Question

At what minimum pH will \(10^{-3}\) M \(-\mathrm{Al}(\mathrm{OH})_{3}\) go into solution \((V=1 \mathrm{~L})\) as \(\mathrm{Al}(\mathrm{OH})_{4}^{-}\) and at what maximum \(\mathrm{pH}\), it will dissolved as \(\mathrm{Al}^{3+}\) ? Given: \(\log 2=0.3\) \(\mathrm{Al}(\mathrm{OH})_{4}^{-} \rightleftharpoons \mathrm{Al}^{3+}+4 \mathrm{OH}^{-} ; K_{\mathrm{eq}}=1.6 \times 10^{-34}\) \(\mathrm{Al}(\mathrm{OH})_{3} \rightleftharpoons \mathrm{Al}^{3+}+3 \mathrm{OH}^{-} ; K_{\mathrm{eq}}=8.0 \times 10^{-33}\) (a) \(9.3,9.7\) (b) \(9.7,9.3\) (c) \(4.3,9.3\) (d) \(4.7,9.3\)

Short answer

The minimum pH at which \(10^{-3}\) M \(-\mathrm{Al(OH)_3}\) will go into solution as \(\mathrm{Al(OH)_4^-}\) is 9.3, and the maximum pH at which it will dissolve as \(\mathrm{Al^{3+}}\) is 9.7.

Step-by-step solution

- Write down the dissociation equations and their equilibrium constants

First, let's identify the reactions and their equilibrium constants: For the formation of \(\mathrm{Al(OH)_4^-}\): \[\mathrm{Al(OH)_4^-} \rightleftharpoons \mathrm{Al^{3+}} + 4 \mathrm{OH^-} ; K_{eq} = 1.6 \times 10^{-34}\]For the dissolution of \(\mathrm{Al(OH)_3}\): \[\mathrm{Al(OH)_3} \rightleftharpoons \mathrm{Al^{3+}} + 3 \mathrm{OH^-} ; K_{eq} = 8.0 \times 10^{-33}\]

- Determine the [OH-] for Al(OH)4-

For the formation of \(\mathrm{Al(OH)_4^-}\), assume that \(x\) moles/L of \(\mathrm{Al(OH)_3}\) goes into solution to form \(\mathrm{Al(OH)_4^-}\). Therefore: \[K_{eq} = \frac{[\mathrm{Al^{3+}}][\mathrm{OH^-}]^4}{[\mathrm{Al(OH)_4^-}]} = 1.6 \times 10^{-34}\]Since \([\mathrm{Al(OH)_4^-}] = 10^{-3}\) M, we can substitute this into the equation, and because the concentration of \(\mathrm{Al^{3+}}\) is equal to \(x\), we get: \[1.6 \times 10^{-34} = \frac{x[\mathrm{OH^-}]^4}{10^{-3}}\]From this, we express \([\mathrm{OH^-}]\) in terms of \(x\): \[1.6 \times 10^{-31} = x[\mathrm{OH^-}]^4\]Since \(\mathrm{pOH} = -\log [\mathrm{OH^-}]\), we have: \[\mathrm{pOH} = 4\log \frac{1}{[\mathrm{OH^-}]}\]Substituting \(x\) from the previous equation gives: \[\mathrm{pOH} = 4\log \frac{1}{(1.6 \times 10^{-31}/x)^{1/4}}\]

- Calculate [OH-] and pH for Al(OH)3 dissolution

For the dissolution of \(\mathrm{Al(OH)_3}\), using the equilibrium constant, we get: \[K_{eq} = \frac{[\mathrm{Al^{3+}}][\mathrm{OH^-}]^3}{[\mathrm{Al(OH)_3}]} = 8.0 \times 10^{-33}\]Again, assume that \(y\) moles/L dissolves, therefore, \([\mathrm{Al^{3+}}] = y\) and the undissolved \(\mathrm{Al(OH)_3}\) concentration is negligible: \[8.0 \times 10^{-33} = y[\mathrm{OH^-}]^3\]Thus, express \([\mathrm{OH^-}]\) in terms of \(y\): \[[\mathrm{OH^{-}}]= (8.0 \times 10^{-33}/y)^{1/3}\]\(\mathrm{pOH}\) can then be determined: \[\mathrm{pOH}=3\log \frac{1}{[\mathrm{OH^-}]}\]And the \(\mathrm{pH}\) is calculated from \(\mathrm{pH} + \mathrm{pOH} = 14\)

