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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)鈧(s) at 20掳C is approximately \(4.37 * 10^{-8}\).

Step by step solution

01

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

The balanced chemical equation for the dissolution of nickel(II) hydroxide in water is: Ni(OH)鈧(s) 鈬 Ni虏鈦(aq) + 2OH鈦(aq)
02

Write the equilibrium expression for this reaction

The equilibrium expression, Ksp, for the dissolution of Ni(OH)鈧(s) in water is given by: Ksp = [Ni虏鈦篯 * [OH鈦籡虏 where [Ni虏鈦篯 and [OH鈦籡 represent the equilibrium concentrations of the Ni虏鈦 and OH鈦 ions, respectively.
03

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

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

Determine the equilibrium concentrations of Ni虏鈦 and OH鈦 ions in the saturated solution

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

Calculate Ksp for Ni(OH)鈧(s)

Finally, we can use the equilibrium concentrations of Ni虏鈦 and OH鈦 ions in the saturated solution to calculate Ksp for Ni(OH)鈧(s): Ksp = [Ni虏鈦篯 * [OH鈦籡虏 Ksp = (1.51 * 10鈦宦)(3.02 * 10鈦宦)虏 Ksp = 4.37 * 10鈦烩伕 So, the solubility product constant, Ksp, for Ni(OH)鈧(s) at 20掳C is approximately 4.37 * 10鈦烩伕.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dissolution
Dissolution is the process where a solid substance dissolves in a solvent, forming a solution. For nickel(II) hydroxide, Ni(OH)鈧, this involves mixing with water. The solid compound splits into ions. In the classroom or lab, you'd often see it written as:
  • Ni(OH)鈧(s) 鈬 Ni虏鈦(aq) + 2OH鈦(aq)
During dissolution, the solid's particles disperse and interact with the solvent, in this case, water. The compounds break into smaller particles or ions and evenly mix into the liquid. Having a balanced chemical equation helps understand the composition and the number of ions that result from the process.
Chemical Equilibrium
In dissolution, chemical equilibrium occurs when the rate of the solid dissolving equals the rate of ions recombining into the solid. At this point, the concentrations of ions in solution remain constant. The equilibrium is depicted by consistent forward and reverse reactions, represented by the double arrow (鈬) in chemical equations.
  • This means the dissolution of Ni(OH)鈧 into Ni虏鈦 and OH鈦 ions and their precipitation back to solid form occur at the same rate.
This dynamic balance is crucial for understanding how much of the solid can dissolve and what concentrations ions achieve in a solution at equilibrium.
Molar Solubility
Molar solubility represents the number of moles of a substance that can dissolve per liter of solution at a specific temperature. For Ni(OH)鈧, it's calculated from the mass that dissolves in water. Knowing that 0.14 g dissolves in a liter, you'd convert this mass to moles using its molar mass.
  • Molar solubility of Ni(OH)鈧: 1.51 x 10鈦宦 mol/L
Molar solubility helps predict how much of a substance can dissolve in a given volume of solvent before reaching saturation. Above this point, the substance may not dissolve further without changing conditions.
Equilibrium Concentrations
Equilibrium concentrations indicate the concentrations of ions in a saturated solution at equilibrium. To find them, you relate molar solubility to ion production.
  • From Ni(OH)鈧: [Ni虏鈦篯 equals the molar solubility, which is 1.51 x 10鈦宦 mol/L.
  • Since each mole of Ni(OH)鈧 produces two moles of OH鈦, [OH鈦籡 is 2 * 1.51 x 10鈦宦 mol/L = 3.02 x 10鈦宦 mol/L.
These equilibrium concentrations are vital for calculations involving the solubility product constant, Ksp, which quantifies the solubility of a compound.

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Most popular questions from this chapter

The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is 1 \(\mathrm{mg} \mathrm{F}^{-}\) per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{CaF}_{2}=4.0 \times 10^{-11}\right)\)

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr},\) and \(\mathrm{AgI?}\)

A solution contains \(1.0 \times 10^{-6} M \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) and \(5.0 \times 10^{-7} M\) \(\mathrm{K}_{3} \mathrm{PO}_{4} .\) Will \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) precipitate? \(\left[K_{\mathrm{sp}} \text { for } \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}=1.0 \times 10^{-31} . ] \right.\)

The copper(I) ion forms a chloride salt that has \(K_{\mathrm{sp}}= 1.2 \times 10^{-6} .\) Copper (I) also forms a complex ion with \(\mathrm{Cl}^{-} :\) $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \qquad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in 0.10\(M\) \(\mathrm{NaCl}\).

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