) solution to begin precipitation of \(\mathrm{CaF}_{2}(s) ?\) For $\mathrm{CaF}_{2}, K_{\mathrm{sp}}= 4.0 \times 10^{-11}\( and \)K_{\mathrm{a}}\( for \)\mathrm{HF}=7.2 \times 10^{-4}$ . Assume no volume change on addition of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s).\)
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