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Chapter 17: Problem 109

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14} .\) (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?\) \((\mathbf{b} ) \)The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

Short Answer

Expert verified
(a) The molar solubility of Cd(OH)鈧� without NaBr solution is \(s = 9.47 \times 10^{-6} \mathrm{mol} / \mathrm{L}\). (b) The initial concentration of NaBr needed to increase the molar solubility of Cd(OH)鈧� to \(1.0 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) is \(x = 6.68 \times 10^{-2} \mathrm{mol} / \mathrm{L}\).

Step by step solution

01

(a) Finding the molar solubility of Cd(OH)鈧� without NaBr solution

First, let's write the solubility equilibrium equation for Cd(OH)鈧�: Cd(OH)鈧�(s) 鈬� Cd虏鈦�(aq) + 2OH鈦�(aq) Now, let's denote the molar solubility of Cd(OH)鈧� as 's'. Therefore, the concentration of Cd虏鈦� will be 's' and the concentration of OH鈦� will be '2s' in the solution. The solubility product constant (K鈧涒倸) is given by: K鈧涒倸 = [Cd虏鈦篯[OH鈦籡虏 We have the value of K鈧涒倸 as 2.5 脳 10鈦宦光伌. Substituting the values, we get: \(2.5 \times 10^{-14} = s \times (2s)^2\) Now, solve this equation for 's' to find the molar solubility of Cd(OH)鈧�.
02

(b) Finding the initial concentration of NaBr needed to increase the molar solubility of Cd(OH)鈧�

To find the initial concentration of NaBr, we need to use the complex formation constant (Kf) for the complex ion CdBr鈧劼测伝. The complex formation reaction can be written as: Cd虏鈦�(aq) + 4Br鈦�(aq) 鈬� CdBr鈧劼测伝(aq) Now, let's denote the initial concentration of NaBr required as 'x'. Since NaBr is a strong electrolyte, it will completely dissociate in water and will provide 'x' moles of Br鈦� per liter. The final concentration of Br鈦� should be 4 times the increased molar solubility. Now, we can write the expression for the complex formation constant (Kf) as: Kf = [CdBr鈧劼测伝] / ([Cd虏鈦篯[Br鈦籡鈦�) We have the value of Kf as 5 脳 10鲁 and are given that the increased molar solubility of Cd(OH)鈧� (concentration of Cd虏鈦�) is 1.0 脳 10鈦宦� mol/L, which can be written in terms of 'x': Kf = (1.0 脳 10鈦宦�) / ((1.0 脳 10鈦宦�)(x鈦�)) Now, solve this equation for 'x' to find the initial concentration of NaBr needed to increase the molar solubility of Cd(OH)鈧�.

