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The solubility product for \(\mathrm{Zn}(\mathrm{OH})_{2}\) is \(3.0 \times 10^{-16}\). The formation constant for the hydroxo complex, \(\mathrm{Zn}(\mathrm{OH})_{4}^{2-}\), is \(4.6 \times 10^{-17} .\) What concentration of \(\mathrm{OH}^{-}\) is required to dissolve \(0.015\) mol of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?

Short Answer

Expert verified
To dissolve 0.015 mol of Zn(OH)_2 in a liter of solution, a concentration of \(2.516 \times 10^{-3} \, \text{M}\) OH^- is required.

Step by step solution

01

Write the solubility product expression for Zn(OH)_2

We know that the solubility product constant (K_sp) of Zn(OH)_2 is given by: \[K_{sp} = [Zn^{2+}][OH^-]^2\] Given: \(K_{sp} = 3.0 \times 10^{-16}\)
02

Write the formation constant expression for Zn(OH)_4^(2-)

We know that the formation constant (K_f) of Zn(OH)_4^(2-) is given by: \[K_{f} = \frac{[Zn(OH)_4^{2-}]}{[Zn^{2+}][OH^-]^2}\] Given: \(K_f = 4.6 \times 10^{-17}\)
03

Calculate the required Zn^2+ concentration to dissolve 0.015 mol of Zn(OH)_2

We are given that we need to dissolve 0.015 mol of Zn(OH)_2 in 1 L of solution. Therefore, we require: \[[Zn^{2+}] = \frac{0.015 \, \text{mol}}{1 \, \text{L}} = 0.015 \, \text{M}\]
04

Calculate the concentration of OH^-

We can now use the K_f expression and the required Zn^2+ concentration to find the concentration of OH^- needed to dissolve 0.015 mol of Zn(OH)_2. From the formation constant expression (Step 2), we have: \[[Zn(OH)_4^{2-}] = K_f [Zn^{2+}][OH^-]^2\] Plugging the value of [Zn^2+] from Step 4: \[[OH^-]^2 = \frac{[Zn(OH)_4^{2-}]}{K_f [Zn^{2+}]}\] \[[OH^-]^2 = \frac{0.015}{4.6 \times 10^{-17} (0.015)}\] Calculating the concentration of OH^-: \[[OH^-] = \sqrt{\frac{0.015}{4.6 \times 10^{-17} (0.015)}}\] \[[OH^-] = 2.516 \times 10^{-3} \, \text{M}\] Thus, a concentration of \(2.516 \times 10^{-3}\, \text{M}\) OH^- is required to dissolve 0.015 mol of Zn(OH)_2 in a liter of solution.

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Key Concepts

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

Formation Constant
Understanding the formation constant (also known as stability constant) is pivotal in grasping complex ion chemistry. It helps us determine how strongly a certain complex forms in solution. The formation constant, denoted as \(K_{f}\), quantifies the stability of a complex ion such as \(\mathrm{Zn(OH)_4^{2-}}\). This constant signifies the ratio of products to reactants for the formation of a complex ion at equilibrium.

Let's dissect the concept further. In our example, the formation constant expression is given by:
  • \(K_{f} = \frac{[\mathrm{Zn(OH)}_4^{2-}]}{[\mathrm{Zn^{2+}}][\mathrm{OH}^-]^2}\)
With a \(K_f\) of \(4.6 \times 10^{-17}\), it implies that the \(\mathrm{Zn(OH)_4^{2-}}\) complex doesn't form easily, due to the low value of \(K_f\). The lower the \(K_f\), the less stable the complex is, making the decomposition into its components likely easier.

In conclusions, understanding the formation constant helps in predicting whether a complex ion will remain intact in a solution, thereby influencing the solubility of certain compounds.
Hydroxo Complex
A hydroxo complex arises when metal ions in solution associate with hydroxide ions. In the context of our exercise, \(\mathrm{Zn(OH)_4^{2-}}\) serves as our primary focus. Hydroxo complexes are characterized by their metal center being coordinated to hydroxide ligands from an aqueous medium.

Here's a closer look:
  • The central metal in our example is Zn, or zinc, forming the core of the hydroxo complex.
  • The hydroxide ions \((\mathrm{OH}^-)\), acting as ligands, form bonds with the \(\mathrm{Zn^{2+}}\).
Such associations are heavily influenced by the pH of the solution. Altering the concentration of \(\mathrm{OH}^-\) ions can shift the equilibrium between different zinc species in the solution. In our exercise, the hydroxo complex stabilization in the solution necessitates the precise concentration of \(\mathrm{OH}^-\) to facilitate the solubility of \(\mathrm{Zn(OH)_2}\).

Understanding these dynamics provides insight into the behavior of these complexes in various chemical and environmental settings.
Concentration Calculation
Calculating the required concentration of ions in a solution is fundamental to solving many chemistry problems. Here, we needed to determine the concentration of hydroxide ions \((\mathrm{OH}^-)\) necessary to dissolve 0.015 mol of \(\mathrm{Zn(OH)_2}\) completely in one liter of solution.

Steps taken in this calculation included:
  • Recognizing that \([\mathrm{Zn^{2+}}]\) was set at 0.015 M, based on the dissolution of 0.015 mol \(\mathrm{Zn(OH)_2}\) into a 1L solution.
  • Rearranging the formation constant expression \(K_f\) to isolate \([\mathrm{OH}^-]^2\).
  • Substituting the known values, and solving the equation to derive \([\mathrm{OH}^-]\).
The resulted calculation yielded a \([\mathrm{OH}^-]\) concentration of \(2.516 \times 10^{-3} \text{ M}\) for successful dissolution. With this mathematical approach, you can compute the necessary ion concentration for solubility processes.

These calculations integrate aspects of equilibrium chemistry, illustrating how concentrations translate into practical applications, such as solution preparation in a laboratory setting.

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Most popular questions from this chapter

(a) Explain the difference between solubility and solubility-product constant. (b) Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2}\), and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\) \(\left(K_{s p}=4 \times 10^{-10}\right)\) in \(1.0 \mathrm{~L}\) of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) because the solution will be quite basic.)

Suppose you want to do a physiological experiment that calls for a \(\mathrm{pH} 6.5\) buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{X}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a \(1.0 M\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH} 6.50 ?\) (lgnore any volume change.)

A sample of \(0.2140 \mathrm{~g}\) of an unkown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M} \mathrm{NaOH}\). The acid required \(27.4 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

(a) Precipitation of the group 4 cations (Figure 17.22) requires a basic medium. Why is this so? (b) What is the most significant difference between the sulfides precipitated in group 2 and those precipitated in group \(3 ?\) (c) Suggest a procedure that would serve to redissolve the group 3 cations following their precipitation.

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