- Calculate the minimum pH for Al(OH)4- formation

Using the defined relationship for \(\mathrm{pOH}\) in Step 2 and \(\mathrm{pOH} + \mathrm{pH} = 14\), calculate the minimum \(\mathrm{pH}\) for the formation of \(\mathrm{Al(OH)_4^-}\).

- Calculate the maximum pH for Al(OH)3 dissolution

Using the expression for \(\mathrm{pOH}\) obtained in Step 3 and the relationship \(\mathrm{pOH} + \mathrm{pH} = 14\), find the maximum \(\mathrm{pH}\) at which \(\mathrm{Al(OH)_3}\) will dissolve.

Key concepts

Physical Chemistry and pH Calculations

Physical chemistry is the branch of chemistry concerned with understanding the behavior of chemicals and materials at the molecular and atomic levels. A crucial aspect of physical chemistry involves pH calculations, which relate to the acidity or basicity of a solution.

pH is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It is inversely related to the concentration of hydrogen ions \(H^+\) in the solution, given by the formula \( pH = -\log[ H^+ ] \). To solve pH calculation problems, one must understand the concept of chemical equilibrium and be familiar with constants such as the solubility product \(K_{sp}\) and equilibrium constants \(K_{eq}\) for dissociation reactions. These problems often involve hydrolysis of salts, where salts react with water to produce an acidic or basic solution.

Understanding the relationship between the concentrations of different species in a reaction, their respective equilibrium constants, and how to manipulate logarithmic expressions is pivotal in performing accurate pH calculations in physical chemistry.

Chemical Equilibrium

Chemical equilibrium is reached in a reversible chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. In the context of pH calculation problems, the equilibrium constant \(K_{eq}\) provides the necessary link to connect the concentration of the ions in solution with the degree of dissociation of a compound.

When dealing with equilibrium problems, it is essential to write the equilibrium expression correctly and substitute known values such as the initial concentrations and the equilibrium constant to find the unknown concentrations. The concept of equilibrium is tightly connected to the pH of a solution, especially when the reaction involves \(H^+\) or \(OH^-\) ions, making it essential for predicting the extent of a reaction in solution and calculating the subsequent pH.

Solubility Product

The solubility product, represented by the symbol \(K_{sp}\), is a type of equilibrium constant that applies to the dissolution of sparingly soluble compounds. It quantifies the solubility of a solute under equilibrium conditions and is defined as the product of the concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient.

In pH calculation problems, the solubility product becomes particularly important when determining the solubility of salts, which can potentially affect the pH of the solution. For a given salt, the \(K_{sp}\) value is used to calculate the maximum concentration of ions that can exist in a solution before precipitation occurs. The dissolution of \(Al(OH)_3\) and the formation of \(Al(OH)_4^{-}\) in the given exercise show the practical application of \(K_{sp}\) in finding the pH at which these species will dissolve or precipitate.

Hydrolysis of Salts

Hydrolysis of salts involves the reaction of a salt with water to produce an acidic or basic solution. When a salt derives from a weak acid or a weak base, the ions react with water to generate \(OH^-\) or \(H^+\) ions, thus affecting the pH of the solution.

Consider the salt \(Al(OH)_4^-\) in the given exercise; it can hydrolyze to form \(Al^{3+}\) and \(OH^-\) ions. By calculating the \(K_{eq}\) for this reaction, we can determine the concentration of \(OH^-\), which in turn allows us to find the pOH and finally the pH of the solution. The hydrolysis of \(Al(OH)_3\) as \(Al^{3+}\) in a similar manner contributes to the understanding of acidity of the solution at different pH levels. Knowing how to calculate and interpret the hydrolysis of salts is essential for solving many pH problems in physical chemistry.