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Chapter 17: Problem 108

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

Short Answer

Expert verified
To dissolve \(0.015 mol\) of \(\mathrm{Zn(OH)_{2}}\) in 1L of solution, an OH鈦� ion concentration of approximately \(3.44 \times 10^{-18} M\) is required.

Step by step solution

01

Write the balanced dissolution equation for Zn(OH)鈧�

First, we need to write the balanced dissolution equation for Zn(OH)鈧� in water: \[Zn(OH)_{2}(s) \rightleftharpoons Zn^{2+}(aq) + 2OH^-(aq)\]
02

Write the solubility product expression for Zn(OH)鈧�

Now, the solubility product expression for Zn(OH)鈧� can be written as: \[K_{sp} = [Zn^{2+}][OH^-]^2 = 3.0 \times 10^{-16}\]
03

Write the formation constant expression for Zn(OH)鈧劼测伝

The formation constant expression for the hydroxide complex, Zn(OH)鈧劼测伝, can be written as: \[K_f = \frac{[Zn(OH)_{4}^{2-}]}{[Zn^{2+}][OH^-]^4} = 4.6 \times 10^{17}\]
04

Relate the moles of Zn(OH)鈧� to the concentration of Zn(OH)鈧劼测伝

Suppose 'x' moles of Zn(OH)鈧� dissolve to form Zn(OH)鈧劼测伝. Since 0.015 moles of Zn(OH)鈧� are given, the concentration of Zn(OH)鈧劼测伝 is x/1L = x M. Therefore, both the concentration of Zn虏鈦� and the concentration of OH鈦� ions need to be in terms of 'x'. From the dissolution equation, for each mole of Zn(OH)鈧� that dissolve, one mole of Zn虏鈦� ions will be produced, so the Zn虏鈦� concentration will also be x M. Since the dissolution equation shows a 1:2 stoichiometry between Zn虏鈦� and OH鈦�, the concentration of OH鈦� ions will be 2x M.
05

Substitute the expressions in Ksp and Kf formulae

Now substitute the expression for the concentration of Zn虏鈦� and OH鈦� ions in Ksp and Kf formulae: For Ksp: \[3.0 \times 10^{-16} = (x)(2x)^2\] For Kf: \[4.6 \times 10^{17} = \frac{x}{(x)(2x)^4}\]
06

Solve the Kf equation for x

Rearrange the Kf equation to solve for x: \[x = 4.6 \times 10^{17} \times (x)(2x)^4\] Since we need only a numerical value for x according to the question, we can assume anything multiplied or divided by x is equal to 1 to simplify the equation: \[x = 4.6 \times 10^{17}\]
07

Use x value in Ksp equation and solve for x

Now plug the value of x in the Ksp equation: \[3.0 \times 10^{-16} = (4.6 \times 10^{17}) (2x)^2\] Solve for x: \[x = \sqrt{\frac{3.0 \times 10^{-16}}{4.6 \times 10^{17} \times 4}}\] \[x \approx 1.72 \times 10^{-18}\]
08

Calculate the concentration of OH鈦� ions

The concentration of OH鈦� ions can be obtained by multiplying x by 2: \[OH^- = 2x = 2 \times 1.72 \times 10^{-18} \approx 3.44 \times 10^{-18} M\] Thus, a concentration of approximately \(3.44 \times 10^{-18} M\) OH鈦� ions is required to dissolve 0.015 mol of Zn(OH)鈧� in 1L of solution.

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Key Concepts

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

Dissolution Equation
Understanding the dissolution equation is crucial when dealing with solubility in chemistry. It represents the process where a solid compound, in this case, \(\mathrm{Zn}(\mathrm{OH})_{2}\), breaks down into its ionic constituents in a solvent such as water. The balanced dissolution equation is written as:
\[\mathrm{Zn}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Zn}^{2+}(aq) + 2\mathrm{OH}^-(aq)\]
This equation is reversible, indicating that the solid can re-form from the ions in solution under certain conditions. The stoichiometry, or the proportion of ions released per mole of solid dissolved, is evident from the coefficients of the products.
Formation Constant
The formation constant (also known as the stability constant) is a special type of equilibrium constant that measures the propensity of a complex ion to form from its separate ions in solution. For our example with the zinc hydroxide complex \(\mathrm{Zn}(\mathrm{OH})_{4}^{2-}\), the formation constant is denoted as \(K_f\) and expressed mathematically by the relationship:
\[K_f = \frac{[\mathrm{Zn}(\mathrm{OH})_{4}^{2-}]}{[\mathrm{Zn}^{2+}][\mathrm{OH}^-]^4} = 4.6 \times 10^{17}\]
A large \(K_f\) value implies a highly stable complex ion, meaning it readily forms in solution. Conversely, a low \(K_f\) suggests the complex is less stable and less likely to form.
Stoichiometry
Stoichiometry deals with the quantitative relationships between the reactants and products in a chemical reaction. It arises from the law of conservation of mass and the fixed composition of compounds. For instance, in the dissolution of \(\mathrm{Zn}(\mathrm{OH})_{2}\), the ratio of zinc ions (\(\mathrm{Zn}^{2+}\)) to hydroxide ions (\(\mathrm{OH}^-\)) is 1:2 as illustrated in the dissolution equation.
Thus, knowing the moles of one species can help us calculate the moles 鈥� and subsequently the concentration 鈥� of another species. This stoichiometric relationship is fundamental when solving for the unknown concentrations in solubility problems.
Concentration Calculation
Calculating the concentration of ions in solution is an essential step in solving solubility problems. Concentration is typically expressed in molarity (M), which is the number of moles of solute per liter of solution. The concentration calculation ties together the stoichiometry of the dissolution equation with the known solubility product (\(K_{sp}\)) and formation constant (\(K_f\)).
In our exercise, we use these relationships to derive the concentration of hydroxide ions necessary to dissolve a set amount of \(\mathrm{Zn}(\mathrm{OH})_{2}\). These calculations are built upon the dissolution and formation reactions and their respective constants to find the concentration of one species when given the quantity of another.

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Most popular questions from this chapter

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{\mathrm{s}}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\), and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the "Chemistry and Life" box on page 755, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}_{\text {k }}\) whose \(K_{p}=1.0 \times 10^{-40}\), (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Calculate the \(\mathrm{pH}\) at the equivalence point for titrating \(0.200 \mathrm{M}\) solutions of each of the following bases with \(0.200 \mathrm{M} \mathrm{HBr}\) : (a) sodium hydroxide \((\mathrm{NaOH})\), (b) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right)\), (c) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{2}\right)\).

An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered off, the \(\mathrm{pH}\) is adjusted to about 1 and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled in; a precipitate again forms. After filtering off this precipitate, the pH is adjusted to 8 and \(\mathrm{H}_{2} \mathrm{~S}\) is again added; no precipitate forms. No precipitate forms upon addition of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). (See Figure 7.23.) The remaining solution shows a yellow color in a flame test (see Figure 7.22). Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent: CdS, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{HgO}_{3}, \mathrm{ZnSO}_{4}, \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\), and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) ?

A buffer contains 0.15 mol of propionic acid $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)$ and 0.10 mol of sodium propionate \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COONa}\right)\) in 1.20 \(\mathrm{L}\) . (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.01 \(\mathrm{mol}\) of \(\mathrm{NaOH}\) ? (c) What is the pH of the buffer after the addition of 0.01 \(\mathrm{mol}\) of \(\mathrm{HI} ?\)

Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.
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