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Key Concepts

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

Ksp
The concept of the solubility product constant, often denoted as \(K_{sp}\), is crucial for understanding solubility equilibrium in ionic compounds. \(K_{sp}\) refers to the equilibrium constant for a solid substance dissolving in an aqueous solution. When an ionic compound dissolves, it dissociates into its constituent ions. For \[ \text{Cd(OH)}_2 \] it dissociates as follows:
  • \[ \text{Cd(OH)}_2(s) \rightleftharpoons \text{Cd}^{2+}(aq) + 2\text{OH}^{-}(aq) \]
Here, the product of the concentrations of the ions, each raised to the power of its coefficient in the balanced equation, gives you the \(K_{sp}\). For \[ \text{Cd(OH)}_2, \]\(K_{sp} = [\text{Cd}^{2+}][\text{OH}^{-}]^2\).This value, \(2.5 \times 10^{-14}\), is very tiny, indicating that \[ \text{Cd(OH)}_2 \] is not very soluble. The primary use of this constant is to help calculate the molar solubility of a compound, as shown in step 1 of the solution process.
Complex ion formation
Complex ion formation is a process where metal ions form a complex with certain ligands. In this context, \(\text{Cd}^{2+}\) ions combine with \(\text{Br}^-\) ions to form the complex ion \(\text{CdBr}_4^{2-}\). This greatly affects the solubility of the original compound because the formation of the complex ion reduces the concentration of free metal ions in solution, effectively increasing the solubility of the salt.In the context of \(\text{Cd(OH)}_2\), when this compound is exposed to an excess of \(\text{Br}^-\) ions from NaBr, it can form a stable complex given by the equation:
  • \[ \text{Cd}^{2+}(aq) + 4\text{Br}^- (aq) \rightleftharpoons \text{CdBr}_4^{2-} (aq) \]
The formation of this complex ion \(\text{CdBr}_4^{2-}\) hence increases the solubility of \(\text{Cd(OH)}_2\) by shifting the equilibrium of the dissolution reaction. This is an essential concept when considering how different ions and chemicals might affect the solubility of compounds in various scenarios.
Molar solubility
Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution until the solution becomes saturated. It's an important concept allowing us to quantify how much of a compound can dissolve in solution, given its \(K_{sp}\). For \(\text{Cd(OH)}_2\), the molar solubility is the concentration of \(\text{Cd}^{2+}\) ions, 's', when equilibrium is reached:
  • \[2.5 \times 10^{-14} = s \times (2s)^2\] simplifies to \(s = \sqrt[3]{\frac{2.5 \times 10^{-14}}{4}}\)
This calculates the molar solubility without any complex formation.However, the presence of complex ions can change this solubility drastically. As in the case above, adding \(\text{NaBr}\) leads to complex ion formation, thereby increasing the molar solubility significantly to \(1.0 \times 10^{-3} \text{ mol/L}\). Knowing how to calculate and adjust molar solubility is useful in various chemical applications and is key to handling reactions in the lab.
Complex ion stability constant (Kf)
The stability constant of a complex ion, known as \(K_f\), reflects the stability of the complex ion in solution. Specifically, it measures the propensity of the complex to form from the central metal ion and its ligands. A larger \(K_f\) value indicates a more stable complex.For the complex ion \(\text{CdBr}_4^{2-}\), the equation:
  • \[ K_f = \frac{[\text{CdBr}_4^{2-}]}{[\text{Cd}^{2+}][\text{Br}^-]^4} \]
Utilizes the given \(K_f\) value of \(5 \times 10^3\), showing how increased concentrations of \(\text{Br}^-\) (provided by NaBr) push the formation of more \(\text{CdBr}_4^{2-}\), thus enhancing solubility.Understanding \(K_f\) is essential for predicting and manipulating the solubility of salts in complex solutions. When designing experiments or in industrial processes, knowing the stability constants allows for precise control of compound dissolution and reactions.

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Most popular questions from this chapter

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is 0.010\(M\) in \(\mathrm{Ba}^{2+}(a q)\) and 0.010\(M\) in \(\mathrm{Sr}^{2+}(a q) .\) (a) What concentration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. \(\mathrm{BaSO}_{4} : K_{s p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}\): \(K_{s p}=3.2 \times 10^{-7} . )\) (b) Which cation precipitates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) when the second cation begins to precipitate?

Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically. the F- ion concentration is adjusted to about 1 ppm. Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is 8 ppm. Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)

Which of the following solutions is a buffer? (a) 0.10\(M\) \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(0.10 \mathrm{MCH}_{3} \mathrm{CONa},(\mathbf{b}) 0.10 \mathrm{MCH}_{3} \mathrm{COOH}\) (c) 0.10 \(\mathrm{M} \mathrm{HCl}\) and \(0.10 \mathrm{M} \mathrm{NaCl},(\mathbf{d})\) both a and \(\mathrm{c},(\mathbf{e})\) all of a, \(\mathrm{b},\) and \(\mathrm{c} .\)

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{p} \mathrm{H}\) of a 0.050 M solution of \(\mathrm{CaCl}_{2}\) is adjusted to 8.0? (b) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when 100 mL of 0.050 M \(\mathrm{AgNO}_{3}\) is mixed with 10 mL of \(5.0 \times 10^{-2} \mathrm{MNa}_{2} \mathrm{SO}_{4}\) solution?

Use values of \(K_{s p}\) for AgI and \(K_{f}\) for \(A g(C N)_{2}^{-}\) to (a) calculate the molar solubility of Agl in pure water, (b) calculate the equilibrium constant for the reaction \(\operatorname{AgI}(s)+2 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{I}^{-}(a q), \quad(\mathbf{c})\) determine the molar solubility of AgI in a 0.100 \(\mathrm{MNaCN}\) solution.
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